# 5. Network Theorems and Applications – Network Analysis and Synthesis

## 5. Network Theorems and Applications

CHAPTER OBJECTIVES

After carefully studying this chapter, you should be able to do the following:

State and explain Superposition theorem with an example.

Solve circuit problems using Superposition theorem.

State and explain Thevenin’s theorem.

Apply Thevenin’s theorem to calculate current flowing through any branch of an active electric network.

Explain with the help of an example the procedure for applying Thevenin’s theorem to an electric network.

State and explain Millman’s theorem.

Solve network problems using Millman’s theorem.

State and explain maximum power transfer theorem.

Establish the condition for maximum power transfer in a complex impedance circuit.

State and explain, with the help of an example, the Reciprocity theorem.

State Tellegen’s theorem and apply the theorem to solve network problem.

State and explain Compensation theorem.

Simplify electrical circuits using star-delta transformations.

Establish transforming relations of star-delta transformation of resistance.

Solve complex network problem by using network theorem suitably.

### 5.1 INTRODUCTION

In a previous chapter, we had described the methods of solving network problems using mesh current analysis and nodal voltage analysis. The procedure involves solving a number of equations. For a complex network, the number of equations becomes large. However, many networks often require only restricted analysis, for example, finding the current through a particular branch or a particular circuit element.

A number of theorems for circuit analysis have been developed to solve circuit problems with ease. When all the theorems and techniques will be known, we will be in a position to apply a particular theorem or a technique that reduces the time for solving a given problem. In this chapter, the following theorems and techniques have been discussed.

1.Superposition theorem

2.Thevenin’s theorem

3.Norton’s theorem

4.Millman’s theorem

5.Maximum power transfer theorem

6.Reciprocity theorem

7.Tellegen’s theorem

8.Compensation theorem

9.Star-delta transformation

### 5.2 SUPERPOSITION THEOREM

An electrical network may contain more than one source of supply. The sources may be voltage sources or current sources or a combination of both. In solving circuit problems having multiple sources of supply, the effect of each source is considered separately with all other sources replaced by their internal resistances every time. The combined effect of all the sources is then taken into consideration to calculate the current in any branch.

The superposition theorem can be stated as follows: In a linear network containing more than one source, the current in any branch or the potential difference across any two points can be found by considering each source separately and then by adding their individual effects. While considering each source, the other sources are replaced by their internal resistances. If the value of internal resistances of the sources are not given, the voltage sources are short-circuited and the current sources are open-circuited.

The procedure is illustrated through a few examples.

Example 5.1 Using superposition theorem, calculate the currents in the network shown in Figure 5.1.

Solution: Let us consider separately the effect of each voltage source by short-circuiting the other source as shown in Figure 5.2.

Figure 5.1

Figure 5.2

Now, we will combine the effect of the two voltage sources.

Current supplied by the 24V source = 4 A

Current through branch AB = 4 A

Current supplied by the 12V source = 2 A

Current through branch CB = 2 A

Current through branch BD = 4 + 2 = 6 A.

Figure 5.3

Example 5.2 Calculate the current through the 6 Ω resistor shown in Figure 5.4 using superposition theorem.

Solution: We will first consider the effect of voltage source by open-circuiting the current source, as shown in Figure 5.5.

Figure 5.4

Now, we will consider the effect of current source by short-circuiting the voltage source, as shown in Figure 5.6.

By applying current divider rule, we get the following form:

Figure 5.5

Current through 6 Ω resistor across BD = I1I2 = 2.4 − 0.8 = 1.6 A

Figure 5.6

### 5.3 THEVENIN’S THEOREM

It may sometimes be required to calculate the current through any circuit element in an active electrical network when the value of the circuit element is changed to different values, keeping all other circuit elements unchanged. We can use Thevenin’s theorem to determine the current flowing through any circuit element in an active network.

According to Thevenin’s theorem, current I through a resistor R connected across any two points A and B of an active network containing one or more sources of emf is as follows:

where Voc is the potential difference across the terminals A and B with R disconnected; and r is the resistance of the network between A and B with R disconnected and the sources of emfs replaced by their internal resistances (if the value of internal resistance of the voltage source is not provided, the source terminals have to be shown short-circuited.)

Thevenin’s theorem can be stated as follows: Any two terminals A and B of an active network can be replaced by a constant voltage source having an emf E and an internal resistance r. The value of E is equal to the open-circuit potential difference between terminals A and B, that is, Voc and r is the resistance of the entire network measured or calculated between A and B with the resistance R between A and B disconnected and sources of emfs replaced by their internal resistances.

#### 5.3.1 Procedure for Applying Thevenin’s Theorem

The procedure is illustrated through an example. Consider a network as shown in Figure 5.7.

Figure 5.7

Let us assume that we are required to calculate the current through the 3 Ω resistor. Let this resistance be called the load resistance. We will remove the load resistance and place the rest of the network in a dotted box as shown is Figure 5.8.

Figure 5.8

Now, remove the load resistance of 3 Ω temporarily and find the open-circuit voltage across the terminals A and B.

From Figure 5.8(b), by applying KVL in the loop, we obtain the following:

Further,

Figure 5.9

Now, short-circuit the voltage sources and calculate the resistance of the network across terminals A and B. The Thevenin’s equivalent circuit consists of a voltage source VTh, which is 8 V, in series and a resistance RTh, which is 1.33 Ω. Thevenin’s equivalent circuit has been shown in Figure 5.9.

The current through the load resistance is calculated by connecting the load resistance across terminals A and B of the Thevenin’s equivalent circuit. The load current is calculated as

Example 5.3 Applying Thevenin’s theorem, calculate the current flowing through the 10 Ω resistor in the circuit shown in Figure 5.10.

Figure 5.10

Solution: We will convert the current source into an equivalent voltage source and remove temporarily the resistance of 10 Ω from terminals AB as shown in Figure 5.11. Then, we will calculate open-circuit voltage VTh across AB. Then, we will calculate the resistance of the network, that is, RTh across A and B by short-circuiting the voltage sources.

Applying Kirchhoff’s voltage law in the loop, the value of I is calculated as follows:

Figure 5.11

Note that since the terminals AB are open-circuited no current can flow through the 3 Ω resistor. The voltage across AB is the same as the voltage across terminals PQ. VTh = VAB = VPQ = 4 + 2I = 4 + 2 × 1 = 6V.

RTh is calculated as shown in Figure 5.12.

Figure 5.12

Now, we can draw the Thevenin’s equivalent circuit and connect the load resistance of 10 Ω across terminals AB to determine the load current as shown in Figure 5.13.

Figure 5.13

Example 5.4 Applying Thevenin’s theorem, calculate the current through the load resistance RL = 10 Ω in the circuit shown in Figure 5.14.

Figure 5.14

Solution: We temporarily remove the resistance RL from the terminals AB. To calculate VTh, that is, VAB, we will apply superposition theorem and calculate the total the current flowing through the 8 Ω resistor and then calculate the voltage drop across it. The voltage drop across the 8 Ω resistor is equal to the open-circuit voltage VAB, which is the VTh.

We will then calculate RTh across terminals AB by open-circuiting the current source and short-circuiting the voltage sources. First, we consider the 24V source, then the 12V source, and then the 2A source. Later, we calculate the current through the 8 Ω resistor. Keeping in view the current direction in each case, we calculate the total current through the 8 Ω resistor, as shown in Figure 5.15.

1.Considering the 24V source, we open-circuit 2A current source and short-circuit the 12V voltage source.

2.Considering the 12V source, we open-circuit the 2A current source and short-circuit the 24V voltage source.

3.Considering the 2A current source, and short-circuiting the 12V and 24V voltage sources.

Figure 5.15

Following the principle of the superposition, we consider the combined effect of all the sources at which the current flows through the 8 Ω resistor.

RTh is calculated by calculating the equivalent resistance of the circuit across terminals A and B and by shorting the voltage sources and keeping open the current source as shown in Figure 5.16.

Figure 5.16

Thus, the Thevenin’s equivalent circuit and current through the load resistance is shown in Figure 5.17.

Figure 5.17

### 5.4 NORTON’S THEOREM

In Thevenin’s theorem, we have seen that a two-terminal active network is converted into a voltage source and an equivalent series resistance which is connected across the load through which the current is to be calculated.

Another method of analysing a network is provided by Norton’s theorem. In applying Norton’s theorem, a two-terminal network with current and voltage sources is converted into a constant current source and a parallel resistance and is connected across the load through which the current is to be calculated.

Norton’s theorem is stated as follows: Any two-terminal linear network consisting of voltage and/or current sources can be converted into a constant current source and a parallel resistance. The value of the current source is the current that will flow if the two terminals are short-circuited. The value of the parallel resistance is the equivalent resistance of the whole network viewed from the open-circuited terminals after all the sources are replaced by their internal resistance. (Note that if the internal resistances are not given, the voltage sources are short-circuited and current sources are open-circuited.)

Example 5.5 Using Norton’s theorem, calculate the current through the 10 Ω resistor in the network, as shown in Figure 5.18.

Figure 5.18

Solution: We short-circuit the terminals AB as in Figure 5.19(a) and calculate the short-circuit current. Then, we calculate the resistance across the open-circuited terminals A and B by short-circuiting the voltage sources as shown in Figure 5.19(b).

Figure 5.19

Resistance R across the open-circuited branch across terminal A and B is given as in the following:

Norton’s equivalent current source with parallel resistance along with the 10 Ω resistance across terminals A and B is shown in Figure 5.20. Using current divider rule, current through the load is calculated as follows:

Figure 5.20

Example 5.6 Using Norton’s theorem, calculate the current through the 10 Ω resistor in the network shown in Figure 5.21.

Figure 5.21

Solution: We short-circuit terminals AB and calculate the short-circuit current. Then, we calculate the resistance across the open-circuited terminals A and B.

Since terminals AB are shorted, no current will flow through the branch containing 6 Ω and 4 Ω resistors. See Figure 5.22.

Using current divider rule, the current is calculated as follows:

Figure 5.22

The open-circuit resistance across terminals A and B is calculated as shown in Figure 5.23.

Figure 5.23

Now, we draw the Norton’s equivalent circuit as a current source with a parallel resistance and connect the 10 Ω resistor across its terminals as shown in Figure 5.24.

Figure 5.24

### 5.5 MILLMAN’S THEOREM

When a number of voltage sources form parallel branches, the common voltage across their terminals can be calculated by applying Millman’s theorem.

This theorem is illustrated through an example. Let us assume that three voltages V1, V2 and V3 are connected in parallel across terminals A and B. R1, R2 and R3 are, respectively, their internal resistances as shown in Figure 5.25(a). We convert the voltage sources into equivalent current sources as shown in Figure 5.25(b). By combining the current sources and the resistances, the equivalent circuit will be as shown in Figure 5.26.

Figure 5.25

Figure 5.26

Resultant voltage across A and B, that is,

Millman’s theorem states that in any network, if there are a number of voltage sources V1, V2, V3, … Vn in parallel with their internal resistances R1, R2, R3, … Rn, respectively, then, these sources can be replaced by a single voltage source with its internal resistance as V′ and R′, respectively, where

Example 5.7 Three voltage sources of 12 V, 13V and 14V having internal resistances of 1 Ω, 2 Ω and 3 Ω, respectively, are connected in parallel as shown in Figure 5.27. What will be the voltage available across their terminals?

Figure 5.27

Solution:

Substituting the values, we calculate the voltage as follows:

### 5.6 MAXIMUM POWER TRANSFER THEOREM

The power is supplied from the source to the load. Let the internal resistance of the source be Ri and the load resistance be RL.

Figure 5.28

Applying maximum power transfer theorem, we find out at what value of load (load resistance or impedance) maximum power will be transferred from the source to the load. Let us assume that a source having an emf E and internal resistance Ri is connected to a load of resistance RL as shown in Figure 5.28.

The current flowing in the circuit is given as follows:

Power delivered is equal to power consumed assuming no line loss. Power delivered P is expressed as in the following:

To determine the value of RL at which P will be maximum, we differentiate P with respect to RL and equate to zero.

This shows that the maximum power will be transferred from the source to the load when the value of load resistance becomes equal to the internal resistance of the source.

The maximum power transfer theorem is stated as follows: Maximum power is transferred to the load, when the load resistance is equal to the source resistance.

When a complex network is analysed for maximum power transfer, the circuit is first converted into a voltage source with an internal resistance by applying Thevenin’s theorem.

The value of maximum power is calculated as follows:

In power systems, maximum power transfer from the generator to the load is not tried because of poor efficiency of transmission and poor voltage regulation.

In the field of electronics, maximum power transfer from the source to the load is achieved through impedance matching.

In electronics, we often deal with small power. For example, maximum power transfer is desirable from the output amplifier to the speaker of an audio sound system. A TV antenna receives power from radio waves. The power collected by the antenna is very small. The TV receiver circuit is designed to make maximum use of the power delivered by the antenna.

Example 5.8 A 8V battery is supplying power through a network to a load, RL, as shown in Figure 5.29. Calculate the value of RL at which the power transfer will be maximum.

Figure 5.29

Solution: The circuit is converted into a Thevenin’s equivalent circuit across terminals AB as shown in stages in Figures 5.30, 5.31 and 5.32.

Figure 5.30

Note that when terminals AB is open-circuited, no current will flow through the 1 Ω resistors and hence there will be no voltage drop in these resistors. The equivalent resistance RTh is given as follows:

Figure 5.31

Thus, the Thevenin’s equivalent circuit is represented as shown in Figure 5.32.

Figure 5.32

For maximum power transfer, resistance can be calculated as in the following:

The value of maximum power transfer is as follows:

Example 5.9 In the circuit shown in Figure 5.33, calculate the value of load resistance RL at which the maximum power will flow through the load. Further, calculate the transmission efficiency when maximum power transfer occurs.

Figure 5.33

Solution: We redraw the circuit as shown in Figure 5.34, and convert the current source into its equivalent voltage source.

Figure 5.34

VTh and RTh are calculated as shown in the following. As in Figure 5.35, we remove the load resistance RL and then proceed to calculate VAB, that is, VTh.

Applying KVL, we calculate the following;

Figure 5.35

The Thevenin equivalent resistance between terminals A and B is calculated by short-circuiting the voltage sources.

Thevenin’s equivalent circuit is shown in Figure 5.36. In Figure 5.36, for maximum power transfer,

Figure 5.36

Maximum power = I2 RL = 42 × 2.4 = 38.4 W.

### 5.7 MAXIMUM POWER TRANSFER THEOREM FOR COMPLEX IMPEDANCE CIRCUITS

Maximum power transfer theorem can be applied to complex impedance circuits.

For a complex source impedance, maximum power transfer occurs when the load impedance is the complex conjugate of the source impedance. Let us consider a circuit with a source supplying power to a load as shown in Figure 5.37. ZS is the source impedance and ZL is the load impedance.

Figure 5.37

Power delivered to the load, P = I2RL

Power is maximum with constant value of RL when XL = −XS. Then, the maximum power is

However, when RL is variable, maximum power occurs when RL = RS.

Thus, for maximum power transfer, the following conditions are obtained:

This shows that maximum power transfer occurs when

That is, load impedance is equal to the complex conjugate of the source impedance. That is,

### 5.8 RECIPROCITY THEOREM

The reciprocity theorem is stated as follows: In a linear bilateral network, if a voltage source V in a branch A produces a current I in any branch B, then if the same voltage source is removed and inserted in branch B, it will produce a current I in branch A. In other words, we can say that voltage V and I are interchangeable between branch A and B.

This theorem is illustrated through an example.

Example 5.10 Verify reciprocity theorem in the circuit shown in Figure 5.38.

Figure 5.38

Solution: The 24V voltage source in branch AB is causing a current I flowing through branch CD.

We have to show that same current I will flow in branch AB when 24V source is removed and inserted in branch CD.

We calculate the total current supplied by the 24V battery as shown in Figure 5.39.

Figure 5.39

The current through branch CD is calculated by using current divider rule as in Figure 5.40(a).

Figure 5.40(a)

The current flowing in branch CD has been calculated as 1.5 A.

Now, we place the 24V source in branch CD, as shown in Figure 5.40(b).

Figure 5.40(b)

The total current supplied by the 24V battery is as follows:

The same amount of current, that is, 1.5 A is flowing through branch AB.

Thus, the reciprocity theorem is established.

### 5.9 TELLEGEN’S THEOREM

Tellegen’s theorem states that the algebraic sum of powers in all branches in a network at any instant is zero. This theorem is valid for any network that may be linear or non-linear, active or passive and time varying or time invariant, and all branch currents and voltages in the network must satisfy Kirchhoff’s laws.

According to Tellegen’s theorem, the rate of supply of energy by the active elements of a network equals the rate of energy dissipated or stored by the passive elements of the network. Tellegen’s theorem is stated mathematically as follows:

Vk and Ik should satisfy KVL and KCL, respectively. In the above expession b indicates the number of branches.

Example 5.11 In the network shown in Figure 5.41, the branch voltages and currents have the following values. Verify Tellegen’s theorem for the network shown in the Figure.

Figure 5.41

Solution: We will first verify if the data provided satisfy Kirchhoff’s laws.

1.By applying KVL for loop ABCA, + V0V1V3 = 0
or + 20 − 16 − 4 = 0

2.For loop BDECB, + V2V5 + V3 = 0
or + 2 − 6 + 4 = 0

3.For loop ABDECA, −V1 + V2V5 + V0 = 0
or − 16 + 2 − 6 + 20 = 0

4.For loop ABDA, − V1 + V2 + V4 = 0
or −16 + 2 + 14 = 0

To verify the applicability of KCL, the values at nodes A, B and C are calculated as follows: at node A, −I0 = I1 + I4

or + 16 = 12 + 4

At node B, I1 + I2 = I3

or 12 + 2 = 14

At node C, I3 + I5 = −Io

or 14 + 2 = 16

Now, we apply Tellegen’s theorem to show that the sum of instantaneous power of all the branches is zero.

### 5.10 COMPENSATION THEOREM

This theorem is useful in determining the changes in current or voltage when the value of resistance gets changed in the circuit. If a small change in resistance in a network takes place from R to R + ΔR, this will cause a change in current in all branches.

According to Compensation theorem, the change in current in all the branches is equal to the current produced by a voltage source of magnitude (I ΔR) placed in series with the resistance whose value has changed.

Example 5.12 In the circuit shown in Figure 5.42, the resistance of the branch having 5 Ω resistance has changed to 6 Ω due to the connection of an ammeter having an internal resistance of 1 Ω. Determine the value of compensation source voltage and verify results.

Figure 5.42

Solution: Total current delivered by the source can be given as follows:

Current through the 5 Ω resistor is I = 2 A.

When resistance has changed to 6 Ω, current through the branch is calculated as in the following:

Figure 5.43

Compensation voltage VC = I ΔR = 2 × 1 = 2 V

The circuit is drawn as in Figure 5.43 with the compensation voltage source. The compensation source is inserted in series with R + Δ R.

Now, the current through the 5 Ω branch is calculated with the compensation voltage source in place and replacing the voltage source by its internal resistance.

Thus, the result is verified.

### 5.11 STAR-DELTA TRANSFORMATION

In any network, resistance elements may be seen as connected in series, in parallel, in star formation or in delta formation. To simplify such circuits, star connected resistances can be converted into equivalent delta connected resistances and vice-versa. In such transformation, the circuit conditions are not changed.

#### 5.11.1 Transforming Relations from Delta to Star

Let us consider three resistances RAB, RBC and RCA connected in delta formation between the terminals AB, BC and CA, respectively, as shown in Figure 5.44(a). These three resistances can be converted into equivalent star forming resistances RA, RB and RC as shown in Figure 5.44(b).

Figure 5.44

For the purpose of equivalence, we will equate the resistances of the two networks across terminals AB, BC and CA as shown in equations (5.1), (5.2) and (5.3), respectively.

Subtracting equation (5.2) from equation (5.1), we obtain the following:

Adding equation (5.4) and equation (5.3), we get the following form:

Similarly, the values of RB and RC can be calculated.

When delta connected resistances are changed to star connected resistances, their values are given as follows:

Example 5.13 Three resistance of values 1 Ω, 2 Ω and 3 Ω are connected in delta formation between terminals AB, BC and CA, respectively, as shown in Figure 5.45. Calculate the equivalent star connected resistances.

Figure 5.45

Solution:

#### 5.11.2 Transforming Relations from Star to Delta

Now, let us consider three resistances RA, RB and RC connected in star. The equivalent delta connected resistances are RAB, RBC and RCA as shown in Figure 5.46(a) and (b).

The basic equations guiding the transformation will be the same, that is, equations (5.1), (5.2) and (5.3).

From the basic equations, we derived equations (5.5), (5.6) and (5.7). We will use equation (5.5), (5.6) and (5.7) to determine RAB, RBC and RCA in terms of RA, RB and RC.

Figure 5.46

Multiplying equation (5.5) by equation (5.6), we get the following:

Multiplying equation (5.5) by equation (5.7), the following form is obtained:

Multiplying equation (5.6) by equation (5.7), we write the equation as follows:

Adding equation (5.8), (5.9) and (5.10), we obtain the equation as follows:

Earlier, we had equation (5.7) as in the following:

By substituting the equation, we get the following form:

Dividing both sides by RC, we write RAB as follows:

Similarly, RBC and RCA can be calculated. When star connected resistances are changed to delta connected resistances, their values are given as follows:

Example 5.14 Three resistances each of 3 Ω value are connected in star formation. Calculate their equivalent delta.

Solution: The three star connected resistances and their equivalent delta forming resistances between the terminals A, B and C are shown in Figure 5.47.

Figure 5.47

Using the transforming equations, the following can be obtained:

Example 5.15 Calculate the equivalent resistance of the network shown in Figure 5.48 across terminals A and B using star-delta transformation where necessary.

Figure 5.48

Solution: From Figure 5.48, it is seen that point D and E in the network is joined together and hence considered as the same point. Between E and F, the two 4 Ω resistors are in parallel. Therefore, the network is redrawn and shown in Figure 5.49(a). The same is redrawn by arranging the resistance elements as in Figure 5.49(b).

Figure 5.49

Now converting the delta forming resistances of 2 Ω values across terminals CBD into equivalent star the network is redrawn as shown in Figure 5.50(a). Through series-parallel conversion, the equivalent resistance between terminals AB is calculated as shown in Figure 5.50(b), (c) and (d).

Figure 5.50

Example 5.16 Six resistances, each of value R are connected as shown in Figure 5.51. Calculate the equivalent resistance of the network across terminals BC.

Figure 5.51

Solution: There are many ways of solving this problem. However, we will convert the star forming resistances across terminals ABC with star point at N and then make series-parallel conversions.

Figure 5.52

As shown in Figure 5.52, between terminals AB, resistance R and RAB are in parallel. Similarly, between terminals AC, resistances R and RAC are in parallel. Again, between terminals BC, resistances R and RBC are in parallel. The equivalent resistance between terminals BC is calculated as shown in Figure 5.53.

Figure 5.53

Note: The time taken to solve this problem will be less if the delta forming resistances are converted to equivalent star.

### 5.12 NUMERICALS ON NETWORK THEOREMS

Having studied all the network theorems, we will now solve some more network problems.

Example 5.17 Using Thevenin’s theorem, calculate the current flowing through the 8 Ω resistor connected across the terminals A and B in Figure 5.54.

Figure 5.54

Solution: We will first convert the current source of 5A with the parallel resistance of 1 Ω into an equivalent voltage source and then redraw the circuit as shown in Figure 5.55.

To apply Thevenin’s theorem, we will temporarily remove the 8 Ω resistor through which the current is to be calculated. Then we will calculate open-circuit voltage VAB, which is called VTh. We will then calculate RTh, that is, the resistance of the entire network across terminals AB. Then, we will draw the Thevenin’s equivalent circuit and calculate current through the load resistance of 8 Ω. We draw the circuit with 8 Ω resistor removed, as shown in Figure 5.56.

Figure 5.55

Figure 5.56

By applying KVL in the loop, as shown in the Figure, we get the loop equation as follows:

Since, no current will flow through the 2 Ω resistor, there will be no voltage drop across it. Voltage across PQ will be equal to VAB. To calculate VPQ, we move from P to Q. The battery voltage is taken as positive and voltage drop in the 1 Ω resistor is also taken as positive. Thus,

Point P is at higher potential than point Q. The equivalent resistance of the whole network across terminal AB with the voltage sources short-circuited is calculated as:

The Thevenin’s equivalent circuit is drawn and the load maintenance of 8 Ω is placed across AB, as shown in Figure 5.57.

Figure 5.57

Example 5.18 Find the current in the 3 Ω resistor in the circuit of Figure 5.58 using Thevenin’s theorem.

Figure 5.58

Solution: Step 1: Remove the branch through which the current is to be calculated. Therefore, we remove the 3 Ω resistor and redraw the circuit as shown in Figure 5.59.

Figure 5.59

Step 2: Let us find the voltage across open-circuited terminals, that is, A and B. From Figure 5.59, it is clear that i1 =10A and i2 = 0.

Step 3: Let us find (ZTh), that is, the equivalent impedance of the network removing all the sources, as seen from open-circuited terminals A and B. The current source has been kept open as shown in Figure 5.60.

Between terminal BB′, we can get the following form:

Figure 5.60

Step 4: Therefore, the Thevenin’s equivalent circuit will be as shown in Figure 5.61.

Now, the current through 3 Ω resistance is calculated as follows:

Figure 5.61

Example 5.19 Find the current through (5 + j4) Ω impedance shown in Figure 5.62 using Thevenin’s theorem.

Solution: Step 1: Remove the branch (5 + j4) Ω impedance, as shown in Figure 5.63(a).

The same circuit is redrawn as shown in Figure 5.63(b).

Figure 5.62

Figure 5.63

Now, equivalent impedance of the network is as follows:

Now, by using current divider rule, we calculate the current as follows:

and

Now

By substituting the value of I1 in the equation,

Figure 5.64

Step 2: Let us find ZTh across terminals A and B by short-circuiting the voltage source, as shown in Figure 5.64.

Step 3: The Thevenin’s equivalent circuit is shown in Figure 5.65.

Figure 5.65

Current through the (5 + j4 Ω) impedance is = I = 9.61∠56.45° A.

Example 5.20 Find the current through (5 + j4) Ω impedance in the network shown in Figure 5.66 using Norton’s theorem.

Figure 5.66

Solution: Note that this problem has already been solved using Thevenin’s theorem. Here, we will use Norton’s theorem and verify the result.

Figure 5.67

Step 1: Short-circuit the branch in which response is to be determined as shown in Figure 5.67.

This diagram can be redrawn as shown in Figure 5.68.

Figure 5.68

Applying KVL in closed circuit CDEHC, the following form is obtained:

Applying KVL in closed circuit CFGHC, the equation can be written as follows:

Applying KVL in closed circuit DFBAD, we get the equation as in the following:

Step2: Let us solve equations 1, 2 and 3 for IN using Cramer’s rule

Step 3: Let us find ZN (Norton equivalent impedance) as shown in Figure 5.69.

Figure 5.69

Step 4: let us draw Norton’s equivalent circuit as shown in Figure 5.70.

Figure 5.70

Now, by current divider rule, the current through the impedance (5 + j4) is given as follows:

Example 5.21 Find the current in 5 Ω resistance in the network shown in Figure 5.71 using superposition theorem and then verify the result using Thevenin’s theorem.

Figure 5.71

Solution: In the given circuit, there are two sources. We will consider one source at a time.

Let us first see the effect of 9A current source.

The circuit is redrawn with 9A current source and the voltage source short-circuited, as shown in Figure 5.72.

Let us find current in 5 Ω resistor due to 9A current source, acting alone using the mesh analysis.

Current through 5 Ω resistor = i3i2 (from left to right)

Now, let us find i2 and i3

Applying KVL in mesh II, we get the following form:

Figure 5.72

Substituting i1 = 9, we get the equation as follows:

Applying KVL in mesh III, we write the equation as in the following:

Applying KVL in mesh IV, the equation can be calculated as follows:

Let us solve equations (5.18), (5.19) and (5.20) using Cramer’s rule for i2 and i3

Therefore

Therefore, the current due to the 9A current source alone through the 5 Ω resistor is equal to (i2i3), that is, (3.82 − 0.8588)A. That is, a current of 2.9612 A will flow from right to left.

Now, we will have to find current through 5 Ω resistor due to 12V source acting alone. We keep the 12V source in the circuit and open-circuit the current source, as shown in Figure 5.73(a).

Figure 5.73

Applying KVL in mesh I, we get the following form:

Applying KVL in mesh II, we can write the equation as follows:

Applying KVL in mesh III, the equation can be calculated as follows:

Let us solve equations (5.21), (5.22) and (5.23) for i1 and i2.

By superposition theorem, we can say the following:

Using Thevenin’s theorem, the following steps are to be followed:

Step 1: Open-circuit the branch containing 5 Ω resistance, as shown in Figure 5.74.

Step 2: Let us find VTh (voltage across AB)

The circuit can be redrawn as in Figure 5.75.

From mesh I, the current can be given as follows

Applying KVL in mesh II, we get the following form:

Substituting i1 = 9, we can write the equation as follows:

Figure 5.74

Applying KVL in mesh III, the equation can be calculated as in the following form:

Solving equation (5.24) and (5.25), we get the equation as follows:

Figure 5.75

Therefore, the value of i2 and i3 can be calculated as in the following:

Step 3: Let us find RTh using Figure 5.76(a) and (b).

Figure 5.76

Now, let us draw Thevenin’s equivalent circuit as shown in Figure 5.77.

Current through 5 Ω resistance

Figure 5.77

Example 5.22 Find the current through the load resistance RL in the circuit shown in Figure 5.78 using Thevenin’s theorem and Norton’s theorem.

Figure 5.78

Solution: Using Thevenin’s theorem, the following steps are to be followed:

Step 1: Open-circuit the terminals AB, as shown in Figure 5.79.

Step 2: Let us find voltage (VTh) across open-circuited terminals

From mesh I, we get i1 = 25 A

From mesh II, we obtain −j3(i2ii) + j3i2 + 4i2 + (5 + j5) i2 = 0

or + j3i1 + (9 + j5) i2 = 0

By substituting i = 25, we obtain the following:

Figure 5.79

Step 3: Now, let us find ZTh from circuits shown in Figure 5.80.

Figure 5.80

Let us draw Thevenin’s equivalent circuit, as shown in Figure 5.81.

Figure 5.81

By using Norton’s theorem, the following steps are performed.

Step 1: Short-circuit the terminals AB as shown in Figure 5.82 and draw an equivalent circuit.

Step 2: Let us use mesh analysis to find IN

From mesh I, i1 = 25 A

and from mesh II, we get the following:

Figure 5.82

Step 3: To find ZN refer to Figure 5.83

Figure 5.83

The Norton’s equivalent circuit is shown in Figure 5.84.

Current through RL is IL and it can be expressed as follows:

Figure 5.84

Figure 5.85

Example 5.23 Find the current through ZL in the circuit shown in Figure 5.85 using Thevenin’s theorem and Norton’s theorem

Solution: Using Thevenin’s theorem, the following steps are to be followed:

Step 1: We remove the load impedance ZL and redraw the circuit as shown in Figure 5.86.

Figure 5.86

Step 2: Let us find VTh using mesh analysis

From mesh I, i1=10 A

From mesh II and by applying KVL, we get the following equations:

Substituting i1 = 10, the current and voltage are calculated as follows:

We find ZTh by open-circuiting the current source as shown in Figure 5.87.

Figure 5.87

The Thevenin’s equivalent circuit is shown in Figure 5.88.

Figure 5.88

Using Norton’s theorem, the following steps are to be performed.

Step 1: We short-circuit the load terminals A and B as shown in Figure 5.89. The whole of current 10A will flow through the short-circuited path, so that IN = 10A.

Step 2: To find ZN across terminals A and B, we open-circuit the current source as shown in Figure 5.90. Across terminals A and B, 3 Ω resistor and

Figure 5.89

Figure 5.90

Norton’s equivalent circuit is shown in Figure 5.91.

Using current divider rule, current can be calculated as follows:

Figure 5.91

Example 5.24 Determine the current through (4 − j8) Ω branch in the circuit shown in Figure 5.92 using Thevenin’s theorem and Norton’s theorem.

Figure 5.92

Solution: Using Thevenin’s theorem, the steps to be followed are given as follows:

Step 1: Open-circuit the branch through which current is to be determined as shown in Figure 5.93.

Step 2: To find VTh, we find the current through the circuit as follows:

Figure 5.93

To find ZTh, we short-circuit the voltage sources as shown in Figure 5.94.

Figure 5.94

Figure 5.95

Thevenin’s equivalent circuit will be as shown in Figure 5.95.

Figure 5.96

Using Norton’s theorem, the following steps are to be followed:

Step 1: Short-circuit the branch through which current is to be determined, as shown in Figure 5.96.

Let us find IN. From mesh I, 100 − 3i1 = 0

From mesh II, −j4i2j50 = 0

Norton’s equivalent circuit is shown in Figure 5.97.

By current divider formula, we can calculate IL as in the following:

Figure 5.97

Example 5.25 Verify the reciprocity theorem for the network shown in Figure 5.98.

Solution: Let us find current in branch CD, that is, I due to the voltage source in branch AB as in Figure 5.99.

Applying KVL in mesh I, we get the following:

Figure 5.98

Figure 5.99

Applying KVL in mesh II, the following form can be obtained:

Applying KVL in mesh III, the equation can be written as follows:

Let us solve equations (5.26), (5.27) and (5.28) using Cramer’s rule for i3

and

Now, let us insert the voltage source in branch CD and calculate the current through branch AB, as shown in Figure 5.100.

Figure 5.100

Applying KVL in mesh I, we get the following:

Applying KVL in mesh II, the following form is obtained:

Applying KVL in mesh III, the equation can be written as follows:

Now, let us solve equations (5.30), (5.31) and (5.32) for i3

and

Therefore,

From equations (5.29) and (5.34), it is clear that ratio in both the cases is same, and hence the reciprocity theorem is verified.

Example 5.26 Determine the maximum power delivered to the load in the circuit shown in Figure 5.101.

Figure 5.101

Solution: Firstly, let us find the Thevenin’s equivalent voltage source across terminals A and B.

Open-circuit the terminals AB as in Figure 5.102.

Figure 5.102

To find ZTh, we consider the circuit with short-circuiting the voltage source as in Figure 5.103.

Figure 5.103

Therefore, the given circuit can be redrawn as shown in Figure 5.104.

Figure 5.104

Now, to get the maximum power delivered to the load impedance, the load impedance must be the complex conjugate of the source impedance.

Therefore, maximum power transferred to the load is

Example 5.27 Determine the maximum power delivered to the load in the circuit, as shown in Figure 5.105.

Solution: Firstly, let us find the Thevenin’s equivalent circuit across terminals AB. We will calculate VTh and RTh. Figure 5.106 shows the diagrammatic representation of the circuit with load impedance removed.

Applying KVL in mesh I, we get the following form:

Figure 5.105

Figure 5.106

To find ZTh, we short-circuit the voltage sources and find equivalent impedance of the circuit across the terminals AB as in Figure 5.107.

Figure 5.107

The Thevenin’s equivalent circuit is shown in Figure 5.108.

Now, to get the maximum power delivered to load impedance, the load must be equal to complex conjugate of the source impedance.

Therefore, for maximum power to be delivered to load, ZL is calculated as follows:

Figure 5.108

Example 5.28 Determine the value of load impedance, ZL, for which maximum power will be delivered to this load from the source in the circuit shown in Figure 5.109.

Solution: Open-circuit voltage across terminals A and B after removing the load impedance is calculated using Figure 5.110.

Figure 5.109

Applying KVL in the loop in the circuit of Figure 5.110, we get the following form:

Figure 5.110

To find ZTh, we consider the circle across open-circuited terminals A and B by short-circuiting the voltage source as shown in Figure 5.111.

Figure 5.111

Therefore, Thevenin’s equivalent circuit is shown in Figure 5.112.

Figure 5.112

Now, according to the maximum power transfer theorem, ZL = Complex conjugate of ZTh

Example 5.29 Determine the maximum power delivered to the load in the circuit shown in Figure 5.113.

Figure 5.113

Solution: Let us draw Thevenin’s equivalent circuit of the given circuit. Firstly, we calculate the VTh after removing the load impedance as shown in Figure 5.114.

From mesh I,

Applying KVL in mesh II, we get the following:

Figure 5.114

We substitute the value of i1 in the equation, we obtain the equation as follows:

To find ZTh, we open-circuit the current source as in Figure 5.115.

Figure 5.115

Thevenin’s equivalent circuit of the given circuit can be drawn as shown in Figure 5.116.

According to the maximum power transfer theorem, ZL = complex conjugate of ZTh

Figure 5.116

Maximum power transferred to load, PL=IL2RL

Example 5.30 Calculate the current flowing through the load impedance of the circuit shown in Figure 5.117 applying Thevenin’s theorem.

Solution: We open-circuit the load terminals by removing the load impedance as shown in Figure 5.118 and calculate the VAB, that is, VTh.

We apply KVL in the loop and calculate i as follows:

Figure 5.117

Figure 5.118

VTh = VAB = voltage drop across PQ + voltage rise across QP + voltage drop across SA.

Note that no voltage will drop across the 3 Ω resistor as no current is flowing.

To calculate ZTh, we short-circuit the voltage sources and calculate the equivalent impedance from the circuit shown in Figure 5.119.

Figure 5.119

The Thevenin’s equivalent circuit is shown in Figure 5.120.

Figure 5.120

REVIEW QUESTIONS

1.State and explain Thevenin’s theorem

2.With a simple example, show how by applying Thevenin’s theorem, current flowing through a branch of an electrical network can be calculated.

3.Write the steps of application of Thevenin’s theorem.

4.State and explain Norton’s theorem.

5.Distinguish between Thevenin’s theorem and Norton’s theorem.

6.What is maximum power transfer theorem? Prove the theorem.

7.Explain reciprocity theorem with the help of a suitable example.

8.Explain Tellegen’s theorem with an example.

9.State and explain Millman’s theorem.

10.Explain superposition theorem with an example.

11.Write the conversion formula for delta to star conversion of three resistors.

12.Write the relationship of star-delta transformation of the resistors.

Numerical Problems

1.Calculate the current flowing through the 5 Ω resistor as shown in Figure 5.121

Figure 5.121

[Ans. 0.663A]

2.Calculate the current flowing through the 2 Ω resistor connected across terminals A and B in the network shown in Figure 5.122 by applying the following:

 (i) Kirchhoff’s laws (ii) Thevenin’s theorem (iii) Nodal voltage analysis

Compare the time taken in each case.

Figure 5.122

[Ans. I = 0.817 A; applying Kirchhoff’s laws takes maximum time.]

3.Apply Norton’s theorem to calculate the current through the 5 Ω resistor in the circuit shown in Figure 5.123. Further, verify by applying Thevenin’s theorem.

Figure 5.123

[Ans. I = 1A]

4.Calculate the value of RL for which maximum power will be transferred from the source to the load in the network shown in Figure 5.124. Further, calculate the value of maximum power transferred.

Figure 5.124

[Ans. RL = 7.33 Ω; Pmax = 0.545 W]

5.By using superposition theorem, calculate the current flowing through the 10 Ω resistor in the network shown in Figure 5.125.

Figure 5.125

[Ans. 0.054 A]

6.Apply Thevenin’s theorem to calculate the current flowing through the 30 Ω resistor connected across terminals A and B in the network shown in Figure 5.126.

Figure 5.126

[Ans. IAB = 1.25A]

7.For the circuit shown in Figure 5.127, determine the load impedence which will dissipate maximum power. Also, claculate the maximum power.

Figure 5.127

[Ans. ZL = (5.4 + j7.8) Ω Pmax = 6.0 W]