Example (3):

With reference to the network of Fig. (1), (where the number against resistances indicate their values in ohms and the internal resistance of the battery is given (1 Ω) 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 15Ω

Solution:

(i)  Current in the network after load resistance has been removed [Fig. (1(b))]=24/(12+3+1)=1.5A

Then voltage across terminals AB=V=12*1.5=18V

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 [Fig. (2(a)). Imagine that battery of 24V is removed but not its internal resistance. Then equivalent resistance of the circuit as looked into from points A and B is

R=12*4/(12+4)=3 Ω

(iii) When load resistance of 15Ω is connected across the terminals, then the network is reduced to the structure shown in Fig. (2(b)). Then     I=18/(15+3)=1A.