Question

Use Thevenin’s theorem to find the current flowing through the 6 Ω resistor of the network shown in Fig. . All resistances are in ohms.

Answer 1

When 6 Ω resistor is removed [Fig.], whole of 2 A current flows along DC producing a drop of (2 × 2) = 4 V with the polarity as shown. As we go along BDCA, the total voltage is

= − 4 + 12 = 8 V —with A positive w.r.t. B. 

Hence, V_oc = V_th = 8 V

For finding R_i or R_th 18 V voltage source is replaced by a short-circuit and the current source by an open-circuit, as shown in Fig. . The two 4 Ω resistors are in series and are thus equivalent to an 8 Ω resistance. However, this 8 Ω resistor is in parallel with a short of 0 Ω. Hence, their equivalent value is 0 Ω. Now this 0 Ω resistance is in series with the 2 Ω resistor. Hence, Ri = 2 + 0 = 2 Ω. The Thevenin’s equivalent circuit is shown in Fig.. 

∴ I = 8/(2 + 6) = 1 Amp