A 100 W bulb $B_{1}$ , and two $60 W$ bulbs $B_{2}$ and $B_{3}$ , are connected to a $250 V$ source as shown in the figure. Now $W_{1}, W_{2}$ and $W_{3}$ are the output powers of the bulbs $B_{1}, B_{2}$ and $B_{3}$ respectively. Then,

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$W_{1} < W_{2} < W_{3}$

$W_{1} > W_{2} = W_{3}$

$W_{1} > W_{2} > W_{3}$

$W_{1} < W_{2} = W_{3}$

answer is D.

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Detailed Solution

$P = \frac{V^{2}}{R}$ so, $R = \frac{V^{2}}{P}$

$∴$ $R_{1} = \frac{V^{2}}{100}$ and $R_{2} = R_{3} = \frac{V^{2}}{60}$

Now,

$W_{1} = \frac{(250)^{2}}{{(R_{1} + R_{2})}^{2}} \cdot R_{1}$

$W_{2} = \frac{(250)^{2}}{{(R_{1} + R_{2})}^{2}} \cdot R_{2}$ and $W_{3} = \frac{(250)^{2}}{R_{3}}$

$W_{1} : W_{2} : W_{3} = 15 : 25 : 64$ or $W_{1} < W_{2} < W_{3}$

$∴$ The correct option is (d).

Note We have used $W = i^{2} R$ for $W_{1}$ and $W_{2}$ and $W = \frac{V^{2}}{R}$ for $W_{3}$ .

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