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 R₁, R₂ and R₃.

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 R₂ and E_TH will be the voltage across R₃.

E_TH = Voltage across R₃

= Current through R₃ × Resistance R₃

= [V / (R₁ + R₃)] × R₃

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 R₁ and R₃ are in parallel and this parallel combination is in series with the resistance R₂ .

So, R_TH = R₂ + [(R₁ ×R₃ ) /(R₁+R₃)]

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.