Basic Concepts & Network Theorems - AMIE

A PREPARATORY COURSE FOR J.E. LEVEL EXAMS BASIC CONCEPTS AND NETWORK THEOREMS

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Basic Concepts & Network Theorems

OHM’S LAW Ohm’s law gives a relationship between current and potential difference. According to Ohm’s law : At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. If I is the current flowing through a conductor and V is the potential difference (or voltage) across its ends, then according to Ohm’s law is given by following equation

V = IR

Example

An electric iron draws a current of 3.4 A from the 220 V supply line. What current will this electric iron draw when connected to 110 V supply line ?

Solution

First of all we will calculate the resistance of electric iron. Now, in the first case, the electric

iron draws a current of 3.4 A from 220 V supply line. So,

Potential difference, V = 220 V

Current, I = 3.4 A

And, Resistance, R = ? (To be calculated)

Now from V = IR, we get R = 64.7

Thus, the resistance of electric iron is 64.7 ohms. This resistance will now be used to find out the current drawn when the electric iron is connected to 110 V supply line. So,

V/I = R or 110/I = 64.7

I = 1.7 A

Thus, the electric iron will draw a current of 1.7 amperes from 110 volt supply line.

FACTORS AFFECTING THE RESISTANCE OF A CONDUCTOR The electrical resistance of a conductor (or a wire) depends on the following factors :

length of the conductor It has been found by experiments that on increasing the length of a wire, its resistance increases; and on decreasing the length of the wire, its resistance decreases.

area of cross-section of the conductor (or thickness of the conductor) It has been found by experiments that the resistance of a conductor is inversely proportional to its area of cross-section. Since the resistance of a wire (or conductor) is inversely proportional to its area of cross-section, therefore, when the area of cross-section of a wire is doubled, its resistance gets halved; and if the area of cross section of wire is halved, then its resistance will get doubled. The thickness of a wire is usually represented by its diameter. It can be shown by calculations that the resistance of a wire is inversely proportional to the square of its diameter. Thus, when the diameter of a wire is doubled (made 2 times), its resistance becomes one-fourth (1/4).




nature of the material of the conductor The electrical resistance of a conductor (say, a wire) depends on the nature of the material of which it is made. Some materials have low resistance whereas others have high resistance. For example, if we take two similar wires, having equal lengths and diameters, of copper metal and nichrome alloy, we will find that the resistance of nichrome wire is about 60 times more than that of the copper wire. This shows that the resistance of a conductor depends on the nature of the material of the conductor.

temperature of the conductor It has been found that the resistance of all pure metals increases on raising the temperature; and decreases on lowering the temperature. But the resistance of alloys like manganin, constantan and nichrome is almost unaffected by temperature.

RESISTIVITY Resistivity,

RA

l

where (rho) is a constant known as resistivity of the material of the conductor. Resistivity is also known as specific resistance. R = resistance of the conductor; A = area of cross-section of the conductor and l = length of the conductor.

Example

A 6 resistance wire is doubled up by folding. Calculate the new resistance of the wire.

Solution

Suppose the length of 6 resistance wire is l, its area of cross-section is A and its resistivity

is .

Then 6xl

A

(1)

Now, when this wire is doubled up by folding, then its length will become half, that is, the length will become 1/2. But on doubling the wire by folding, its area of cross-section will become double, that is, the area of cross-section will become 2A. Suppose the new resistance of the doubled up wire (or folded wire) is R.

So, 62 2 4

xl l

x xA A

(2)

Now, dividing equation (2) by equation (1) and solving, we get

R = 1.5




D.C. SERIES AND PARALLEL CIRCUITS In D.C. series circuit, if R1, R2 and R3 are the resistances connected in series, then same current flows in each of these resistances and the total resistance of the combination is given by

1 1 2 3R R R R

Current in each resistance will be same but voltage will be different.

If the above three resistors are connected in parallel combination, then same voltage acts across each of these resistors and total resistance is

1 2 3

1 1 1 1

R R R R

where, 1/R is the inverse of net resistance of parallel combination.

Here, Voltage will be same in each resistance but current will be different.

Series Vs. Parallel Circuits In series circuit various parameters are connected end to end, whereas in parallel

circuit, elements are connected across each other.

In series circuit vector sum of voltages across each element is equal to the applied voltage, whereas in parallel circuit voltage across all the elements are equal.




In series circuit, the value of current is same in all the elements whereas the current in each element is different through each element in parallel circuit. Total current is the

vector sum of all the currents.

In the series circuit equivalent impedance is equal to the vector sum of all the impedances, whereas in parallel circuit the reciprocal of equivalent impedance is

equal to the sum of reciprocals of each impedance.

ELECTRIC POWER When an electric current flows through a conductor, electrical energy is used up and we say that the current is doing work. We know that the rate of doing work is called power, so electric power is the electrical work done per unit time. That is,

Power = Work done/ Time taken

or P = W/t

We have calculated the power by dividing work done by time taken. Now, the unit of work is “joule” and that of time is “second”. So, the unit of power is “joules per second”. This unit of power is called watt. Thus, the SI unit of electric power is watt which is denoted by the

letter W. The power of 1 watt is a rate of working of 1 joule per second.

Actually, watt is a small unit, therefore, a bigger unit of electric power called kilowatt is used

for commercial purposes. It is obvious that :

1 kilowatt = 1000 watts

or 1 kW = 1000 W

Power formulas are

First formula

P = VI

Second formula

P = I2R

Third formula

P = V2/R

ELECTRICAL ENERGY Electrical energy = Power × Time

or E = P × t

In the formula : Electrical energy = Power × Time, if we take the power in ‘watts’and time in ‘hours’ then the unit of electrical energy becomes ‘Watt-hour’ (Wh). One watt-hour is the amount of electrical energy consumed when an electrical appliance of 1 watt power is used for 1 hour.




COMMERCIAL UNIT OF ELECTRICAL ENERGY : KILOWATT-HOUR The SI unit of electrical energy is joule and we know that “1 joule is the amount of electrical energy consumed when an appliance of 1 watt power is used for 1 second”. Actually, joule represents a very small quantity of energy and, therefore, it is inconvenient to use where a large quantity of energy is involved. So, for commercial purposes we use a bigger unit of electrical energy which is called “kilowatt-hour”. One kilowatt hour is the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour. Since a kilowatt means 1000 watts, so we can also say that one kilowatt-hour is the amount of electrical energy consumed when an electrical appliance of 1000 watts is used for 1 hour. In other words, one kilowatt-hour is the energy dissipated by a current at the rate of 1000 watts for 1 hour. From this discussion we conclude that the commercial unit of electrical energy is kilowatt-hour which is written in short form as kWh.

One kilowatt-hour is equal to 3.6 × 106 joules of electrical energy.

Example

A radio set of 60 watts runs for 50 hours. How much electrical energy is consumed?

Solution

Electrical energy = Power × Time

or E = P × t (1)

We want to calculate the electrical energy in kilowatt-hours, so first we should convert the

power of 60 watts into kilowatts by dividing it by 1000. That is :

Power, P = 60 watts = 60/1000 = 0.06 kW

And, Time, t = 50 hours

Now, putting P = 0.06 kW and, t = 50 hours in equation (1), we get :

Electrical energy, E = 0.06 × 50 = 3 kilowatt-hours (or 3 kWh)

Thus, electrical energy consumed is 3 kilowatt-hours

It can be converted to Joules.

3 kWh = 3.6 × 106 × 3 J = 10.8 × 106 J

Problem

A current of 4 A flows through a 12 V car headlight bulb for 10 minutes. How much energy

transfer occurs during this time?

Answer: 0.008 kWh

HOW TO CALCULATE THE COST OF ELECTRICAL ENERGY CONSUMED Kilowatt-hour is the “unit” of electrical energy for which we pay to the Electricity Supply Department of our City. One unit of electricity costs anything from rupees 3 to rupees 5 (or even more). The rates vary from place to place and keep on changing from time to time.




Now, by saying that 1 unit of electricity costs say, 3 rupees, we mean that 1 kilowatt-hour of electrical energy costs 3 rupees. The electricity meter in our homes measures the electrical

energy consumed by us in kilowatt-hours.

Example

A refrigerator having a power rating of 350 W operates for 10 hours a day. Calculate the cost of electrical energy to operate it for a month of 30 days. The rate of electrical energy is Rs. 3.40 per kWh.

Solution

Electrical energy E = P x t = (350/1000 kW) x (10 x 30 hours) = 105 kWh

Thus, the electrical energy consumed by the refrigerator in a month of 30 days is 105

kilowatt-hours.

Now, Cost of 1 kWh of electricity = ₹. 3.40

So, Cost of 105 kWh of electricity = ₹ 3.40 × 105 = ₹ 357

Problem

A bulb is rated at 200 V-100 W. What is its resistance ? Five such bulbs burn for 4 hours. What is the electrical energy consumed? Calculate the cost if the rate is ₹ 4.60 per unit.

Answer: R = 400 ohm; E = 2 kWh; Cost = ₹ 9.20.

HEATING EFFECT OF CURRENT When an electric current is passed through a high resistance wire, like nichrome wire, the resistance wire becomes very hot and produces heat. This is called the heating effect of current. The heating effect of current is obtained by the transformation of electrical energy into heat energy.

Heat produced, H = I2 × R × t joules

This formula gives us the heat produced in joules when a current of I amperes flows in a wire of resistance R ohms for time t seconds. This is known as Joule’s law of heating.

Example

Calculate the heat produced when 96,000 coulombs of charge is transferred in 1 hour through a potential difference of 50 volts.

Solution

First of all we will calculate the current by using the values of charge and time. We know that:

Current I = Q/t = 96000/60 x 60 = 26.67 A

R = V/I = 50/26.67 = 1.87




Heat produced will be

H = I2Rt = (26.67)2 x 60 x 60 = 4788400 J = 4788.4 kJ

IMPORTANT TERMS IN CIRCUITS

Loop & Mesh A loop is any closed path of a circuit, while mesh is a loop which does not contain any other loop within it. Therefore, all the meshes are loops and a loop is not necessarily a mesh. Loop may have other loops or meshes inside it.

Node and Junction

A point where two or more branches meet is called a node, while a junction is a point at which three or more branches are joined together. Therefore, all the junctions are nodes and a node is not necessarily a junction.

Sign Convention An electrical element having two terminals A and B known as a two-terminal element, is shown in figure.

Current i flows in the direction shown, from A to B. Voltage v drops from A to B i.e. A is at a higher potential than B. Hence A is marked positive (+) and B negative (-).

ENERGY SOURCES There are basically two types of energy sources ; voltage source and current source. These are classified as i) Ideal source and ii) Practical source.

Let us see the difference between ideal and practical sources.

Voltage Source Ideal voltage source is defined as the energy source which gives constant voltage across its terminals irrespective of the current drawn through its terminals. The symbol for ideal voltage source is shown in the Fig. (a). This is connected to the load as shown in Fig. (b). At any time the value of voltage at load terminals remains same. This is indicated by V-1 characteristics shown in the Fig. (c).

But practically, every voltage source has small internal resistance shown in series with voltage source and is represented by Rse as shown in the Fig.




Because of the Rse, voltage across terminals decreases slightly with increase in current and it is given by expression.

Voltage sources are further classified as follows,

Time Invariant Sources: The sources in which voltage is not varying with time are known as time invariant voltage sources or D.C. sources. These are denoted by capital letters.

Time Variant Sources: The sources in which voltage is varying with time are known as time variant voltage sources or A.C sources. These are denoted by small letters.

Current Source Ideal current source is the source which gives constant current at its terminals irrespective of the voltage appearing across its terminals. The symbol for ideal current source is shown in the Fig. (a). This is connected to the load as shown in the Fig. (b). At any time, the value of the current flowing through load IL is same i.e. is irrespective of voltage appearing across its terminals. This is explained by V-I characteristics shown in the Fig. (c).

But practically, every current source has high internal resistance, shown in parallel with current source and it is represented by Rsh. This is shown in the Fig.

Because of Rsh current through its terminals decreases slightly with increase in voltage at its terminals.

Similar to voltage sources, current sources are classified as follows :

Time Invariant Sources: The sources in which current is not varying with time are known as time invariant current sources or D.C. sources. These are denoted by capital letters.

Time Variant Sources: The sources in which current is varying with time are known as time variant current sources or A.C sources. These are denoted by small letters.

KIRCHOFF’S LAW There are two laws given by Kirchhoff:




First law According to this law “ In any network of wires carrying current, the sum of currents flowing towards the junction point(or Node Point) is equal to the sum of currents flowing away from the points, i.e. the algebraic sum of currents meeting at a point is zero”.

Consider a junction point in a complex network as shown in following figure.

At this junction point if I1 = 2A, I2 = 4A and I3 = 1A then to determine I4 we write, total current entering is 2 + 4= 6A while total current leaving is 1+ 14 A

And hence, I4 = 5 A.

This analysis of currents entering and leaving is nothing but the application of Kirchhoff s Current Law. The law can be stated as,

The total current flowing towards a junction point is equal to the total current flowing away from that junction point.

Another way to stale the law is,

The algebraic sum of all the current meeting at a junction point is always zero.

The word algebraic means considering the signs of various currents.

Signs of current moving towards the point O are taken as +ve. While that of the currents moving away from O has been taken as -ve.

Second Law According to this law, the algebraic sum of voltage drops(products of currents and resistances) and the e.m.f. in any closed circuit(or loop) is zero. [Algebraic here means a voltage rise due to e.m.f.’s being assigned (+) sign and a voltage drop due to flow of current in a resistance being assigned (-ve) sign.

Sign Convention Potential rise i.e. travelling from negative to positively marked terminal must be

considered as Positive.

Potential drop i.e. travelling from positive to negatively marked terminal considered as Negative.

While tracing a closed path, select any one direction clockwise or anticlockwise. This selection is totally independent of the directions of the currents and voltages of various, branches of that closed path.

This selection is totally independent of the directions of currents and voltages of

APPLICATION OF KVL TO A CLOSED PATH Consider a closed path of a complex network with various branch currents assumed as shown in the Fig. (a).




As the loop is assumed to be a part of complex network, the branch currents are assumed to be different from each other.

Due to these currents the various voltage drops taken place across various resistances are marked as shown in the Fig. (b).

The polarity of voltage drop along the current direction is to be marked as positive (+) to negative (-).

(a) (b)

Let us trace this closed path in clockwise direction i.e. A-B-C-D-A.

Across R1 there is voltage drop I1 R1 and as getting traced from +ve to -ve, it is drop and must be taken as negative while applying KVL.

Battery E1 is getting traced from negative to positive i.e. it is a rise hence must be considered as positive.

Across R2 there is a voltage drop I2R2 and as getting traced from +ve to -ve, it is drop and must be taken negative.

Across R3 there is a drop I3R3 and as getting traced from +ve to -ve, it is drop and must be taken as negative.

Across R4 there is drop I4R4 and as getting traced from +ve to -ve, it is drop must be taken as negative.

Battery E2 is getting traced from -ve to +ve, it is rise and must be taken as positive.

We can write an equation by using KVL around this closed path as,

1 1 1 2 2 3 3 4 4 2 0I R E I R I R I R E [KVL Equation]

or 1 2 1 1 2 2 3 3 4 4E E I R I R I R I R

Steps to Apply Kirchhoff’s Laws to Got Network Equations The steps are stated based en the branch current method.

Step 1 : Draw the circuit diagram from the given information and insert all the values of sources with appropriate polarities and all the resistances.

Step 2 : Mark all the branch currents with some assumed directions using KCL at various nodes and junction points. Kept the number of unknown currents minimum as far as possible to limit the mathematical calculations required to solve them later on.

Assumed directions may be wrong, in such case answer of such current will be mathematically negative which indicates the correct direction of the current. A particular




current leaving a particular source has some magnitude, then same magnitude of current should enter that source after travelling through various branches of the network.

Step 3 : Mark all the polarities of voltage drops and rises as per directions of the assumed branch currents flowing through various branch resistances of the network. This is necessary for application of KVL to various closed loops.

Step 4 : Apply KVL to different closed paths in the network and obtain the corresponding equations. Each equation must contain some clement which is not considered in any previous equation.

Stop 5 : Solve the simultaneous equations for the unknown currents. From these currents unknown voltages and power consumption in different resistances can be calculated.

What to do if current source exists ? If there is current source in the network then complete the current distribution considering the current source. But white applying KVL, the loops should not be considered involving current source. The loop equations must be written to those loops which do not include any current-source. This is because drop across current source is unknown.

For example, consider the circuit shown in the following figure.

The current distribution is completed in terms of current source value. Then KVL must be applied to the loop bcdeb, which does not include current source. The loop abefa should not be used for KVL application, as it includes current source. Its effect is already considered at the time of current distribution.

Example

Apply Kirchhoff’s current law and voltage law to the circuit shown in the following figure.

Indicate the various branch currents. Write down the equations relating the various branch currents. Solve these equations to find the values of these currents. Is the sign of any of the calculated currents negative ? If yes, explain the significance of the negative sign.

Solution

Step 1 and 2 : Draw the circuit with all the values which are same as the given network. Mark all the branch currents starting from +ve of any of the source, say +ve of 50 V source.




Step 3 : Mark all the polarities for different voltages across the resistances. This is combined with step 2 shown in the network below in Fig. (a).

Step 4 : Apply KVL to different loops.

Loop 1: A-B-E-F-A., - 15 I1 - 20 I2 + 50 = 0

Loop 2 : B-C-D-E-D, - 30 (I1 – I2) - 100 + 20I2

Rewriting all the equations, taking constants on one side.

15 I1 + 20 I2 = 50 (1)

and - 30 I1 + 50 I2 = 100 (2)

Solving I1 = 0.37 A, I2 = 2.22 A (both positive hence assumed directions were correct)

Also I1 – I2 = 0.37 – 2.22 = - 1.85 A

Negative sign Indicates assumed direction is wrong.

Hence I1 – I2 = 1.85 A flowing in opposite direction to that of the assumed direction.

Example

Determine the current supplied by each battery in the circuit shown in the Fig. by using Kirchhoff’s laws.

Solution

Show all the branch currents and the polarities of voltage drops across the resistances due to the respective currents as shown in following figure.




Applying KVL to various loops :

For loop 1, ABGHA

15 3 5 20 0I I

For loop 2, BCFGB

1 2 14( ) 5 2 5 5 3 0I I I I

For loop 3, CDEFC

1 2 28( ) 30 5 2 0I I I I

Solving (1), (2) and (3)

I = 2.558 A, I1 = 0.7357 A, I2 = 4.9581 A

Hence the current supplied by various batteries can be calculated as below :

Current supplied by B1 = I = 2.558 A

Current supplied by B2 = I1 = 0.7357 A

Current supplied by B3 = I2 = 4.9581 A

Current supplied by B4 = (I – I1) = (2.558 - 0.7357) = 1.8223 A

Current supplied by B5 = (I – I1 – I2) = (2.558 - 0.7357 - 4.9581) = -3.1358 A

(- ve sign means opposite direction)

Example

Using Kirchhoff’s laws, calculate the current delivered by the battery shown in Fig.

Solution

The various branch currents are shown in the Fig.

Now do yourself.

I1 = 1.3852 A




THEVENIN'S & NORTON'S THEOREM

Thevenin's Theorem Consider an active linear network (circuit) N, represented as a black box as shown in figure (a). Let A and B be any pair of terminals of N with respect to which an equivalent circuit is desired.

Thevenin's theorem states that:

"With respect to terminal pair AB, the network N may be replaced by a voltage source Vth in series with an internal impedance Zth. The voltage source Vth called the Thevenin's voltage, is the potential difference (VA - VB) between the terminals A and B, and Zth is the internal impedance of the network N as seen from the terminals A and B with all the sources set to zero, i.e., with all the voltage sources shorted and all the current sources open circuited leaving behind their internal. resistances or impedances".

The series combination of Vth and Zth, as shown in figure(b), is known as Thevenin's equivalent of the network N with respect to terminal pair AB.

Hence, it follows that:

If a load impedance ZL, be connected across terminals A and B of network, the current IL delivered to the load may be calculated from the equivalent circuits by connecting this impedance across A and B and shall be given as

thL

th L

VI

Z Z

Norton's Theorem Norton's theorem states that:

"With respect to the terminal pair AB the network N may be replaced with a current source IN in parallel with an internal impedance ZN. The current source IN, called the Norton's current, is the current that would flow from A to B when the terminals A and B are shorted together and ZN is the same internal impedance is defined earlier in context of Thevenin's theorem".

The parallel combination of IN, and ZN, as shown in figure (b), is known as Norton's equivalent of the network TV with respect to terminal pair AB.

(a) (b)

Hence, it follows that:

If a load impedance ZL, be connected across terminals A and B of network, the current IL delivered to the load may be calculated from the equivalent circuits by connecting this impedance across A and B and shall be given as




. NL N

N L

ZI I

Z Z

[where ZN = Zth]

Procedure to Obtain Vth and Zth or IL and ZN

Step 1: Remove the portion of the network across which the Thevenin's or Norton's equivalent circuit is to be found.

Step 2: Mark the terminals AB of the remaining 2-terminal circuit.

Step 3: (a) Obtain the open circuit voltage at the terminals AB, i.e., Vth keeping all the sources at their normal values.

(b) Short the terminals AB. Obtain the short circuit current i.e. IN flowing from A to B keeping all the sources of their normal values.

Step 4: Calculate Zth = ZN

Case I: If circuit having only independent sources : In this case, set all the sources at zero values, i.e. with all the voltage sources shorted and all the current sources open circuited leaving behind their internal resistances or impedances.

Case II: If circuit having independent and dependent sources both : In this case, we calculate Vth and IN, then the internal impedance of the network N is obtained as

thth N

N

VZ Z

I

Case III: If circuit having only dependent sources: In this case, we apply a voltage V at the terminal pair AB. A current I will flow due to application of V, then the internal impedance of the network N is obtained as

th N

VZ Z

I

Step 5: Draw the equivalent circuit across AB

(a) Thevenin's voltage Vth in series with Zth

(b) Norton's current IN in parallel with ZN.

Example

For the circuit shown in figure, determine the current I through the 10 resistance by Thevenin's theorem.

Solution

For Vth

From the circuit shown in following figure




50 5 15 100th o oV I I

where I0 is the current flowing in the loop in clockwise direction, and is given by

0

502.5

20I A

Therefore 50 5( 2.5) 62.5thV V

For Rth/Zth


5 15

5 ||15 3.755 15th

xR

Therefore, current in 10 resistance (from figure below)

62.5

4.54510 3.75 10

th

th

VI A

R

Example

For the circuit shown in figure, determine the current I through the 10 resistance by Norton's theorem.

Solution

For IN





50 100 50

5 15 3NI A

and 3.75N thR R

Therefore, current in 10 resistance from following figure

50 3.75

. . 4.54510 3 3.75 10

NN

N

RI I A

R

Example

Find the Norton's equivalent circuit across terminals AB of the circuit shown in given figure.

Solution

For IN:

(5 resistance is short circuited) from the circuit shown in figure below.

Applying KVL

012 10i

or 0 1.2i A

And applying KCL at A

0 0 02 3NI i i i

or 3.6NI A

For Vth:

Vth = voltage across 5 resistance




= 5 x (current through the 5 resistance)

= 5 x 3i0

Now applying KVL in figure given in question, we have

0 0 0 012 10 5( 2 ) 12 / 25i i i i

or 12

15 7.225thV x V

Therefore 7.2

23.6

thN

N

VR

I

Hence, Norton's equivalent circuit is shown in following figure.

Example

Calculate the current in 6 resistor of the circuit of following figure by Thevenin's theorem.

Solution

For Vth:

From following figure

18 2 0x x thV V V

and 3 1 3xV x V

Therefore 18 3 27th xV V V

For IN or Isc:

From following figure




18 2 0 6x x xV V V V

and 1(3 ) 9x sc scV I I A

Therefore 27

39

thth

sc

VR

I

Using Thevenin's theorem circuit reduces as shown in figure.

Therefore 6

273

3 6I A

Example

Find the current in the 14 resistor of the circuit of figure by Thevenin's theorem.

Solution

For Vth:

We open circuit AB.

Thus 10 5( 0.1 ) 10 0.5x th x xV V V V

Therefore 20thV V

Next by short circuited AB resulting in Vx = 0 and 0.1Vx = 0. Thus circuit reduces as shown in following figure.

10 10

5 8 13scI A

So 20

26(10 /13)

thth

sc

VR

I




and 14

200.5

14 26 14th

th

VI A

R

Example

Find the Norton's equivalent of the network shown in figure.

Solution

As the circuit does not have any independent source, hence, ( ) 0sc NI I

For RN:

In this case, a voltage V is applied and let current I, which would flow due to V as shown in figure below, then we know that RN = V/I.

Let v be the node voltage at M,, then KCL at node a and M gives

1 1

100 50

V V vI

(1)

and 110.1

50 200

V v vV

(2)

and also 1V V (3)

From eq. 1 1 1100 2 2I V V v

1100 3 2I V v 1(a)

From eq. 2 1 14 4 20V v v V

15 16v V

1

16

5v V 2(a)

From eq. 1(a) and 2 (a)

1 1 1

16 47100 3 2

5 5I V V V

or 1 500

47

V

I




So 1N

VVR

I I (from eq. 3)

or RN = 10.63

Problem

Find the Thevenin's and Norton's equivalents of following circuit.

Answer: Vth = 0, Zth = 0.6; IN = 0, ZN or Zth = 0.6

Problem

Find the Thevenin's and Norton's equivalents of following circuit.

Answer: Vth = 0, Zth = 192.3; IN = 0, ZN = Zth = 192.3

SUPERPOSITION THEOREM This theorem finds use in solving a network where two or more sources are present and connected not in series or in parallel.

Statement If a number of voltage or current sources are acting simultaneously in a linear network, the resultant current in any branch is the algebraic sum of the currents that would be produced in it, when each source acts alone replacing all other independent sources by their internal resistances.

Step 1: Take only one independent source of voltage/current and deactivate the other independent voltage/current sources. (For voltage sources, remove the source and short circuit the respective circuit terminals and for current sources, just delete the source keeping the respective circuit terminals open). Obtain branch currents.

Step 2. Repeat the above step for each of the independent sources.

Step 3. To determine the net branch current utilising superposition theorem, just add the currents obtained in step 1 and step 2 for each branch. If the currents obtained in step 1 and step 2 are in same direction, just -add them; on the other hand, if the respective currents are directed opposite in each step, assume the direction of the clockwise current to be +ve and subtract the current obtained in the next step from the original current. The net current in each branch is then obtained.

Example




Find v in the circuit of given figure using superposition theorem.

Solution

Let us first take 2 V source deactivating the current sources (see figure below)

1

21

2 21

2 2

i Ax

v1 (drop across rL due to 2 V source) = 1 x 1 = 1 V

Next, taking the lower current source only (see figure below)

2

1 3 15( 5) ( 5)

1 1 (2 / 3) 8 8i A

3

15 2 15 2 5.

8 1 2 8 3 4i A

This gives 2

5(1) (5 / 4)

4v V

In following figure




4

15 15 / 5 3

(2 / 3) 1i A

This gives 2

3. 22 1rLi A

3 2 1 2v x V

By superposition

1 2 3 1 5 / 4 2 7 / 4v v v v V

Example

Find v by superposition theorem.

Solution

Taking the 10 V source only, with reference to following figure.

1

1010 10 6.67

5 10v xi V

Taking the 5A source only, with reference to figure given below

10

55. 1.67

5 10i A

2 1.67 10 16.70v x V

By superposition theorem

1 2 6.67 16.70 23.37v v v V

Example

Find i0 and i from the circuit given below using superposition theorem.




Solution

Assuming only 6 V source to be active, with reference to figure given below

0 0 06 (1 5) ' 2 ' 0 ' 3 / 4i i i A

0 ' ' (3 / 4)i i A

Next, assuming 1 A source active only, with reference to figure given below

00 0

" 2 ""1 " " 1.2 " 0.4 "

1 5

v ivi i v i

But 0 " "/1i v

We finally get

0 0 01 1.2 " 0.4 " 0.8 "i i i

i.e. 0 " 1.25i A

and 00

" 2 "" "/ 5 0.25

5

v ii i A

Using the principle of superposition

0 0 0' " (3 / 4) 1.25 0.5i i i A

and 0 ' " (3 / 4) 0.25 0.5i i i A

Problem

For the circuit shown below, find current I through the 10 resistance by super position theorem.




Answer: 4.545A

Date post:	17-Jan-2023
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Basic Concepts & Network Theorems - AMIE

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