## Objective: To Verify Thevenin Theorem.

Its provides a mathematical technique for replacing a given network, as viewed from two terminals, by a single voltage source with a series resistance. It makes the solution of complicated networks quite quick and easy. The application of this extremly useful theorem will be explained with the help of following simple example.

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## Circuit diagram:

Suppose, it is required to find current flowing through load resistance R_{L}, as shown in above figure.

This expression proceed as under:

1) Remove R_{L} from the circuit terminals A and B and redraw the circuit as shown in figure. Obviously, the terminal have become open circuited.

## Circuit diagram:

2) Calculate the open circuit Voltages V_{O.C. }which appears across terminals A and B when they are open .ie. when R_{L} is removed.

As seen, V_{.O.C.}= drop across R_{2}= IR_{2} where I is the circuit current when A and B is open.

It is also called Thevenin voltage(V_{th}).

3) Now, imagine the battery to be removed from the circuit, leaving its internal resistance r behind and redraw the circuit as shown in figure below.

## Circuit diagram:

When viewed inwards from the terminals A and B, the circuit consists of two parallel paths: one containing R_{2} and another containing

(R_{1}+r). The equivalent resistance of the network as viewed from these terminals is given as,

The resistance "R_{th}" is also called Thevenin equivalent resistance.

Consequently , as viewed from terminals A and B, the whole network (excluding R_{1}) can be reduced to single source ( called thevenin's

source) whose e.m.f equal to V_{O.C.} and whose internal reistance equal to R_{th}.

4) R_{L} is now connected back across terminals A and B from where it was temporaily removed earlier. Current flowing through R_{L} is given by,