# Thevenin’s Theorem with Solved Examples

Table of Contents

## Introduction:

**Thevenin’s Theorem with solved Examples-** This theorem was presented by a French engineer M.L. Thevenin in 1893. Through this theorem, complicated networks (especially electronic networks) can be solved very quickly and easily. This theorem can be described as follows;

The current through a resistor R connected across any two points A and B of an active network is obtained by dividing the p.d (potential difference) between A and B, with R disconnected, by (R + r), where r is the resistance of the network measured between points A and B with R disconnected and the sources of e.m.f replaced by their internal resistances.

The elaboration of the aforementioned description is as follows;

Suppose that are two such terminals of any network, which consists of resistances R_{2 }and R_{3} and a battery having an e.m.f of E_{1} and internal resistance R_{1. }This has been illustrated in fig. 2.56 (a). Whereas, the load resistance R located between AB terminals determines the value of current passing through the load resistance R. First, load resistance R is disconnected from the circuit, as has been illustrated in fig. 2.56 (b). In such a situation, the following current will flow through R_{3}.

Current through R_{3} = E_{1} / R_{1} + R_{3}

Potential difference R_{3} = E_{1} R_{3} / R_{1} + R_{3}

No current will pass through R_{2} due to terminals A and B being open. Therefore, the voltage value across AB will be as follows (remember that these voltages are also known as Thevenin voltages, i.e., V_{th})

Potential difference across AB = V = E_{1 }R_{3} / R_{1} + R_{3}

The network which is obtained after isolating load resistance and converting battery into its internal resistance, has been shown in fig. 2.56 ©. Now, if network is viewed from inside the points AB, then total network resistance will be as below;

Resistance of Network between A and B = r = R_{2} + R_{1} R_{3 }/ R_{1} + R_{3}

Remember that “r’ is also known as Thevenin resistance (r_{th})

According to Thevenin theorem, active network shown in fig. 2.56 (a) (which has been depicted inside dotted lines) can be converted to a simple circuit shown within the dotted lines in fig. 2.56 (d). Similarly, source’s e.m.f E equals to the open circuit potential difference found between points A and B and internal resistance “r”, whereas, the values of v and r can be determined with the help of afore- mentioned formulae. Thus,

Current through R, I = E / r + R

(ii). Thevenin theorem can also be described in the following words;

Current flowing through the load resistance R_{L} across any two terminals A and B of any linear active network equals V_{o.c} / (R_{i} + R_{L}), whereas V _{o.c} means open circuit voltages (i.e., voltages found across both terminals A and B after separating R_{L} from the circuit, these voltages are also known as Thevenin voltages). And Ri means network’s internal resistance, which can be obtained through viewing inside the open circuit network from terminals A and B, provided all voltage sources in the network have been changed to their internal resistance

Let us assume that we have to determine the value of current passing through the load resistance R_{L} as shown in the figure 2.57 (a). Its procedure is as follows;

Disconnect the load resistance (R_{L}) connected with the circuit terminals A and B from the circuit and make the circuit once again. After disconnection of R_{L}, AB terminals become an open circuit.

Determine the open circuit voltages V _{o.c} found across A and B. Remember that these voltages are received when R_{L} is dissociated from the circuit, and terminals A and B become open. It is evident from figure (b) that V _{o.c} are equal to the voltage drops i.e., IR_{2} across R_{2, }whereas I is the circuit current which passes through the circuit after the opening of A and B.

I = E / R_{1} + R_{2} + r ∴ V_{OC} = IR_{2} = ER_{2} / R_{1 }+ R_{2} + r

Here “r” means internal resistance of the battery, and V_{o.c }are also called Thevenin voltages (V_{th}).

Now, assume that battery has been removed from the circuit. And instead of the battery, its internal resistance “r” has been included. Now, redraw figure according to 2.57 (c). When viewed inside the circuit from terminals A and B (as has been indicated by an arrow sign in fig. c), circuit seems to consist of two parallel paths. The resistance of one of these parallel paths is R_{2} and that of the other parallel path is (R_{1} + r). The following equivalent circuit resistance is obtained after viewing the circuit through these terminals.

R = R_{2} (R_{1} + r) / R_{2} + (R_{1} + r)

This resistance is known as Thevenin resistance (R_{th}) and it is also occasionally indicated by R_{i} or R_{o}. Resultantly, when an entire network is viewed from terminals A and B, it converts into a single source (which is known as a Thevenin source), having e.m.f equivalent to V _{O.C} (or V_{th}) and its internal resistance R_{th} (or R_{i}). This has been illustrated by fig. 2.58

Fig. 2.57

Fig. 2.58

Now, RL is reconnected once again parallel to these terminals A and B, where from it had been temporarily disconnected. In such a situation, the value of current passing through RL will be as follows;

I = V_{th }/ R _{th} + R _{L}

It is obvious from the above- mentioned detail that any network composed of resistors and voltage sources (and even current sources) when viewing back from two points A and B can be replaced to a single voltage source, and a single resistance (in case of DC circuits) parallel to this voltage source in a series, and in a situation of AC circuits, converted to impedance.

After replacing the circuit into a single voltage source set in a series, the determination of value of any current flowing through any resistance parallel to terminals A and B, becomes easy. Thus, Thevenin theorem being used in the DC circuits, may be described in the following words;

The current flowing through a load resistance R_{L }connected across any two terminals A and B of a network is given by (V_{th} / R_{th} + R_{L}) where V_{th} or V_{o.c} is the open circuit voltage (i.e., voltage across the two terminals when RL is removed) and R_{th} or R_{i} is the internal resistance of the network as viewed back into the open circuit network from terminals A and B with all voltage sources replaced by their internal resistance (if any) and current sources by infinite resistance.

## Method for finding Thevenin’s Equivalent of a Given Circuit

1). First of all, temporarily remove this resistance (which is known as load resistance) from the circuit, the current of which is desired to be determined.

2). Then determine the open circuit voltage V _{o.c} (which are also known as Thevenin voltages V _{th}). Remember that these voltages are found between those two points, wherefrom the load resistance has been removed.

3). After replacing entire voltage sources with their internal resistances, find out the total network resistance viewing inward towards the circuit through these points or terminals. This total resistance of the circuit is called Thevenin resistance (R _{th}).

4). Replace a complete network to a single Thevenin source, having a voltage V_{th} or V _{o.c}, and an internal resistance R_{th} or R_{i}.

5). Then fix the load resistance between these two terminals once again, wherefrom it had been temporarily removed.

6). Finally, determine the value of current flowing through the load resistance, with the help of the following formula;

I = V_{th} / R _{th} + R _{L }or I = V _{o.c} / (R_{i} + R_{L})

## Calculations About Thevenin Theorem

**Example 1;** Convert the circuit shown in figure 2.59 9a) to a single voltage source in series with a single resistor.

Fig. 2.59 a, b

**Solution;**

Obviously, we have to find equivalent Thevenin circuit. For this purpose, we have to calculate (i) V_{th} or V_{AB}

With terminals A and B open, the two voltage sources are connected in subtractive series because they oppose each other. Net voltage around the circuit is (15 – 10) = 5 V and total resistance is 98 + 4) = 12 Ω. Hence, circuit current is = 5/12 A. Drop across 4 Ω resistor = 4 x 5 / 12 = 5 /3 V with the polarity as shown in fig. 2.59 (a).

V_{AB} = V_{th }= + 10 + 5/3 = 35 / 3 V

Incidentally, we could also find VAB while going along the parallel route BFEA. Drop across 8 Ω resistor = 8 x 5/2 = 10 / 3 V. V_{AB} equals the algebraic sum of voltages met on the way from B to A. Hence, V_{AB} = (-10 / 3) + 15 = 35 / 3 V.

As shown in figure 2.59 9b0, the single voltage source has a voltage of 35 / 3 V.

For finding R_{th}, we will replace the two voltage sources by short – circuits. In that case, R_{th }= R_{AB} = 4 x 8 / 4 + 8 = 32 / 12 = 8/3 Ω

**Example 2;** With the help of Thevenin’s theorem to calculate the current through the 4 Ω resistor of the circuit of fig. 2.60 (a);

**Solution;**

As shown in the fig. 2.60 (b), 4 Ω resistance has been removed, thereby open – circuiting the terminals A and B. we will now find V_{AB} and R_{AB}, which will give us V_{th} and R_{th} respectively. The potential drop across 5 Ω resistor can be found with the help of voltage divider rule. Its value is = 15 x 5 / (5 + 10) = 5V.

Fig. 2.60 (a, b)

For finding V_{AB}, we will go from point B to point A in the clockwise direction and find the algebraic sum of voltages met on the way.

V_{AB} = -6 + 5 = -1 V. It means that point A is negative with respect to point B, or point B is at a higher potential than point A by one volt.

Fig. 2.61

In fig. 2.60 ©, the two voltage sources have been short – circuited. The resistance of the network as viewed from points A and B is the same as viewed from points A and C.

R_{AB} = R_{AC} = 5 x 10 / 5 + 10 = 50 / 15 = 10 / 3 Ω

Thevenin’s equivalent source is shown in fig. 2.61 in which 4 Ω resistor has been joined back across terminals A and B. Polarity of the voltage source is worth noting.

I = 1/ (10 / 3) + 4 = 3/ 22 = 0.136 A from E to A

**Example 3;** With reference to the network of fig. 2.62 (a), by applying Thevenin’s theorem find the following;

(i). The equivalent e.m.f of the network when viewed from terminals A and B.

(ii). The equivalent resistance of the network when looked into from terminals A and B.

(iii). Current in the load resistance R_{L} of 15 Ω

**Solution;**

(i). Current in the network before load resistance is connected = 24 / (12 + 3 + 1) = 1.5 A

∴ Voltage across Terminals AB = V_{oc} = V_{th} = 12 x 1.5 = 18 V

Hence, so far as terminals A and B are concerned, the network has an e.m.f of 18 volt (and not 24V).

(ii). There are two parallel paths between points A and B. Imagine that battery of 24 V is removed but not its internal resistance. Then, resistance of the circuit as looked into from points A and B is fig ©.

R_{i }= R_{th} = 12 x 4 / 12 + 4 = 3 Ω

(iii). When load resistance of 15 Ω is connected across the terminals, the network is reduced to the structure shown in fig. (d).

Fig. 2.62

I = V _{th} / R_{ th} + R_{L}) = 18/ (15 + 3) = 1 / A

**Example 4;** Using Thevenin theorem, calculate the current flowing through the 4 Ω resistor of fig. 2.63 (a).

**Solution;**

(i). Finding V_{th }

If we remove the 4 Ω resistor, the circuit becomes as shown in fig. 2.63 (b). Since full 10 A current passes through 2 Ω resistor, drop across it is 10 x 2 = 20 V. Hence, V_{B} = 20 V with respect to the common ground. The two resistors of 3 Ω and 6 Ω are connected in series across 12 V battery.

Figure 2.63 (a, b, c, d)

Hence, drop across 6 Ω resistor = 12 x 6 / (3 + 6) = 8V

V_{A} = 8 V with respect to the common ground

V_{th} = V_{BA} = V_{B} – V_{A} = 20 – 8 = 12 V – with B at higher potential

(ii). Finding R_{th}

Now, we will find R_{th} i.e., equivalent resistance of the network as looked back into the open circuited terminals A and B. For this purpose, we will replace both the voltage and current sources. Since voltage source has no internal resistance, it would be replaced by a short – circuit i.e., zero resistance. However, current source would be removed and replaced by an open i.e., infinite resistance. In that case, the circuit becomes as shown in fig. 2.63 9c). As seen from fig. 2.63 (d), Rth = 6 x 3 / 6 + 3 + 2 = 4 Ω. Hence, Thevenin’s equivalent circuit consists of a voltage source of 12 V and a series resistance of 4 Ω as shown in fig. 2.64 (a). When 4 Ω resistor is connected across terminals A and B

Fig 2.64 (a, b)

**Example 5;** For the circuit shown in fig. 2.65 (a), calculate the current in the 10 Ω resistance. Use Thevenin’s theorem only.

Figure 2.65 (a, b, c)

**Solution;**

When the 10 Ω resistance is removed, the circuit becomes as shown in fig. (b). Now, we will find the open – circuit voltage V_{AB} = V_{th}. For this purpose, we will go from point B to point A and find the algebraic sum of the voltages met on the way. It should be noted that with terminals A and B open, there is no voltage drop on the 8 Ω resistance. However, the two resistances of 5 Ω and 2 Ω are connected in series across the 20 Ω battery. As per voltage – divider rule, drop on 2 Ω resistance = 20 x 2 / (2 + 5) = 5.71 V with the polarity as shown in the figure. Then,

V_{AB} = V_{th} = + 5.71 – 12 = – 6.29 V

The negative sign shows that point A is negative with respect to point B or which is the same thing, point B is positive with respect to point A.

For finding R_{AB} = R_{th}, we replace the batteries by short – circuits as shown in fig. ©.

∴ R_{AB} = R_{th} = 8 + 2 x 5 / 2+ 5 = 9.43 Ω

Hence, the equivalent Thevenin’s source with respect to terminals A and B is as shown in fig. 2.66. When 10 Ω resistance is reconnected across A and B, current through it is

I = 6.29 / (9.43 + 10).

Fig. 2.66

**Example 6;** The four arms of a Wheatstone bridge have the following resistances;

AB = 100, BC = 10, CD = 4, DA = 50 Ω. A galvanometer when a p.d of 10 V is maintained across AC.

**Solution;**

(i). When galvanometer is removed from fig. 2.67 9a), we get the circuit of fig. 2.67 (b).

(ii). Let us next find the open – circuit voltage V_{oc} (also called Thevenin voltage V_{th}) between points B and D. Remembering that ABC (as well as ADC) is a potential divider on which a voltage drops of 10 Ω takes place, we get;

Potential of B w.r.t C = 10 x 10 / 110 = 10 / 11 = 0.909 V

Potential of D w.r.t C = 10 x 4 /54 = 20 / 27 = 0.741 V

p.d between B and D is V_{oc} or V_{th} = 0.909 – 0.741 = 0.168 V

(iii). Now, remove the 10 – V battery retaining its internal amounts to short -circuiting points A and C as shown in figure 2.67 (d).

Fig. 2.67 (a, b, c, d)

(iv). Next, let us find the resistance of the whole network as viewed from points B and D. It may be easily found by noting that electrically speaking, points A and C have become one as shown in fig 2.68 (a). It is also seen that BA is in parallel with BC and AD is in parallel with CD. Hence, R_{BD} = 10 x 100 / 10 + 100 = 50 x 4 / 50 + 4 = 12.79 Ω

Fig. 2.68

(iv). Now, so far as points B and D are concerned, the network has a voltage source of 0.168 V and internal resistance R_{i }= 12.79 Ω. This Thevenin source is shown in fig. 2.68 ©.

(v). Finally, let us connect the galvanometer (initially removed) to this Thevenin source and calculate the current I flowing through it. As seen from fig. 2.68 (d)

I = 0.168 / (12.79 + 20) = 0.005 A = 5 mA

**Example 7;** Find the current flowing through the 4 Ω resistor in fig. 2.69 (a) when (i) E = 2 V and (ii) E = 12 V. All resistors are in series.

**Solution;**

When we remove E and 4 Ω resistor, the circuit becomes as shown in fig. 2.60 (b). For finding R_{th} (i.e., the circuit resistance as viewed from terminals A and B, the battery has been short – circuited, as shown.

It has been seen from fig. 2.69 (c) that R_{th} = R_{AB} = 15 x 30 / 15 + 30 + 18 x 9 / 18 + 9 = 16 Ω

Fig. 2.69

We will find V_{th = }V_{AB} with the help of fig. 2.70 (a) which represents the original circuit, except with E and 4 Ω resistor removed. Here, the two circuits are connected in parallel across the 36 V battery. The potential of point A equals the drop on 30 Ω resistance, whereas potential of point B equals the drop across 9 Ω resistance. Using the voltage divider rule, we have;

V_{A} = 26 x 30 / 45 = 24 V

V_{B} = 36 x 9 /27 = 12 V

V_{AB} = V_{A }– V_{B} = 24 – 12 = 12 V

Fig. 2.70 (a, b)

In fig. 2.7 (b), the series combination of E and 4 Ω resistors has been reconnected across terminals A and B of the Thevenin’s equivalent circuit.

I = (12 – E) / 20 = (12 – 2) / 20 = 0.5 A (ii). I = (12 – 12) / 20 = 0

**Example 8;** Calculate the value of V_{th} and R_{th} between terminals A and B of the circuit shown in fig. 2.71 (a). All resistance values are in ohms.

**Solution;**

Forgetting about the terminal B for the time being, there are two parallel paths between E and F; one consisting of 12 Ω and the other of (4 + 8 = 12) Ω. Hence,

R_{EF} = 12 x 12 / 12 + 12 = 6 Ω. The source voltage of 48 V drops across two 6 Ω resistance connected in series. Hence, V_{EF} = 24 V. The same 24 V acts across 12 Ω resistor connected directly between E and F and across two series – connected resistances of 4 Ω and 6 Ω connected across E and F.

Drop across 4 Ω resistor = 24 x 4 / 4 + 8) = 8 V as shown in fig. 2.71 ©

Figure 2.71 (a, b, c)

Now, as we go from B to A via point E, there is a rise in voltage of 8 V followed by another rise in voltage of 24 V, thereby giving a total voltage drop of 32 volts. Hence, V _{th} = 32 V with point A positive.

For finding R _{th}, we short – circuit the 48 V source. This short – circuiting, in effect, combines the points A, D and F electrically as shown in fig. 2.72 (a). As seen from fig. 2.72 (b).

R_{th} = V_{AB} = 8 (4 + 4) / * + 4 + 4 = 4 Ω

Figure 2.72

**Example 9;** C and D in figure 2.73 (a) represent the two terminals of an active network. Calculate the current through R.

**Solution;**

With R disconnected as in fig. 2.73 9b)

I_{1} = 6 – 4 / 2 + 3 = 0.4 A

And p.d across CD = E_{1} – I_{1} R_{1}

V = 6 – (0.4 x2) = 5.2 V

When the e.m. fs are removed, as in fig 2.73 (c), total resistance between C and D = 2 x 3 / 2 + 3 = r = 1.2 Ω

Fig. 2.73 (a, b, c, d)

Here, the network AB in fig. 2.73 (a) can be replaced by a single source having an e.m.f of 5.2 V and an internal resistance of 1.2 Ω as in fig. 2.73 (d); consequently,

I = 5.2 / 1.2 + 10 = 0.4643 A

**Example 10;** The resistance of the various arms of an unbalanced Wheatstone bridge are given in fig. 2.74. The cell has an e.m.f of 2 V and a negligible internal resistance. Determine the value and direction of the current in the galvanometer circuit BD, using Thevenin’s theorem.

**Solution;**

Since we require to find the current in the 40 Ω resistor between B and D, the first step is to remove this resistor, as in fig. 2.74 (a). Then,

p.d between A and B = 2 x 10 / 10 + 30 = 0.5 V

and p.d between A and D = 2 x 20 / 20 + 15 = 1.143 V

p.d between B and D = 0.643 V

B being positive relative to D, consequently, current in the 40 Ω resistor, when connected between B and D, will flow from B to D.

The next step is to replace the cell by a resistor equal to its internal resistance. Since the latter is negligible in this problem, junctions A and C can be short – circuited as in fig. 2.74 (b).

Fig. 2.74 (a, b, c)

Equivalent resistance of BA and BC = 10 x 30 / 10 + 30 = 7.5 Ω

And equivalent resistance of AD and CD = 20 x 15 / 20 + 15 = 8.57 Ω

∴ Total resistance of network between B and D = 7.5 + 8.57 = 16.07 Ω

Hence, the network of fig. 2.74 (a) is equivalent to a source having an e.m.f of 0.643 V and an internal resistance of 16.07 Ω as in fig. 2.74 (c)

∴ Current through BD = 0.643 / 16.07 + 40 = 0.0115 A

=11.5 mA from B to D

**Example 11;** Calculate the current in the 10 Ω resistor in the network shown in fig. below;

Fig. 2.75

**Solution;**

Open circuiting the branch containing the 10 Ω resistor gives the network shown in fig. 2.76

Fig. 2.76

I_{1} = 20 / 5 + 15 = 1.0 A ∴ V _{AC} = 1.0 x 15 = 15 V

I_{2} = 10 / 2 + 8 = 1.0 A ∴ V_{BC} = 1.0 x 8 = 8 V

V_{AB} = V_{AC} – V_{BC} = 15 – 8 = 7 V

Resistance between A and B = Ro = 5 x 15 / 5 + 15 + 8 x2 / 8 + 2 = 3.75 + 1.6 = 5.35 Ω

The current then reduces to that shown in fig. 2.77

Fig. 2.77

Current in 10 Ω resistor across AB = 7 / 5.35 + 10 = 0.46 A from A to B

**Example 12;** Use Thevenin theorem to calculate the current in branch CF of the network below;

Fig. 2.78

**Solution;**

To use Thevenin’s theorem, CF is removed. Then the resistance between C and F, R_{CF} is calculated with the battery short – circuited and the battery voltage E_{CF} between C and F is found.

The current in CF is E_{CF }/ E_{CF }+ 2 since the resistance of the link removed is 2 Ω

Fig. 2.79

With the battery short – circuited R_{CF} = 2 x 5 / 2 + 5 = 10 / 7 Ω

The current in the circuit BCDEG = 10 / 7 A

E_{CF} = 10 – 2 (10 / 7) = 50 / 7 V

∴ Current in CF = 50 / 7 / (10 / 7) + 2 = 50 / 24 = 2.08 A

**Example 13;** For the circuit shown, calculate the current in XY by;

Thevenin theorem (b) Kirchhoff’s law

Figure 2.80

**Solution;**

(a). Thevenin Theorem

XY is removed; then the resistance R_{XY} between X and Y is found with the battery short – circuited. The voltage E_{XY} between X and Y is also found with XY removed. The current in XY is E_{XY} / (R_{XY} + 3)

Fig. 2.81

Since the resistance of the link removed is 3 Ω

R_{XY} = 1 x 8 / 1 + 8 = 8 / 9 Ω

Current in circuit = 20 / 9 A

E_{XY} = 8 x (20 / 9) = 160 / 9

Current in XY = 160 / 9 ÷ (8 / 9) + 3 = 4.57 A.

(b). Kirchhoff’s Laws

Let the currents be as shown

Fig. 2.82

For loop ABCD; I + 8 (I – I_{1}) = 20 … (1)

For loop AXY; I + 3I_{1} = 20 … (2)

From (1) and (2) I1 = 4.57 A

**Example 14;** Use Thevenin’s theorem to calculate the current flowing in branch XY of the circuit shown, and check the result by Kirchhoff’s laws.

Fig. 2.83

**Solution;**

To use Thevenin’s theorem, XY is removed. Then the resistance R_{XY} between X and Y is found with the batteries short- circuited. The voltage between X and Y, E_{XY} is also calculated.

The current in XY is E_{XY} / (3 + R_{XY}) since the resistance of the link removed is 3 Ω

Fig. 2.84

R_{XY} = 8 x 1 / 8 + 1 = 8 / 9

Current I = 20 / 9 A

E_{XY }= 8I – 20 = 160 / 9 – 20 = -20 / 9 V

Current in XY = – 20 / 9 / (8/9) + 3 = – 4/7 A = – 0.57 A, i.e., current flows from Y to X.

Check by Kirchhoff’s laws;

Let the currents be as shown in figure 2.84

For loop AXYD;

I_{1} + 3(I_{1} – I_{2}) + 20 – 20 = 0

i.e., 4I_{1} = 3I_{2} … (1)

For loop ABCD;

I_{1 }+ 8I_{2 }– 20 = 0 … (2)

From (1) and (2)

I_{1}= 1_{2} / 7 A and I_{2 }= 16 / 7 A

I_{1} – I_{2} = 4/7 A as before

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