Chapter one: DC circuit theory & Single-phase series AC ...

1

Chapter one

DC circuit theory amp Single-phase series AC circuits

2

Current and Voltage

Current is the flow of electrons It is the time rate of change of

electrons passing through a defined area such as the cross-section of a

wire Because electrons are negatively charged the positive direction of

current flow is opposite to that of electron flow The mathematical

description of the relationship between the number of electrons (called

charge q ) and current i is

dqi

dt or q t i t d

The unit of charge is the coulomb (C) (in recognition of Charles

Augustin Coulomb French physicist and mathematician 1736-1806)

which represents 18624 10 electrons

The unit of current is the ampere or simply amp (in recognition of

Andre‟ Marie Ampere French physicist and mathematician 1775-1836)

which is defined as a coulomb per second

Ampere = coulomb second

Thus 1 amp is 18624 10 electrons moving from one body to another in 1

second

Energy is required to move a charge between two points in a circuit The

work per unit charge required to do this is called voltage The voltage

difference between two points in a circuit is a measure of the energy

required to move charge from one point to the other

3

The unit of voltage is volt (V) (in recognition of the Italian physicist

Alessandro Volta 1745-1827) which is defined as a charge of 1 joule of

energy per coulomb of charge

A joule (named in recognition of the English physicist James Joule

1818-1889) is a unit of energy or work and has the units of Newton X

meter Thus

volt = joule coulomb joule = Newton x meter

Current Source and Voltage Source

A voltage source is a device that causes a specified voltage to exist

between two points in a circuit The voltage may be time varying or time

invariant (for a sufficiently long time) Often the voltage is denoted by

E or V A battery is an example of this type of voltage

A current source causes a specified current to flow through a wire

containing this source

(a) Voltage source (b) constant voltage source (c) current source

AC DC and Electrical Signals

AC means Alternating Current and DC means Direct Current AC and

DC are also used when referring to voltages and electrical signals which

are not currents For example a 12V AC power supply has an

4

alternating voltage (which will make an alternating current flow) An

electrical signal is a voltage or current which conveys information

usually it means a voltage The term can be used for any voltage or

current in a circuit

Direct Current (DC)

Direct Current (DC) always flows in the same directionbut it may

increase and decrease A DC voltage is always positive (or always

negative)but it may increase and decrease Electronic circuits normally

require a steady DC supply which is constant at one value or a smooth

DC supply which has a small variation called ripple Cells batteries and

regulated power supplies provide steady DC which is ideal for electronic

circuits

Power supplies contain a transformer which converts the mains AC

supply to a safe low voltage AC Then the ACis converted to DC by a

bridge rectifier but the output is varying DC which is unsuitable for

electronic circuits Some power supplies include a capacitor to provide

smooth DC which is suitable for less-sensitive electronic circuits

including most of the projects on this website Lamps heaters and

motors will work with any DC supply

5

Alternating Current (AC)

Alternating Current (AC) flow one way then the other way continually

reversing direction An AC voltage is continually changing between

positive (+) and negative (-)

The rate of changing direction is called the frequency of the AC and it is

measured in hertz (Hz) which is the number of forwards-backwards

cycles per second

Mains electricity in the USA and Canada has a frequency 60 Hz while

other counties have a frequency of 50 Hz

An AC supply is suitable for powering some devices such as lamps and

heaters but almost all electronic circuits require a steady DC supply

13 Properties of electrical signals

6

Amplitude is the maximum voltage reached by the signal

It is measured in volts V

Peak voltage is another name for amplitude

Peak-peak voltage is twice the peak voltage (amplitude) When

reading an oscilloscope trace it is usual to measure peak-peak

voltage

Time period is the time taken for the signal to complete one cycle

It is measured in seconds (s) but time periods tend to be short so

milliseconds (ms)= 10 -3

s and microseconds (micros)10 -6

s

Frequency is the number of cycles per second

It is measured in hertz (Hz)

1 kilohertz (kHz) =1000 Hz and megahertz (MHz) =106 Hz

Frequency = 1 time periodamptime period = 1 frequency

Mains electricity in the Egypt has a frequency of 50Hz

so it has a time period of 150 = 002s = 20ms

14 Root Mean Square (RMS) Values

The value of an AC voltage is

continually changing from zero up to the

positive peak through zero to the

negative peak and back to zero again

Clearly for most of the time it is less than

the peak voltage so this is not a good

measure of its real effect

Instead we use the root mean square voltage (VRMS) which is 07 of the

peak voltage (Vpeak)

VRMS = 07 times Vpeak and Vpeak = 14 times VRMS

7

These equations also apply to current They are only true for sine

waves (the most common type of AC) because the 07 and 14 are

different values for other shapes

The RMS value is the effective value of a varying voltage or current

It is the equivalent steady DC (constant) value which gives the same

effect

What do AC meters show is it the RMS or peak voltage

AC voltmeters and ammeters show the RMS value of the voltage or

current DC meters also show the RMS value when connected to varying

DC providing the DC is varying quickly if the frequency is less than

about 10Hz you will see the meter reading fluctuating instead

Electrical Resistance

Electrical resistance is a measure of the degree to which an object

opposes an electric current through it Its reciprocal quantity is electrical

conductance (provided the electrical impedance is real) measured in

siemens Assuming a uniform current density an objects electrical

resistance is a function of both its physical geometry and the resistivity

of the material it is made from

A

lR

where

is the length

A is the cross sectional area and

ρ is the resistivity of the material

Electrical resistance shares some conceptual parallels with the

mechanical notion of friction The SI unit of electrical resistance is the

ohm symbol Ω

The electrical resistivity ρ (rho) of a material is given by

8

l

RA

Where ρ is the static resistivity (measured in ohm metres Ω-m)

R is the electrical resistance of a uniform specimen of the

material (measured in ohms Ω)

is the length of the piece of material (measured in metres m)

A is the cross-sectional area of the specimen (measured in

square metres msup2)

Electrical resistivity can also be defined as

J

E

Where E is the magnitude of the electric field (measured in volts per

metre Vm)

J is the magnitude of the current density (measured in amperes

per square metre Amsup2)

The conductivityσ (sigma) is defined as the inverse of the electrical

resistivity of the material or

1

Example

What is the value of the resistance of a copper wire with length 250m

and cross section area equals 25 mmsup2

Solution

Plug into the formula and using the value of the resistivity of the copper

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

Ohms lawstates that in an electrical circuit the current passing through

a conductor between two points is directly proportional to the potential

difference (ie voltage drop or voltage) across the two points and

inversely proportional to the resistance between them and for the same

temperature

The mathematical equation that describes this relationship is

R

VI

Where I is the current in amperes (A) V is the potential difference

between two points of interest in volts (V) and R is the resistance it

measured in ohms (which is equivalent to volts per ampere)

The potential difference is also known as the voltage drop and is

sometimes denoted by U E or emf (electromotive force) instead of V

Example

Suppose the ammeter reads 52 A and the voltmeter indicates 233 kV

What is the resistance

Solution

Convert to amperes and volts getting I = 0000052 A and E = 2330 V

Then plug into the formula

R = 23300000052 = 45000000 = 45 M

Example

Find V when I = 120nA and R = 25 k

10

Solution

Convert to amperes and ohms getting 910120 x A and R = 31025 x

mVxxxxRIV 3103102510120 339

Example

Find I when V = 22kV and R = 440

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

The electric power P used in any part of a circuit is equal to the current I

in that part multiplied by the voltage V across that part of the circuit Its

formula is

P = VI

Where P = power W

V = voltage V

I = current A

If we know the current I and the resistance R but not the voltage V we

can find the power P by using Ohm‟s law for voltage so that

substituting V = IR

RIP 2

In the same manner if we know the voltage V and the resistance R but

not the current I we can find the power P by using Ohm‟s law for

current so that substitutingR

VI

11

R

VP

2

Example

The current through a 100Ω resistor to be used in a circuit is 020 A

Find the power rating of the resistor

Solution

WxRIP 410020 22

Example

If the voltage across a 25000 Ω resistor is 500 V what is the power

dissipated in the resistor

Solution

WR

VP 10

25000

500 22

19 Electrical energy

Electrical energy in any part of the circuit is that power of that part

multiplied of the time of consumption and it can be found from the

formula

Electrical energy U = Power x time

U =VIt (Joules)

Kilowatt hour (kWh)= 1000 watt hour

= 1000 x 3600 watt seconds or joule = 3 600 000 J

Example

A source emf of 5 V is supplies a current of 3 A for 10 minutes How

much energy is provided in this time

12

Solution

Energy = power x time power = voltage x current

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

An electric heater consumes 18 MJ when connected to a 250 V supply

for 30 minutes Find the power rating of the heater and the current taken

from the supply

Solution

Power rating of heater = kWWx

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

An electric bulb has a power 100 W If it works 10 hours per day find

the energy consumed in 1 month

Solution

Since energy = power x time

Then Energy U = P x t = 100 x 10 x 30

= 30000 Wh= 30 kWh

Series and Parallel Networks

Series circuits

TheFigure shows three resistors R1 R2 and R3 connected end to end

ie in series with a battery source of V volts Since the circuitis closed

13

a current I will flow and the pd across eachresistor may be determined

from the voltmeter readings V1 V2 and V3

In a series circuit

(a) the current I is the same in all parts of the circuit and hence the same

reading is found on each of the two ammeters shown and

(b) the sum of the voltages V1 V2 and V3 is equal to the total applied

voltage V ie

From Ohm‟s law

where R is the total circuit resistance

Since V = V1 + V2 + V3

then IR = IR1 + IR2 + IR3

Dividing throughout by I gives

Thus for a series circuit the total resistance is obtained by adding

together the values of the separate resistances

Example 113 For the circuit shown in the illustrated figure determine

(a) the battery voltage V

(b) the total resistance of the circuit and

(c) the values of resistance of resistors R1 R2 and R3 given

that the pd‟s across R1 R2 and R3 are 5 V 2 V and 6 V respectively

14

Solution

Voltage divider

The voltage distribution for the circuit shown in the Figure is given by

15

Example

Two resistors are connected in seriesacross a 24 V supply and a current

of 3 A flows in thecircuit If one of the resistors has a resistance of 2 Ω

determine (a) the value of the other resistor and

(b) the difference across the 2 Ω resistor

Solution

(a) Total circuit resistance

Value of unknown resistance

(b) Volt Difference across 2 Ω resistor V1 = IR1 = 3 x 2 = 6 V

Alternatively from above

Parallel networks

The Figure shows three resistorsR1 R2 and R3 connected across each

other iein parallel across a battery source of V volts

16

In a parallel circuit

(a) the sum of the currents I1 I2 and I3 is equal to the total

circuitcurrent I ie I = I1 + I2 + I3 and

(b) the source pdis the same across each of the resistors

From Ohm‟s law

Since I = I1 + I2 + I3

Dividing throughout by V gives

This equation must be used when finding the total resistance R of a

parallel circuit

For the special case of two resistors in parallel

17

Example

For the circuit shown in the Figure find

(a) the value of the supply voltage V

(b) the value of current I

Solution

(a) Vd across 20 Ω resistor = I2R2 = 3 x 20 = 60 V hencesupply

voltage V = 60 V since the circuit is connected in parallel

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

Another solution for part (b)

18

Current division

For the circuit shown in the Figure the total circuit resistance RT

isgiven by

Example

For the circuit shown in the Figure find the current Ix

Solution

19

Commencing at the right-hand side of the arrangement shown in the

Figure below the circuit is gradually reduced in stages as shown in

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

From the above Figure

Kirchhoffrsquos laws

Kirchhoffrsquos laws state

20

(a) Current Law At any junction in an electric circuit the total

current flowing towards that junction is equal to the total current

flowing away from the junction ie sumI = 0

Thus referring to Figure 21

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

(b) Voltage Law In any closed loop in a network the algebraic sum

of the voltage drops (ie products ofcurrent and resistance) taken

around the loop is equal to the resultant emf acting in that loop

E1 - E2 = IR1 + IR2 + IR3

(Note that if current flows away from the positive terminal of a source

that source is considered by convention to be positive Thus moving

anticlockwise around the loop of Figure 22 E1 is positive and E2 is

negative)

21

Example

Find the unknown currents marked in the following Figure

Solution

Applying Kirchhoff‟s current law

For junction B 50 = 20 + I1 hence I1 = 30 A

For junction C 20 + 15 = I2 hence I2 = 35 A

For junction D I1 = I3 + 120 hence I3 = minus90 A

(the current I3 in the opposite direction to that shown in the Figure)

For junction EI4 + I3 = 15 hence I4 = 105 A

For junction F 120 = I5 + 40 hence I5 = 80 A

Example

Determine the value of emf E in the following Figure

22

Solution

Applying Kirchhoff‟s voltage law

Starting at point A and moving clockwise around the loop

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Since I = 2 A hence E = 9 V

Example

Use Kirchhoff‟s laws to determine the currents flowing in each branch

of the network shown in Figure 24

23

Solution

1 Divide the circuit into two loops The directions current flows from

the positive terminals of the batteries The current through R is I1 + I2

2 Apply Kirchhoff‟s voltage law foreach loop

By moving in a clockwise direction for loop 1 of Figure 25

(1)

By moving in an anticlockwise direction for loop 2 of Figure 25

(2)

3 Solve equations (1) and (2) for I1 and I2

(3)

(4)

(ie I2 is flowing in the opposite direction to that shown in Figure 25)

From (1)

Current flowing through resistance R is

Note that a third loop is possible as shown in Figure 26giving a third

equation which can be used as a check

24

Example

Determine using Kirchhoff‟s laws each branch current for the network

shown in Figure 27

Solution

1 The network is divided into two loops as shown in Figure 28 Also

the directions current are shown in this Figure

2 Applying Kirchhoff‟s voltage law gives

For loop 1

(1)

25

For loop 2

(2)

(Note that the opposite direction of current in loop 2)

3 Solving equations (1) and (2) to find I1 and I2

(3)

(2) + (3) give

From (1)

Current flowing in R3 = I1-I2 = 652 ndash 637 = 015 A

Example

For the bridge network shown in Figure 29 determine the currents in

each of the resistors

26

Solution

- Let the current in the 2Ω resistor be I1 and then by Kirchhoff‟s current

law the current in the 14 Ω resistor is (I - I1)

- Let thecurrent in the 32 Ω resistor be I2 Then the current in the 11

resistor is (I1 - I2) and that in the3 Ω resistor is (I - I1 + I2)

- Applying Kirchhoff‟s voltage law to loop 1 and moving in a clockwise

direction

(1)

- Applying Kirchhoff‟s voltage law to loop 2 and moving in an

anticlockwise direction

(2)

Solve Equations (1) and (2) to calculate I1 and I2

(3)

(4)

(4) ndash (3) gives

Substituting for I2 in (1) gives

27

Hence

the current flowing in the 2 Ω resistor = I1 = 5 A

the current flowing in the 14 Ω resistor = I - I1 = 8 - 5 = 3 A

the current flowing in the 11 Ω resistor = I1 - I2 = 5 - 1 = 4 Aand

the current flowing in the 3 Ω resistor = I - I1 + I2 = 8 - 5 + 1 = 4 A

The Superposition Theorem

The superposition theorem statesIn any network made up of linear

resistances and containing more than one source of emf the resultant

current flowing in any branch is the algebraic sum of the currents that

would flow in that branch if each source was considered separately all

other sources being replaced at that time by their respective internal

resistances‟

Example

Figure 211 shows a circuit containing two sources of emf each with

their internal resistance Determine the current in each branch of the

network by using the superposition theorem

28

Solution Procedure

1 Redraw the original circuit with source E2 removed being replaced

by r2 only as shown in Figure (a)

2 Label the currents in each branch and their directions as shown

inFigure (a) and determine their values (Note that the choice of current

directions depends on the battery polarity which flowing from the

positive battery terminal as shown)

R in parallel with r2 gives an equivalent resistance of

From the equivalent circuit of Figure (b)

From Figure (a)

29

by r1 only as shown in Figure (aa)

inFigure (aa) and determine their values

r1 in parallel with R gives an equivalent resistance of

From the equivalent circuit of Figure (bb)

From Figure (aa)

By current divide

5 Superimpose Figure (a) on to Figure (aa) as shown in Figure 213

30

6 Determine the algebraic sum of the currents flowing in each branch

Resultant current flowing through source 1

Resultant current flowing through resistor R

The resultant currents with their directions are shown in Figure 214

Theveninrsquos Theorem

Thevenin‟s theorem statesThe current in any branch of a network is that

which would result if an emf equal to the pd across a break made in

the branch were introduced into the branch all other emf‟s being

removed and represented by the internal resistances of the sources‟

31

The procedure adopted when using Thevenin‟s theorem is summarized

below

To determine the current in any branch of an active network (ie one

containing a source of emf)

(i) remove the resistance R from that branch

(ii) determine the open-circuit voltage E across the break

(iii) remove each source of emf and replace them by their internal

resistances and then determine the resistance r bdquolooking-in‟ at the break

(iv) determine the value of the current from the equivalent circuit

shownin Figure 216 ie

Example

Use Thevenin‟s theorem to find the current flowing in the 10 Ω resistor

for the circuit shown in Figure 217

32

Solution

Following the above procedure

(i) The 10Ω resistance is removed from the circuit as shown in Figure

218

(ii) There is no current flowing in the 5 Ωresistor and current I1 is given

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

Hence pd across AB ie the open-circuit voltage across the break

E = 8 V

(iii) Removing the source of emf gives the circuit of Figure 219

Resistance

(iv) The equivalent Thacuteevenin‟s circuit is shown in Figure 220

Current

33

Hence the current flowing in the 10 resistor of Figure 217 is 0482 A

Example

A Wheatstone Bridge network is shown in Figure 221(a) Calculate the

current flowing in the 32Ω resistorand its direction using Thacuteevenin‟s

theorem Assume the source ofemf to have negligible resistance

Solution

Following the procedure

(i) The 32Ω resistor is removed from the circuit as shown in Figure

221(b)

(ii) The pd between A and C

34

The pd between B and C

Hence

Voltage between A and B is VAB= VAC ndash VBC = 3616 V

Voltage at point C is +54 V

Voltage between C and A is 831 V

Voltage at point A is 54 - 831 = 4569 V

Voltage between C and B is 4447 V

Voltage at point B is 54 - 4447 = 953 V

Since the voltage at A is greater than at B current must flow in the

direction A to B

(iii) Replacing the source of emf with short-circuit (ie zero internal

resistance) gives the circuit shown in Figure 221(c)

The circuit is redrawn and simplified as shown in Figure 221(d) and (e)

fromwhich the resistance between terminals A and B

(iv) The equivalent Thacuteevenin‟s circuit is shown in Figure 1332(f) from

whichCurrent

35

Hence the current in the 32 Ω resistor of Figure 221(a) is 1 A

flowing from A to B

Example

Use Thacuteevenin‟s theorem to determine the currentflowing in the 3 Ω

resistance of the network shown inFigure 222 The voltage source has

negligible internalresistance

Solution

(i) The 3 Ω resistance is removed from the circuit as shown inFigure

223

(ii) The Ω resistance now carries no current

36

Hence pd across AB E = 16 V

(iii) Removing the source of emf and replacing it by its internal

resistance means that the 20 Ω resistance is short-circuited as shown in

Figure 224 since its internal resistance is zero The 20 Ω resistance may

thus be removed as shown in Figure 225

From Figure 225 Thacuteevenin‟s resistance

(iv)The equivalent Thevenin‟s circuit is shown in Figure 226

Nortonrsquos Theorem

Norton‟s theorem statesThe current that flows in any branch of a

network is the same as that which would flow in the branch if it were

connected across a source of electrical energy the short-circuit current

of which is equal to the current that would flow in a short-circuit across

the branch and the internal resistance of which is equal to the resistance

which appears across the open-circuited branch terminals‟

The procedure adopted when using Norton‟s theorem is summarized

below

To determine the current flowing in a resistance R of a branch AB of an

active network

(i) short-circuit branch AB

37

(ii) determine the short-circuit current ISC flowing in the branch

(iii) remove all sources of emf and replace them by their internal

resistance (or if a current source exists replace with an open circuit)

then determine the resistance rbdquolooking-in‟ at a break made between A

and B

(iv) determine the current I flowing in resistance R from the Norton

equivalent network shown in Figure 1226 ie

Example

Use Norton‟s theorem to determine the current flowing in the 10

Ωresistance for the circuit shown in Figure 227

38

Solution

(i) The branch containing the 10 Ωresistance is short-circuited as shown

in Figure 228

(ii) Figure 229 is equivalent to Figure 228 Hence

(iii) If the 10 V source of emf is removed from Figure 229 the

resistance bdquolooking-in‟ at a break made between A and B is given by

(iv) From the Norton equivalent network shown in Figure 230 the

current in the 10 Ω resistance by current division is given by

39

As obtained previously in above example using Thacuteevenin‟s theorem

Example

resistance of the network shown in Figure 231(a) The voltage source

has negligible internal resistance

Solution

(i) The branch containing the 3 resistance is short-circuited as shown in

Figure 231(b)

(ii) From the equivalent circuit shown in Figure 231(c)

40

(iii) If the 24 V source of emf is removed the resistance bdquolooking-in‟ at

a break made between A and B is obtained from Figure 231(d) and its

equivalent circuit shown in Figure 231(e) and is given by

(iv) From the Norton equivalent network shown in Figure 231(f) the

current in the 3 Ωresistance is given by

As obtained previously in previous example using Thevenin‟s theorem

Extra Example

Use Kirchhoffs Laws to find the current in each resistor for the circuit in

Figure 232

Solution

1 Currents and their directions are shown in Figure 233 following

Kirchhoff‟s current law

41

For node A gives 314 III

For node B gives 325 III

2 The network is divided into three loops as shown in Figure 233

Applying Kirchhoff‟s voltage law for loop 1 gives

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

Arrange the three equations in matrix form

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

Star Delta Transformation

Star Delta Transformations allow us to convert impedances connected

together from one type of connection to another We can now solve

simple series parallel or bridge type resistive networks using

Kirchhoff‟s Circuit Laws mesh current analysis or nodal voltage

analysis techniques but in a balanced 3-phase circuit we can use

different mathematical techniques to simplify the analysis of the circuit

and thereby reduce the amount of math‟s involved which in itself is a

good thing

Standard 3-phase circuits or networks take on two major forms with

names that represent the way in which the resistances are connected a

Star connected network which has the symbol of the letter Υ and a

Delta connected network which has the symbol of a triangle Δ (delta)

If a 3-phase 3-wire supply or even a 3-phase load is connected in one

type of configuration it can be easily transformed or changed it into an

equivalent configuration of the other type by using either the Star Delta

Transformation or Delta Star Transformation process

A resistive network consisting of three impedances can be connected

together to form a T or ldquoTeerdquo configuration but the network can also be

redrawn to form a Star or Υ type network as shown below

T-connected and Equivalent Star Network

44

As we have already seen we can redraw the T resistor network to

produce an equivalent Star or Υ type network But we can also convert

a Pi or π type resistor network into an equivalent Delta or Δ type

network as shown below

Pi-connected and Equivalent Delta Network

Having now defined exactly what is a Star and Delta connected network

it is possible to transform the Υ into an equivalent Δ circuit and also to

convert a Δ into an equivalent Υ circuit using a the transformation

process This process allows us to produce a mathematical relationship

between the various resistors giving us a Star Delta Transformation as

well as a Delta Star Transformation

45

These Circuit Transformations allow us to change the three connected

resistances (or impedances) by their equivalents measured between the

terminals 1-2 1-3 or 2-3 for either a star or delta connected circuit

However the resulting networks are only equivalent for voltages and

currents external to the star or delta networks as internally the voltages

and currents are different but each network will consume the same

amount of power and have the same power factor to each other

Delta Star Transformation

To convert a delta network to an equivalent star network we need to

derive a transformation formula for equating the various resistors to each

other between the various terminals Consider the circuit below

Delta to Star Network

Compare the resistances between terminals 1 and 2

Resistance between the terminals 2 and 3

46

This now gives us three equations and taking equation 3 from equation 2

gives

Then re-writing Equation 1 will give us

Adding together equation 1 and the result above of equation 3 minus

equation 2 gives

47

From which gives us the final equation for resistor P as

Then to summarize a little about the above maths we can now say that

resistor P in a Star network can be found as Equation 1 plus (Equation 3

minus Equation 2) or Eq1 + (Eq3 ndash Eq2)

Similarly to find resistor Q in a star network is equation 2 plus the

result of equation 1 minus equation 3 or Eq2 + (Eq1 ndash Eq3) and this

gives us the transformation of Q as

and again to find resistor R in a Star network is equation 3 plus the

gives us the transformation of R as

When converting a delta network into a star network the denominators

of all of the transformation formulas are the same A + B + C and which

is the sum of ALL the delta resistances Then to convert any delta

connected network to an equivalent star network we can summarized the

above transformation equations as

48

Delta to Star Transformations Equations

If the three resistors in the delta network are all equal in value then the

resultant resistors in the equivalent star network will be equal to one

third the value of the delta resistors giving each branch in the star

network as RSTAR = 13RDELTA

Delta ndash Star Example No1

Convert the following Delta Resistive Network into an equivalent Star

Network

Star Delta transformation is simply the reverse of above We have seen

that when converting from a delta network to an equivalent star network

that the resistor connected to one terminal is the product of the two delta

resistances connected to the same terminal for example resistor P is the

product of resistors A and B connected to terminal 1

49

By rewriting the previous formulas a little we can also find the

transformation formulas for converting a resistive star network to an

equivalent delta network giving us a way of producing a star delta

transformation as shown below

Star to Delta Transformation

The value of the resistor on any one side of the delta Δ network is the

sum of all the two-product combinations of resistors in the star network

divide by the star resistor located ldquodirectly oppositerdquo the delta resistor

being found For example resistor A is given as

with respect to terminal 3 and resistor B is given as

with respect to terminal 2 with resistor C given as

with respect to terminal 1

By dividing out each equation by the value of the denominator we end

up with three separate transformation formulas that can be used to

convert any Delta resistive network into an equivalent star network as

given below

50

Star Delta Transformation Equations

Star Delta Transformation allows us to convert one type of circuit

connection into another type in order for us to easily analyse the circuit

and star delta transformation techniques can be used for either

resistances or impedances

One final point about converting a star resistive network to an equivalent

delta network If all the resistors in the star network are all equal in value

then the resultant resistors in the equivalent delta network will be three

times the value of the star resistors and equal giving RDELTA = 3RSTAR

Maximum Power Transfer

In many practical situations a circuit is designed to provide power to a

load While for electric utilities minimizing power losses in the process

of transmission and distribution is critical for efficiency and economic

reasons there are other applications in areas such as communications

where it is desirable to maximize the power delivered to a load We now

address the problem of delivering the maximum power to a load when

given a system with known internal lossesIt should be noted that this

will result in significant internal losses greater than or equal to the power

delivered to the load

The Thevenin equivalent is useful in finding the maximum power a

linear circuit can deliver to a load We assume that we can adjust the

load resistance RL If the entire circuit is replaced by its Thevenin

equivalent except for the load as shown the power delivered to the load

is

51

For a given circuit V Th and R Th are fixed Make a sketch of the power

as a function of load resistance

To find the point of maximum power transfer we differentiate p in the

equation above with respect to RL and set the result equal to zero Do

this to Write RL as a function of Vth and Rth

For maximum power transfer

ThL

L

RRdR

dPP 0max

Maximum power is transferred to the load when the load resistance

equals the Thevenin resistance as seen from the load (RL = RTh)

The maximum power transferred is obtained by substituting RL = RTh

into the equation for power so that

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

If a resistor connected a source of alternating emf the current through

the resistor will be a function of time According to Ohm‟s Law the

voltage v across a resistor is proportional to the instantaneous current i

flowing through it

Current I = V R = (Vo R) sin(2πft) = Io sin(2πft)

The current passing through the resistance independent on the frequency

The current and the voltage are in the same phase

53

Capacitor

A capacitor is a device that stores charge It usually consists of two

conducting electrodes separated by non-conducting material Current

does not actually flow through a capacitor in circuit Instead charge

builds up on the plates when a potential is applied across the electrodes

The maximum amount of charge a capacitor can hold at a given

potential depends on its capacitance It can be used in electronics

communications computers and power systems as the tuning circuits of

radio receivers and as dynamic memory elements in computer systems

Capacitance is the ability to store an electric charge It is equal to the

amount of charge that can be stored in a capacitor divided by the voltage

applied across the plates The unit of capacitance is the farad (F)

V

QCVQ

where C = capacitance F

Q = amount of charge Coulomb

V = voltage V

54

The farad is that capacitance that will store one coulomb of charge in the

dielectric when the voltage applied across the capacitor terminals is one

volt (Farad = Coulomb volt)

Types of Capacitors

Commercial capacitors are named according to their dielectric Most

common are air mica paper and ceramic capacitors plus the

electrolytic type Most types of capacitors can be connected to an

electric circuit without regard to polarity But electrolytic capacitors and

certain ceramic capacitors are marked to show which side must be

connected to the more positive side of a circuit

Three factors determine the value of capacitance

1- the surface area of the plates ndash the larger area the greater the

capacitance

2- the spacing between the plates ndash the smaller the spacing the greater

the capacitance

3- the permittivity of the material ndash the higher the permittivity the

greater the capacitance

d

AC

where C = capacitance

A = surface area of each plate

d = distance between plates

= permittivity of the dielectric material between plates

According to the passive sign convention current is considered to flow

into the positive terminal of the capacitor when the capacitor is being

charged and out of the positive terminal when the capacitor is

discharging

55

Symbols for capacitors a) fixed capacitor b) variable capacitor

Current-voltage relationship of the capacitor

The power delivered to the capacitor

dt

dvCvvip

Energy stored in the capacitor

2

1vCw

Capacitors in Series and Parallel

Series

When capacitors are connected in series the total capacitance CT is

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

Capacitive Reactance

Capacitive reactance Xc is the opposition to the flow of ac current due to

the capacitance in the circuit The unit of capacitive reactance is the

ohm Capacitive reactance can be found by using the equation

fCXC

2

1

Where Xc= capacitive reactance Ω

f = frequency Hz

C= capacitance F

57

For DC the frequency = zero and hence Xc = infin This means the

capacitor blocking the DC current

The capacitor takes time for charge to build up in or discharge from a

capacitor there is a time lag between the voltage across it and the

current in the circuit We say that the current and the voltage are out of

phase they reach their maximum or minimum values at different times

We find that the current leads the capacitor voltage by a quarter of a

cycle If we associate 360 degrees with one full cycle then the current

and voltage across the capacitor are out of phase by 90 degrees

Alternating voltage V = Vo sinωt

Charge on the capacitor q = C V = C Vo sinωt

58

Current I =dq

dt= ω C Vo cosωt = Io sin(ωt +

π

2)

Note Vo

Io=

1

ω C= XC

The reason the current is said to lead the voltage is that the equation for

current has the term + π2 in the sine function argument t is the same

time in both equations

Example

A 120 Hz 25 mA ac current flows in a circuit containing a 10 microF

capacitor What is the voltage drop across the capacitor

Solution

Find Xc and then Vc by Ohms law

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

An inductor is simply a coil of conducting wire When a time-varying

current flows through the coil a back emf is induced in the coil that

opposes a change in the current passing through the coil The coil

designed to store energy in its magnetic field It can be used in power

supplies transformers radios TVs radars and electric motors

The self-induced voltage VL = L dI

dt

Where L = inductance H

LV = induced voltage across the coil V

59

dI

dt = rate of change of current As

For DC dIdt = 0 so the inductor is just another piece of wire

The symbol for inductance is L and its unit is the henry (H) One henry

is the amount of inductance that permits one volt to be induced when the

current changes at the rate of one ampere per second

Induced emf in the coil VL = L dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

ω LVo cos(ωt) = Io sin(ωt minus

π

2)

60

Note Vo

Io= ω L = XL

In an inductor the current curve lags behind the voltage curve by π2

radians (90deg) as the diagram above shows

Power delivered to the inductor

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

Inductive reactance XL is the opposition to ac current due to the

inductance in the circuit

fLX L 2

Where XL= inductive reactance Ω

f = frequency Hz

L= inductance H

XL is directly proportional to the frequency of the voltage in the circuit

and the inductance of the coil This indicates an increase in frequency or

inductance increases the resistance to current flow

Inductors in Series and Parallel

If inductors are spaced sufficiently far apart sothat they do not interact

electromagnetically with each other their values can be combined just

like resistors when connected together

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

A choke coil of negligible resistance is to limit the current through it to

50 mA when 25 V is applied across it at 400 kHz Find its inductance

Solution

Find XL by Ohms law and then find L

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

(a) Calculate the reactance of a coil of inductance 032H when it is

connected to a 50Hz supply

(b) A coil has a reactance of 124 ohm in a circuit with a supply of

frequency 5 kHz Determine the inductance of the coil

Solution

(a)Inductive reactance XL = 2πfL = 2π(50)(032) = 1005 ohm

(b) Since XL = 2πfL inductance L = XL 2πf = 124 2π(5000) = 395 mH

Problem

A coil has an inductance of 40 mH and negligible resistance Calculate

its inductive reactance and the resulting current if connected to

(a) a 240 V 50 Hz supply and (b) a 100 V 1 kHz supply

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

RC connected in Series

By applying Kirchhoffs laws apply at any instant the voltage vs(t)

across a resistor and capacitor in series

Vs(t) = VR(t) + VC(t)

However the addition is more complicated because the two are not in

phase they add to give a new sinusoidal voltage but the amplitude VS is

less than VR + VC

Applying again Pythagoras theorem

RCconnected in Series

The instantaneous voltage vs(t) across the resistor and inductor in series

will be Vs(t) = VR(t) + VL(t)

64

And again the amplitude VRLS will be always less than VR + VL

RLC connected in Series

The instantaneous voltage Vs(t) across the resistor capacitor and

inductor in series will be VRCLS(t) = VR(t) + VC(t) + VL(t)

The amplitude VRCLS will be always less than VR + VC + VL

65

Problem

In a series RndashL circuit the pd across the resistance R is 12 V and the

pd across the inductance L is 5 V Find the supply voltage and the

phase angle between current and voltage

Solution

66

Problem

A coil has a resistance of 4 Ω and an inductance of 955 mH Calculate

(a) the reactance (b) the impedance and (c) the current taken from a 240

V 50 Hz supply Determine also the phase angle between the supply

voltage and current

Solution

R = 4 Ohm L = 955mH = 955 x 10_3 H f = 50 Hz and V = 240V

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

The phase angle between current and voltage

tanϕ = XL

R=

3

5 ϕ = 3687 degree lagging

67

Problem

A coil takes a current of 2A from a 12V dc supply When connected to

a 240 V 50 Hz supply the current is 20 A Calculate the resistance

impedance inductive reactance and inductance of the coil

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

XL = Z2 minus R2 = 1039 ohm

Since XL = 2πfL so L =XL

2πf= 331 H

This problem indicates a simple method for finding the inductance of a

coil ie firstly to measure the current when the coil is connected to a

dc supply of known voltage and then to repeat the process with an ac

supply

Problem

A coil consists of a resistance of 100 ohm and an inductance of 200 mH

If an alternating voltage v given by V = 200 sin 500 t volts is applied

across the coil calculate (a) the circuit impedance (b) the current

flowing (c) the pd across the resistance (d) the pd across the

inductance and (e) the phase angle between voltage and current

Solution

Since V = 200 sin 500 t volts then Vm = 200V and ω = 2πf = 500 rads

68

ac
Heading3

Page 2: Chapter one: DC circuit theory & Single-phase series AC ...

2

Current and Voltage

Current is the flow of electrons It is the time rate of change of

electrons passing through a defined area such as the cross-section of a

wire Because electrons are negatively charged the positive direction of

current flow is opposite to that of electron flow The mathematical

description of the relationship between the number of electrons (called

charge q ) and current i is

dqi

dt or q t i t d

The unit of charge is the coulomb (C) (in recognition of Charles

Augustin Coulomb French physicist and mathematician 1736-1806)

which represents 18624 10 electrons

The unit of current is the ampere or simply amp (in recognition of

Andre‟ Marie Ampere French physicist and mathematician 1775-1836)

which is defined as a coulomb per second

Ampere = coulomb second

Thus 1 amp is 18624 10 electrons moving from one body to another in 1

second

Energy is required to move a charge between two points in a circuit The

work per unit charge required to do this is called voltage The voltage

difference between two points in a circuit is a measure of the energy

required to move charge from one point to the other

3

meter Thus

4

Direct Current (DC)

circuits

5

cycles per second

6

voltage

s

7

effect

A

lR

where

is the length

ohm symbol Ω

8

l

RA

J

E

metre Vm)

1

Example

Solution

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 3: Chapter one: DC circuit theory & Single-phase series AC ...

3

meter Thus

4

Direct Current (DC)

circuits

5

cycles per second

6

voltage

s

7

effect

A

lR

where

is the length

ohm symbol Ω

8

l

RA

J

E

metre Vm)

1

Example

Solution

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 4: Chapter one: DC circuit theory & Single-phase series AC ...

4

Direct Current (DC)

circuits

5

cycles per second

6

voltage

s

7

effect

A

lR

where

is the length

ohm symbol Ω

8

l

RA

J

E

metre Vm)

1

Example

Solution

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 5: Chapter one: DC circuit theory & Single-phase series AC ...

5

cycles per second

6

voltage

s

7

effect

A

lR

where

is the length

ohm symbol Ω

8

l

RA

J

E

metre Vm)

1

Example

Solution

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 6: Chapter one: DC circuit theory & Single-phase series AC ...

6

voltage

s

7

effect

A

lR

where

is the length

ohm symbol Ω

8

l

RA

J

E

metre Vm)

1

Example

Solution

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 7: Chapter one: DC circuit theory & Single-phase series AC ...

7

effect

A

lR

where

is the length

ohm symbol Ω

8

l

RA

J

E

metre Vm)

1

Example

Solution

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 8: Chapter one: DC circuit theory & Single-phase series AC ...

8

l

RA

J

E

metre Vm)

1

Example

Solution

from the table

A

lR

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 9: Chapter one: DC circuit theory & Single-phase series AC ...

9

7211052

25010 x 1726

-8

x

xR

Ohms law

temperature

R

VI

Example

Solution

R = 23300000052 = 45000000 = 45 M

Example

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 10: Chapter one: DC circuit theory & Single-phase series AC ...

10

Solution

mVxxxxRIV 3103102510120 339

Example

Solution

Using Ohms Law Eq

Ax

R

VI 50

440

1022 3

Electrical Power

formula is

P = VI

Where P = power W

V = voltage V

I = current A

substituting V = IR

RIP 2

VI

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 11: Chapter one: DC circuit theory & Single-phase series AC ...

11

R

VP

2

Example

Solution

WxRIP 410020 22

Example

Solution

WR

VP 10

25000

500 22

formula

U =VIt (Joules)

Example

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 12: Chapter one: DC circuit theory & Single-phase series AC ...

12

Solution

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example

from the supply

Solution

x

t

UP 11000

6030

1081 6

Thus AV

PI 4

250

1000

Example

Solution

= 30000 Wh= 30 kWh

Series circuits

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 13: Chapter one: DC circuit theory & Single-phase series AC ...

13

In a series circuit

voltage V ie

From Ohm‟s law

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 14: Chapter one: DC circuit theory & Single-phase series AC ...

14

Solution

Voltage divider

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 15: Chapter one: DC circuit theory & Single-phase series AC ...

15

Example

Solution

Parallel networks

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 16: Chapter one: DC circuit theory & Single-phase series AC ...

16

From Ohm‟s law

parallel circuit

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 17: Chapter one: DC circuit theory & Single-phase series AC ...

17

Example

Solution

Current I = I1 + I2 + I3 = 6 + 3 + 1 = 10 A

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 18: Chapter one: DC circuit theory & Single-phase series AC ...

18

Current division

isgiven by

Example

Solution

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 19: Chapter one: DC circuit theory & Single-phase series AC ...

19

Figure (a)ndash(d)

From Figure (d)

From Figure (b)

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 20: Chapter one: DC circuit theory & Single-phase series AC ...

20

I1 + I2 = I3 + I4 + I5 I1 + I2 - I3 - I4 - I5 = 0

E1 - E2 = IR1 + IR2 + IR3

negative)

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 21: Chapter one: DC circuit theory & Single-phase series AC ...

21

Example

Solution

Example

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 22: Chapter one: DC circuit theory & Single-phase series AC ...

22

Solution

3 + 6 + E - 4 = (I)(2) + (I)(25) + (I)(15) + (I)(1)

5 + E = I (7)

Example

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 23: Chapter one: DC circuit theory & Single-phase series AC ...

23

Solution

(1)

(2)

(3)

(4)

From (1)

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 24: Chapter one: DC circuit theory & Single-phase series AC ...

24

Example

shown in Figure 27

Solution

For loop 1

(1)

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 25: Chapter one: DC circuit theory & Single-phase series AC ...

25

For loop 2

(2)

(3)

(2) + (3) give

From (1)

Example

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 26: Chapter one: DC circuit theory & Single-phase series AC ...

26

Solution

direction

(1)

(2)

(3)

(4)

(4) ndash (3) gives

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 27: Chapter one: DC circuit theory & Single-phase series AC ...

27

Hence

resistances‟

Example

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 28: Chapter one: DC circuit theory & Single-phase series AC ...

28

Solution Procedure

From Figure (a)

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 29: Chapter one: DC circuit theory & Single-phase series AC ...

29

From Figure (aa)

By current divide

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 30: Chapter one: DC circuit theory & Single-phase series AC ...

30

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 31: Chapter one: DC circuit theory & Single-phase series AC ...

31

below

Example

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 32: Chapter one: DC circuit theory & Single-phase series AC ...

32

Solution

218

by

Pd across R2 = I1R2 = 1 x 8 = 8 V

E = 8 V

Resistance

Current

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 33: Chapter one: DC circuit theory & Single-phase series AC ...

33

Example

Solution

221(b)

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 34: Chapter one: DC circuit theory & Single-phase series AC ...

34

Hence

direction A to B

whichCurrent

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 35: Chapter one: DC circuit theory & Single-phase series AC ...

35

flowing from A to B

Example

Solution

223

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 36: Chapter one: DC circuit theory & Single-phase series AC ...

36

below

active network

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 37: Chapter one: DC circuit theory & Single-phase series AC ...

37

and B

Example

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 38: Chapter one: DC circuit theory & Single-phase series AC ...

38

Solution

in Figure 228

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 39: Chapter one: DC circuit theory & Single-phase series AC ...

39

Example

Solution

Figure 231(b)

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 40: Chapter one: DC circuit theory & Single-phase series AC ...

40

Extra Example

Figure 232

Solution

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 41: Chapter one: DC circuit theory & Single-phase series AC ...

41

41 405015 II

)(405015 311 III

31 8183 II

52 302010 II

)(302010 322 III

32 351 II

453 4030250 III

)(40)(30250 31323 IIIII

321 19680 III

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 42: Chapter one: DC circuit theory & Single-phase series AC ...

42

0

1

3

1968

350

8018

3

2

1

I

10661083278568

0183

198

8185

1968

350

8018

xx

183481773196

801

196

353

1960

351

803

1

xx

206178191831

838

190

3118

1908

310

8318

2

xx

12618140368

0181

68

503

068

150

3018

3

xxx

17201066

18311

I A

19301066

20622

I A

01101066

1233

I A

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 43: Chapter one: DC circuit theory & Single-phase series AC ...

43

1610011017204 I A

1820011019305 I A

good thing

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 44: Chapter one: DC circuit theory & Single-phase series AC ...

44

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 45: Chapter one: DC circuit theory & Single-phase series AC ...

45

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 46: Chapter one: DC circuit theory & Single-phase series AC ...

46

gives

equation 2 gives

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 47: Chapter one: DC circuit theory & Single-phase series AC ...

47

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 48: Chapter one: DC circuit theory & Single-phase series AC ...

48

Network

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 49: Chapter one: DC circuit theory & Single-phase series AC ...

49

given below

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 50: Chapter one: DC circuit theory & Single-phase series AC ...

50

is

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 51: Chapter one: DC circuit theory & Single-phase series AC ...

51

ThL

L

RRdR

dPP 0max

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 52: Chapter one: DC circuit theory & Single-phase series AC ...

52

RVpRR

Th

ThL 4

2

max

AC Resistor Circuit

flowing through it

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 53: Chapter one: DC circuit theory & Single-phase series AC ...

53

Capacitor

V

QCVQ

V = voltage V

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 54: Chapter one: DC circuit theory & Single-phase series AC ...

54

Types of Capacitors

capacitance

the capacitance

d

AC

discharging

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 55: Chapter one: DC circuit theory & Single-phase series AC ...

55

dt

dvCvvip

2

1vCw

Series

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 56: Chapter one: DC circuit theory & Single-phase series AC ...

56

nT CCCCC

1

1111

321

Parallel

nT CCCCC 321

fCXC

2

1

f = frequency Hz

C= capacitance F

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 57: Chapter one: DC circuit theory & Single-phase series AC ...

57

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 58: Chapter one: DC circuit theory & Single-phase series AC ...

58

Current I =dq

π

2)

Note Vo

Io=

1

ω C= XC

Example

Solution

513210101202

1

2

16xxxxfC

XC

VxxIXV CCC 31310255132 3

Inductor

dt

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 59: Chapter one: DC circuit theory & Single-phase series AC ...

59

dI

dt

Current I = 1

L V dt =

1

LVo sinωt dt

= 1

π

2)

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 60: Chapter one: DC circuit theory & Single-phase series AC ...

60

Note Vo

Io= ω L = XL

idt

diLvip

Energy stored

2

1iLw

Inductive Reactance

fLX L 2

f = frequency Hz

L= inductance H

Series nT LLLLL 321

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 61: Chapter one: DC circuit theory & Single-phase series AC ...

61

Parallel nT LLLLL

1

1111

3211

Example

Solution

5001050

253xI

VX

L

LL

mHHxxxxf

XL L 20101990

104002

500

2

3

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 62: Chapter one: DC circuit theory & Single-phase series AC ...

62

Problem

Solution

Problem

Solution

XL = 2πfL = 2π(50)(40 x 10-3

) = 1257 ohm

Current I = V XL = 240 1257 = 1909A

XL = 2π (1000)(40 x 10-3) = 2513 ohm

Current I = V XL = 100 2513 = 0398A

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 63: Chapter one: DC circuit theory & Single-phase series AC ...

63

less than VR + VC

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 64: Chapter one: DC circuit theory & Single-phase series AC ...

64

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 65: Chapter one: DC circuit theory & Single-phase series AC ...

65

Problem

Solution

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 66: Chapter one: DC circuit theory & Single-phase series AC ...

66

Problem

voltage and current

Solution

XL = 2πfL = 3 ohm

Z = R2 + XL2 = 5 ohm

I = V

Z= 48 A

tanϕ = XL

R=

3

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 67: Chapter one: DC circuit theory & Single-phase series AC ...

67

Problem

Solution

R = d c voltage

d c current=

12

2= 6 ohm

Z = a c voltage

a c current= 12 ohm

Since

Z = R2 + XL2

2πf= 331 H

supply

Problem

Solution

68

ac
Heading3

Page 68: Chapter one: DC circuit theory & Single-phase series AC ...

68

ac
Heading3

Date post:	23-Mar-2022
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Chapter one: DC circuit theory & Single-phase series AC ...

Documents