We will remove RL and find the equivalent Thevenin’s source for the circuit to the left of terminals A and B. As seen from Fig. Vth equals the drop across the vertical resistor of 3Ω because no current flows through 2 Ω and 1 Ω resistors.
Since 15 V drops across two series resistors of 3 Ω each, Vth = 15/2 = 7/5 V.
Thevenin’s resistance can be found by replacing 15 V source with a short-circuit.
As seen from Fig. , Rth = 2 + (3 || 3) + 1 = 4.5 Ω. Maximum power transfer to the load will take place when RL = Rth = 4.5 Ω.
Maximum power drawn by RL = Vth 2 /4 × RL = 7.52 /4 × 4.5 = 3.125 W.
Since same power in developed in Rth, power supplied by the source = 2 × 3.125 = 6.250W.