# 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 (E_{TH}) in series with a single resistance called Thevenin resistance R_{TH }as shown in Fig.2.

Fig.2

The value of E_{TH }and R_{TH }are determined by using Thevenin’s Theorem.

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

I = E_{TH }/( R_{TH }+ R_{L })

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. E _{TH }in series with a single resistance R_{TH}.**

- The e.m.f. E
_{TH }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 R
_{TH }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 R_{1}, R_{2 }and R_{3}.

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

Fig.4

E_{TH }is the voltage across the terminal AB with R_{L }removed. As R_{L }is removed, there will be no current in R_{2 }and E_{TH }will be the voltage across R_{3}.

E_{TH }= Voltage across R_{3}

= Current through R_{3 }× Resistance R_{3}

= [V / (R_{1 }+ R_{3})] × R_{3}

To find R_{TH}, remove the load resistance R_{L }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 R_{TH}

If we look into the circuit, we can see that R_{1 }and R_{3 }are in parallel and this parallel combination is in series with the resistance R_{2} .

So, R_{TH} = R_{2 }+ [(R_{1} ×R_{3 }) /(R_{1}+R_{3})]

When load resistance R_{L} is connected across terminal AB, the current in R_{L }is given by :

I_{L}= E_{TH }/ (R_{TH }+ R_{L})

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

Fig.5

##### Step 1 : Remove the Load Resistor R_{L}

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 R_{L }is removed, there will be no current in the 3 Ω resistor.

So, Thevenin voltage (E_{TH }) 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 **E _{TH }= Voltage drop across 6Ω resistor = 1.5 A × 6 Ω = 9 V**

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

To find R_{TH}, remove the load resistance R_{L }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 R_{TH}

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, **R _{TH} = 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.