Answer
d) When peak value of voltage is doubled and resistance is reduced by a factor of two , Average power will be maximum
Work Step by Step
Average power is given as
$ \bar P = V_{rms} I_{rms} = \frac{V_{rms}^2}{R}$
But $V_{rms} = \frac{V_0 }{\sqrt 2}$
$ \therefore \bar P = \frac{V_0^2}{2R} $
Thus when peak value of voltage is doubled ($ 2V_0$) and resistance is reduced by a factor of two ($ \frac{1}{2} R$), Average power will be maximum($ \bar P = \frac{4 V_0}{R}$).
