Three equal resistors connected in series across a source of emf together dissipate $$10\,W$$  of power. What will be the power dissipated in watt if the same resistors are connected in parallel across the same source of emf ?

Solution :
Power, $$P = \frac{{\Delta U}}{{\Delta t}} = V\frac{{\Delta q}}{{\Delta t}} = Vi$$
$${\text{or}}\,\,P = Vi = \frac{{{V^2}}}{R}\,\,\left( {\because V = iR} \right)$$
When resistors are in series, then
$$\eqalign{ & {R_1} = R + R + R \cr & = 3R \cr} $$
$$\therefore $$ Power dissipated,
$${P_1} = \frac{{{V^2}}}{{{R_1}}} = \frac{{{V^2}}}{{3R}}\,......\left( {\text{i}} \right)$$
When resistors are in parallel, then
$$\eqalign{ & \frac{1}{{{R_2}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \cr & = \frac{3}{R} \cr & \Rightarrow {R_2} = \frac{R}{3} \cr & \therefore {P_2} = \frac{{{V^2}}}{{{R_2}}} = \frac{{{V^2}}}{{\frac{R}{3}}} = \frac{{3{V^2}}}{R}\,.......\left( {{\text{ii}}} \right) \cr} $$
By taking ratio of Eqs. (i) and (ii)
$$\eqalign{ & \frac{{{P_2}}}{{{P_1}}} = \frac{{3{V^2}}}{R} \times \frac{{3R}}{{{V^2}}} \cr & = 9 \cr & {P_2} = 9{P_1} \cr & = 9 \times 10 \cr & = 90\,W \cr} $$