The below figure shows two resistors X and Y connected in series to a battery. The power dissipated for this combination is $\text P$1. When these resistors are connected in parallel to the same battery then the power dissipated is given by $\text P2$. Find out the ratio $\dfrac{\text P$1}{\text P2}.

From the figure,

Resistance of resistor X ($\text R$\text X) = $\text R$
Resistance of resistor Y ($\text R\text Y$) = $2\text R$

Let, voltage supply be '$\text V$'.

Case 1 : Series connection of X and Y

When X and Y are connected in series then net resistance of the circuit is given by,

$\text R$\text S = \text R\text X + \text R\text Y \\[1em] = \text R + 2\text R \\[1em] \Rightarrow \text R\text S = 3\text R

Now,

$\text P$1 = \dfrac{\text V^2}{\text R\text S} \\[1em] = \dfrac{\text V^2}{3\text R}

Case 2 : Parallel connection of X and Y

When X and Y are connected in parallel then net resistance of the circuit is given by,

$\dfrac {1}{\text R$\text P} = \dfrac {1}{\text R\text X} + \dfrac {1}{\text R\text Y} \\[1em] = \dfrac {1}{\text R} + \dfrac {1}{2\text R} \\[1em] = \dfrac {2 + 1}{2\text R} \\[1em] \Rightarrow \dfrac {1}{\text R\text P} = \dfrac {3}{2\text R} \\[1em] \Rightarrow \text R_\text P = \dfrac {2\text R}{3}

Now,

$\text P$2 = \dfrac{\text V^2}{\text R\text P} \\[1em] = \dfrac{\text V^2}{\dfrac {2\text R}{3}}\\[1em] = \dfrac{3\text V^2}{2\text R}

So,

$\dfrac{\text P$1}{\text P2} = \dfrac{\dfrac{\text V^2}{3\text R}}{\dfrac{3\text V^2}{2\text R}} \\[1em] = \dfrac{2 \times 1}{3 \times 3} \\[1em] \Rightarrow \dfrac{\text P1}{\text P2} = \dfrac{2}{9}

Hence, the ratio $\dfrac{\textbf P$\bold 1}{\textbf P\bold 2} \textbf { is } \dfrac{\bold 2}{\bold 9}.