Question

Convert the circuit shown in Fig, to a single voltage source in series with a single resistor.

Answer 1

Obviously, we have to find equivalent Thevenin circuit. For this purpose, we have to calculate (i) V_th or V_AB and (ii) R_th or R_AB. 

With terminals A and B open, the two voltage sources are connected in subtractive series because they oppose each other. Net voltage around the circuit is (15 − 10) = 5 V and total resistance is (8 + 4) = 12 Ω. Hence circuit current is = 5/12 A. Drop across 4 Ω resistor = 4 × 5/12 = 5/3 V with the polarity as shown in Fig. 2.129 (a).

∴ V_AB = V_th = + 10 + 5/3 = 35/3 V. 

Incidently, we could also find V_AB while going along the parallel route BFEA. 

Drop across 8Ω resistor = 8 × 5/12 = 10/3 V. V_AB equal the algebraic sum of voltages met on the way from B to A. Hence, V_AB = (− 10/3) + 15 = 35/3 V. 

As shown in Fig. , the single voltage source has a voltage of 35/3 V.

For finding R_th, we will replace the two voltage sources by short-circuits. In that case, R_th = R_AB = 4 || 8 = 8/3 Ω.