The given circuit of Fig. would be solved by applying Thevenin’s theorem twice, first to the circuit to the left of point C and D and then to the left of points A and B. Using this technique, the network to the left of CD [Fig.] can be replaced by a source of voltage V1 and series resistance Ri1 as shown in Fig. .


V1 = (12 x 6)/(6 + 1 + 1) = 9 volt and Ri1 = (6 x 2)/(6 + 2) = 1.5Ω
Similarly, the circuit of Fig. reduced to that shown in Fig.

V2 = (9 x 6)/(6 + 2 + 1.5) = 5.68 volt and Ri2 = (6 x 3.5)/9.5 = 2.21