A 100 W bulb B₁ and two 60 W bulbs B₂ and B₃are connected to a 250 V source as shown. Now W₁, W₂and W₃ are the output powers of the bulbs B₁ , B₂ and B₃ 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

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

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

Potential drop across bulbs are

$V_{1} = \frac{R_{1}}{R_{1} + R_{2}} V = \frac{3}{8} V$

$V_{2} = \frac{5}{8} V$

$V_{3} = V$

Output power from bulbs,

$W_{1} = \frac{{V_{1}}^{2}}{R_{1}} = 14.06 W$

$W_{2} = \frac{{V_{2}}^{2}}{R_{2}} = 23.43 W$

$W_{3} = \frac{{V_{3}}^{2}}{R_{3}} = 60 W$

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

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