Two equal resistances R are joined with voltage source V in (i)...

(i) In series: total resistance $=R+R=2 R$ Power consumed $P_1=\frac{V^2}{2 R} \quad \ldots$ (i) (ii) In parallel : potential difference across each resistance will be $V$ So, power consumed in each resistance is $P=\frac{V^2}{R}$ So total power consumed in two resistance is $P_2=\frac{V^2}{R}+\frac{V^2}{R}=\frac{2 V^2}{R}$ \quad \ldots$(ii)$$\begin{array}{r}\text{ Hence, required ratio }=\frac{P_1}{P_2}\\ =\frac{V^2 \times R}{2 R \times 2 V^2}=\frac{1}{4}\end{array}$$$