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Electric Power

A 100W bulb...

Question

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 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.

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

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

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

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

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Solution

The correct option is D $W_{1} < W_{2} < W_{3}$
$∴ R_{1} = \frac{V_{1}^{2}}{100}, R_{2} = \frac{V_{2}^{2}}{60}, R_{3} = \frac{V^{2}}{60}$ ,

$W_{1} = \frac{V_{1}^{2}}{R_{1}} = \frac{V^{2} R_{1}}{(R_{1} + R_{2})^{2}}$ ,

$W_{2} = \frac{V_{2}^{2}}{R^{2}} = \frac{V^{2} R_{2}}{(R_{1} + R_{2})^{2}}$

and $W_{3} = \frac{V^{2}}{R_{3}}$

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

or $W_{3} : W_{2} : W_{1}$ .

$= \frac{(250)^{2}}{V^{2}} \times 60 : \frac{(250)^{2}}{{[\frac{1}{100} + \frac{1}{60}]}^{2} V^{4}} \times \frac{v^{2}}{60} : \frac{(250)^{2} V^{2}}{{[\frac{1}{100} + \frac{1}{60}]}^{2} V^{4} \times 100}$

or $W_{3} : W_{2} : W_{1}$

$= 60 : \frac{100 \times 100 \times 60 \times 60}{160 \times 160 \times 60} : \frac{100 \times 100 \times 60 \times 60}{160 \times 160 \times 100}$

$= 64 : 25 : 15$

Hence, $W_{1} < W_{2} < W_{3}$
Hence, the correct option is $(D)$

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