Right option is (d) 66 W
Best explanation: Drop across V1Ω = 5 × 1 = 5V
Also, \(\frac{V-V_{1Ω}}{10} + \frac{V-20-V_{1Ω}}{2} + \frac{V-V_{OC}}{5}\) = 2
Or, 0.1 V – 0.1V1Ω + 0.5V – 10 – 0.5V1Ω + 0.2 – 0.2VOC = 2
Or, 0.8V – 0.6V1Ω = 12 + 0.2VOC
Or, 0.8 V – 0.2VOC = 12 +3=15 (Putting V1Ω = 5)
Again, \(\frac{V_{OC}-V}{5}\) + 2 = 5
Or, 0.2VOC – 0.2V = 3
Again, RTH = {(10||2) + 1} + 5
= (\(\frac{20}{12+1}\)) + 5 = 7.67 Ω
Following the theorem of maximum power transfer, R = RTH = 7.67 Ω
And PMAX = \(\frac{V_{OC}^2}{4R} = \frac{45^2}{4×7.67}\) = 66 W.