Right answer is (b) 76.68 W
The best I can explain: I2 = \(\frac{V_2}{6}\), I1 = \(\frac{I_2}{5} = \frac{V_2}{30}\)
V1 = 5V2
50 = 400(I1 – 0.04V2) + V1
Or, V2 = 21.45 V
∴ PL = \(\frac{V_2^2}{6} \)
= \(\frac{21.45^2}{6} \)
= \(\frac{460.1025}{6}\) = 76.68 W.