Chapter 11 Alternating Current Circuits

Recommended Problems:

1,3,5,9,11,13,15,17,19,21,23,29,31,33,37,41,43,61,65.

Alternating Current Sources

Generator or dynamo is

a device that operate

according to the

electromagnetic induction

and Faraday's law.

It consists of a loop of wire that can rotate, by an external means,

in a magnetic field. The ends of the loop are connected to slip

rings that rotate with the loop. The slip rings are connected to the

external circuit by two fixed brushes.

Figure 33.1 Schematic diagram of a generator.

It is a device that

produces a current by

converting the

mechanical energy into

electrical energy.

As the loop rotate, the magnetic flux through it changes with time

inducing an emf and so a current in the external circuit.

Suppose that the loop, with area A, rotates with constant angular

speed , and suppose that is the angle between the magnetic

field and the area of the loop. The magnetic flux through the loop

at any time is then

tBABAm coscos

Therefore, the induced emf in the loop is given by

ttABdt

d m sinsin max

This means that the emf varies sinusoidally with

time forming a sinusoidal wave, as shown in the

figure. Because of that a source of AC is

represented in the circuits by the symbol

max

-max

t

I

Io

-Io

t

I

t

(a)

The frequency of commercial

generators in our country and most of

the world is 50 Hz, while it is 60 Hz in

some other countries like the USA and

Canada. Note that the frequency f is

related to the angular speed through

the relation . f 2

Ohm’s law is still hold for AC circuits,

i.e.,

tRR

I

sinmax

tII m sin

The voltage across an element in an

AC circuit is given by

tVV m sin

Im and Vm are called the peak current and the peak voltage,

respectively.

As it is clear from Figure 14.7(b), the average value of an AC is

zero. This doesn't mean, however, that no power is needed or that

no heat is produced in a resistor. Electrons for AC move forth and

back and so produce heat.

It is known that the power dissipated in a resistor is given by

tRIRIP m 222 sin

Since the current is now squared, the power is always positive.

It is easy to show that the average value of 2

1sin 2 t

And so the average value of 2212 is mII

the average power dissipated in a resistor in AC circuits is then

R

VRIP m

m

2

212

21

That is, what is important for calculating the average power is the

mean value of the square of the current or voltage. The square

root of each of these values is known as the rms (root-mean-

square) and is given by

2m

rms

II

2m

rms

VV

The rms values of V and I are sometimes known as the effective

values. They are useful because they can be substituted directly

into the famous power formula to get the average power, that is,

R

VRIVIP rms

rmsrmsrms

22

It is the rms values that are specified or measured. Therefore,

ammeters and voltmeters are designed to read the corresponding

rms values. When we say that the standard voltage in our country

is 220 V, we mean that the rms value of V is 220V.

The peak value of such a voltage is V3112 rmsm VV

Phasor: It is a counterclockwise-rotating vector representing an

AC quantity such that its length represents the maximum value of

the quantity and its projection onto the vertical axis represents the

instantaneous value of the quantity.

I vs t

t

It is clear that the instantaneous value is changing with time as the

phasor rotates while the maximum value is constant.

• Test Your Understanding (1)

Consider the voltage phasor shown. Choose the figure at which

the instantaneous value of the voltage has the largest magnitude.

(a) (b) (c) (d)

Choose the figure at which the instantaneous value of the voltage

has the smallest magnitude.

v

(a)

v

(b)

v

(d)

v

(c)

(a) (b) (c) (d)

Example 33.1 A 900.-W microwave oven is designed to

operate at 220. V. Calculate its resistance and the peak current

when it is operating.

Solution: To find the resistance we have

And for the current we have

8.53

900

22022

P

VR rms

A09.4220

900

rmsrms

V

PI A78.52 rmsm II

Resistors in AC Circuits

Consider the circuit that consists of a

resistor and connected in series with an

AC source.

R

tVv m sin

Let vR, be the voltage at some instant across the resistor. This

voltage must equal to the voltage of the AC source, that is

tVv mR sin

The current passing through the resistor is

tIR

tV

R

vi m

mR

sinsin

As it is clear from the last two equations that

The current i and voltage vR across a resistor in a pure resistive

AC circuits are in phase.

Vm

iR, vR

t

Im

iR

t

iR, vR

vR

Inductors in AC Circuits

Consider the circuit that consists of an

inductor and connected in series with

an AC source.

L

tVv m sin

Let vL, be the voltage at some instant

across the inductor. Again this voltage

must equal to the voltage of the AC

source, that is

tVv mL sin

To find the current passing through the inductor we have

dt

diLvL

)cos()sin(1

tL

Vdtt

L

Vdtv

Li mm

L

Using the identity

)cos(2sin tt

2sin2sin

tItL

Vi m

m

L

mm

X

VI with

LX L and Is called the inductive reactance.

The current i lags behind the voltage vL across an inductor in a

pure inductive AC circuits by 90o .

Vm

iR, vR

t

Im

iR

Im

vR

t

iR, vR

Vm

• Test Your Understanding (2)

Consider the AC circuit shown. The

frequency of the AC source is adjusted

while its voltage amplitude is held

constant. The lightbulb will glow the

brightest at

a) high frequencies b) low frequencies

c) all frequencies. d) It will not glow at all

Capacittors in AC Circuits

Consider the circuit that consists of a

capacitor connected in series with an

AC source.

Let vC, be the voltage at some instant

across the i capacitor. Again this

voltage must equal to the voltage of

the AC source, that is

tVv mC sin

To find the current passing through the capacitor we have

C

tVv m sin

CCvq

)cos( tCVdt

dvC

dt

dqi m

C

Using the identity

)cos(2sin tt

2sin2sin tItVCi mm

C

mm

X

VI with

CXC

1and Is called the capacitive reactance.

The current i leads the voltage vC across a capacitor in a pure

capacitive AC circuits by 90o .

Vm

iR, vR

t

Im

iR

Im

vR

t

iR, vR

Vm

• Test Your Understanding (3)

Consider the AC circuit shown. The

frequency of the AC source is adjusted

while its voltage amplitude is held

constant. The lightbulb will glow the

brightest at

a) high frequencies b) low frequencies

c) all frequencies. d) It will not glow at all

• Test Your Understanding (4)

Consider the AC circuit shown. The

frequency of the AC source is adjusted

while its voltage amplitude is held

constant. The lightbulb will glow the

brightest at

a) high frequencies b) low frequencies

c) all frequencies. d) It will not glow at all

SERIRS RLC AC CIRCUITS L

C R

tVv m sin

Consider the circuit which consists of a

resistor, an inductor, and a capacitor,

connected in series with an AC source. Let

vR, vL, and vC be the voltage at some instant

across each element, respectively.

Since they are connected in series, the voltage across the

combination is CLR vvvv

This voltage must equal to the voltage of the AC source, that is

CLRm vvvtV sin

Note that a voltmeter, when connected across the combination,

does not read (VR)rms + (VL)rms + (VC)rms . What will read then? Let

us see.

To solve this equation we use the phase diagram by letting the

phasor representing I to be along the horizontal axis

where the constant is called the phase

angle between the current and the

applied voltage. Now we have

tRIIRv mR sin

2sin

tXIv LmL

2sin

tXIv CmC

The phase diagram is now

Im

Vm

VR

VL

VC

VR

VL-VC

Vm

Supposing the current passing through each element (which is the

same due their series connection) to be given as

tII m sin

From the diagram we conclude that

22CLRm VVVV

R

CL

V

VV tan

Using the relations RIV mR LmL XIV CmC XIV

ZIXXRIV mCLmm 22

22CL XXRZ

With Z is called the impedance of the circuit with its unit is . The

phase angle is now given as

R

XX CL tan

From the last two equations we conclude that

if the circuit consists only of a resistor with the source, (XL =XC

=0), we have Z=R, and the phase angle is zero, that is I and V are

in phase.

if the circuit consists only of an inductor with the source, (R =XC

=0), we have Z= XL, and the phase angle is 90o, that is I lags

behind V by 90o .

if the circuit consists only of a capacitor with the source, (R =XL

=0), we have Z= XC, and the phase angle is -90o, that is I leads V by 90o .

In general if XL > XC the phase angle is +ve, and I lags behind

V by . In the other hand if XL < XC the phase angle is -ve, and I leadsV by

When XL = XC then phase angle =0. In this case, the

impedance equals the resistance and the current has its peak

value given by RVm

The frequency o at which this occurs is called the resonance

frequency of the circuit. Using the condition XL = XC we get

C

Lo

o

1

LCo

1

• Test Your Understanding (6)

Indicate each part of the figure shown whether

XL = XC , XL >XC , XL < XC .

a) XL < XC b) XL = XC

c) XL >XC

Example 33.4 In a series RLC circuit we have Vm =120 V,

R= 200 , C=4.0 F and f= 60 Hz. Find L such that the voltage

across the capacitor lags the applied voltage by 30o.

Solution

Since the angle is between the

applied voltage and the voltage

across R, and since the angle

between VR and VC is 90o

Im

Vm

VR

VL

VC

30o ooo 609030

663)104)(60(2

11Now

6CXC

200

66373.1tan LCL X

R

XX

Hf

LLX L 83.0120

317

2

317317

Example 33.4 In a series RLC circuit we have Vm =150 V,

R= 425 , C=3.5 F, L=1.25 H, and =377 s-1 .

a) Determine Z.

b) Find the maximum current in the circuit.

c) Find the phase angle.

d) Find the maximum voltage across each element.

Solution

Let us first find the reactances

47125.1377LXL 758105.337711 6CXC

a) The impedance is then

5137584714252222

CL XXRz

b) The maximum current is given by

A292.0513

150

Z

VI mm

c) The phase angle is given by

oCL

R

XX34

.425

758417tan

d) For the maximum voltage across each element we have

Note that since the circuit is more capacitive (XL < XC ), the

current leads the applied voltage by the angle 34o

V124 RIV mR

V138 LmL XIV

V221 CmR XIV

Power in AC Circuits

The instantaneous power delivered by Ac source is

)sin()sin( maxmax tVtIiVP

sin)cos(cossinsinusing ttt

sin)cos()sin(cos)(sin maxmax2

maxmax ttVItVIP

sin)cos()sin(cos)(sin2maxmax avravravr tttVIP

Knowing that 212 )(sin avrt 0)cos()sin( avrtt

cos2

1maxmaxVIPavr cosor rmsrmsavr VIP

The quantity cos is called the power factor given by

max

max

max

cosV

RI

V

VR

RIP rmsavr2

Example 33.5 In a series RLC circuit we have Vm =150 V,

R= 425 , C=3.5 F, L=1.25 H, and =377 s-1 . What is the

power delivered to the circuit.

Solution

From the previous example we have Z=518 .

A206.02Z

VI mrms

WRIP rmsavr 1.18)425()206.0( 22

Resonance in a Series in RLC Circuits

It is known that

22

22

22

CL

rmsrms

rmsrmsavr

XXR

RVRI

z

RVRIP

222222

2 1oCL L

CLXX

222222

22

o

rmsavr

LR

RVP

From the last Equation it is clear that the average power is

maximum when = o (at resonance). At resonance the average

power is given by

R

VP rms

avr

2

The average power versus the

frequency is plotted for two

different values of the frequency.

The sharpness of the curve is

measured by a factor known as

the quality factor defined as

oQ

With is the width of the

resonant power curve at half

maximum and given by

L

R

R

LQ o

• Test Your Understanding (7)

The impedance of a series RLC Circuit is

a) Larger than R

b) Less than R c) Equal to R

d) Impossible to determine.

Example 33.7 In a series RLC circuit we have Vm =20 V,

R= 150 , L=2.0 mH, and =5000 s-1 . What is the value of C for

the current is maximum.

Solution

The maximum current is attained when = o .

FCCLC

o 2102

15000

1

3

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