Question

With reference to the network of Fig., 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 15 Ω.

Answer 1

(i) Current in the network before load resistance is connected [Fig.]

= 24/(12 + 3 + 1) = 1.5 A 

∴ voltage across terminals AB = V_oc = V_th = 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 24 V). 

(ii) There are two parallel paths between points A and B. Imagine that battery of 24 V is removed but not its internal resistance. Then, resistance of the circuit as looked into from point A and B is [Fig.] 

R_i = R_th = 12 × 4/(12 + 4) = 3Ω

 (iii) When load resistance of 15Ω is connected across the terminals, the network is reduced to the structure shown in Fig.

I = V_th/(R_th + R_L) = 18/(15 + 3) = 1A