This can be attempted by Thevenin’s Theorem. As in the circuit, with terminals A and B kept open, from the right hand side, VB (w.r. to reference node 0) can be calculated V4 and V5 will have a net voltage of 2 volts circulating a current of (2/8) = 0.25 amp in clockwise direction.
VB = 10 − 0.25 × 2 = 9.5 volts.
On the Left-hand part of the circuit, two loops are there. VA (w.r. to 0) has to be evaluated. Let the first loop (with V1 and V2 as the sources) carry a clockwise current of i1 and the second loop (with V2 and V3 as the sources), a clockwise current of i 2. Writing the circuit equations.
8i − 4i2 = + 4
− 4i + 8i2 = + 4
This gives i1 = 1 amp, i2 = 1 amp
Therefore, VA = 12 + 3 × 1 = 15 volts.
Thevenin − voltage, VTH = VA −VB = 15 − 9.5 = 5.5volts
Solving as shown in Fig. .
RTH = 3 ohms
For maximum power transfer,
RL = 3 ohms
Current = 5.5/6 = 0.9167 amp
Power transferred to load 0.91672 × 3 = 2.52 watts