A 100 W bulb and two 60 W bulbs and , are connected to a 250 V source, as shown in figure. Now , and are the output powers of the bulbs , and , respectively. Then [2002]

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 [2002]

Smart Achievers · Accepted Answer

(4)

As we know, $P = \frac{V^{2}}{R}$ $∴$ $R = \frac{V^{2}}{P}$

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

$W_{1} = \frac{V_{1}^{2}}{R_{1}} = \frac{V^{2} R_{1}}{{(R_{1} + R_{2})}^{2}},$ $W_{2} = \frac{V_{2}^{2}}{R_{2}} = \frac{V^{2} R_{2}}{{(R_{1} + R_{2})}^{2}}$ and $W_{3} = \frac{V^{2}}{R_{3}}$

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

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

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

$= 64 : 25 : 15$

$∴$ $W_{1} < W_{2} < W_{3}$

Solution

Q.
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 [2002]

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

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

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

4 $W_{1} < W_{2} < W_{3}$

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Solution

Q. A 100 W bulb B1 and two 60 W bulbs B2 and B3, are connected to a 250 V source, as shown in figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3, respectively. Then [2002]

1 W1>W2=W3

2 W1>W2>W3

3 W1<W2=W3