Answer
Please see the work below.
Work Step by Step
Given that the resistors are connected in parallel,
$ power \space dissipated \space in \space R_1\space P_1=\frac{V^2}{R_1}$
$ power \space dissipated \space in \space R_2\space P_2=\frac{V^2}{R_2}$
Thus, the net power is $ P_1+P_2=\frac{V^2}{R_1}+\frac{V^2}{R_2}=V^2(\frac{R_2+R_1}{R_1R_2})$
Now, the equivalent resistance of the circuit when the resistors are connected in parallel is given as
$ R_{eq}=(\frac{1}{R_1}+\frac{1}{R_2})^{-1}$
$ R_{eq}=(\frac{R_1+R_2}{R_1R_2})^{-1}$
Thus, the net power is
$ P_{total}=V^2(\frac{R_1+R_2}{R_1R_2})$
Thus, in both cases the net power is the same.
