# Thevenin’s Theorem

In this article we are going to analyse the Thevenin’s Theorem along with an example.

#### Thevenin’s Theorem

Fig.1 shows a network enclosed in a box with two terminal A and B brought out. The network in the box may consist of any number of resistors and e.m.f. sources connected in any manner.

Fig.1

According to Thevenin, the entire circuit enclosed in the box can be replaced by a single source of e.m.f. called Thevenin voltage (ETH) in series with a single resistance called Thevenin resistance RTH  as shown in Fig.2.

Fig.2

The value of ETH  and RTH are determined by using Thevenin’s Theorem.

Once Thevenin’s equivalent circuit is obtained, then current thorough the load resistance RL connected across AB can be found out by using the following equation :

I = ETH /( RTH + RL )

Thevenin’s theorem as applied to d.c. circuit and may be stated as :

Any linear, bilateral network having terminals A and B can be replaced by a single source of e.m.f. ETH in series with a single resistance RTH.

•  The e.m.f. ETH is the voltage obtained across the terminal A and B, with no load connected to it i.e. It is open circuited voltage between terminal A and B.
• The resistance RTH is the resistance of the network measured between terminal A and B with load removed and voltage sources replaced by their internal resistances. Ideal voltage sources are replaced with short circuits and ideal current sources are replaced with open circuits.

Let’s understand this by considering a circuit  as shown in Fig 3 below:

Fig.3

Here we have replaced the circuit inside the enclosed box with with a circuit having one voltage source V and three resistances R1, R2 and R3.

According to Thevenin, the circuit behind the terminal A and B can be replaced by a single voltage source ETH in series with a single resistance RTH  as shown in Fig.4.

Fig.4

ETH is the voltage across the terminal AB with RL removed. As RL is removed, there will be no current in R2 and ETH will be the voltage across R3.

ETH = Voltage across R3

= Current through R3 × Resistance R3

= [V / (R1 + R3)] × R3

To find  RTH, remove the load resistance RL and replace the voltage source by a short circuit as it is assumed that the voltage source is ideal one, hence it has no  internal resistance.

Now the resistance measured between A and B is equal to RTH

If we look into the circuit, we can see that R1 and R3 are in parallel and this parallel combination is in series with the resistance R2 .

So, RTH = R2 + [(R1 ×R3 ) /(R1+R3)]

When load resistance RL is connected across terminal AB, the current in RL is given by :

IL= ETH / (RTH + RL)

Example: Determine Thevenin Equivalent circuit for the below circuit shown in Fig.5.

Fig.5

Fig.6

##### Step 2 : Calculate the Thevenin Voltage

We can see from the Fig.6 that it is a simple circuit with a voltage source and three resistances. As RL is removed, there will be no current in the  3 Ω resistor.

So, Thevenin voltage (ETH ) will be equal to the the voltage drop across the 6Ω resistor.

To find out the voltage drop across the 6Ω resistor, we need to find out the current through it.

So, current through 6Ω resistor =  12 V / (2Ω +6Ω ) = 1.5 A

Hence  ETH = Voltage drop across 6Ω resistor = 1.5 A × 6 Ω = 9 V

##### Step 3:  Calculate the Thevenin Resistance

To find  RTH, remove the load resistance RL and replace the voltage source by a short circuit as it is assumed that the voltage source is ideal one.

Fig.7

Now the resistance measured between A and B is equal to RTH

We can see from the Fig.7 that the 2 Ω resistor  and 6 Ω resistor are in parallel and this parallel combination is in series with the resistance 3 Ω resistor .

So, RTH = 3+ [(2×6 ) /(2+6)] = 4.5 Ω

##### Step 4 : Draw the Thevenin Equivalent Circuit

So, the Thevenin Equivalent circuit will be :

Fig.8

Many times while analyzing power systems and other circuits we come across some circuits where one particular resistor in the circuit i.e. the load resistor, is subject to change. In such cases Thevenin’s theorem comes as  a great rescue.

As, in such circuits, re-calculation of the circuit is needed with each trial value of load resistance to determine the voltage across it and the current through it. Hence, Converting the power system circuit to its Thevenin equivalent can greatly simplify this problem.