Two resistances $R_{1}$ and $R_{2}$ when connected is series and parallel with 120 V line, power consumed will be 25 W and 100 W respectively. Then the ration of power consumed by $R_{1}$ to that consumed by $R_{2}$ will be

1 : 1

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1 : 2

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2 : 1

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1 : 4

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Solution

The correct option is D $1 : 1$
when connected in series $P_{s} = 25 W$
when connected in series $P_{p} = 100 W$
According to question:
$P = \frac{V^{2}}{R} = \frac{P_{P}}{P_{S}} = \frac{R_{S}}{R_{P}} = \frac{(R_{1} + R_{2})}{R_{1} R_{2} / (R_{1} + R_{2})} = \frac{{(R_{1} + R_{2})}_{1}^{2}}{R_{1} R_{2}}$

$\Rightarrow \frac{100}{25} = \frac{{(R_{1} + R_{2})}_{1}^{2}}{R_{1} R_{2}}$ $\Rightarrow \frac{R_{1}}{R_{2}} = \frac{1}{1}$

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