First slide

Thermal effect of current

Question

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

Easy

A

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

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

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

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

Solution

$L e t t h e g i v e n p o w e r r a t i n g s b e d e f i n e d a t a n o p e r a t i n g v o l t a g e V . T h e r e s i s \tan c e s o f t h e t h r e e b u l b s a r e g i v e n b y R_{1} = \frac{V^{2}}{100}, R_{2} = \frac{V^{2}}{60}, R_{3} = \frac{V^{2}}{60} . I n t h e g i v e n c o n f i g u r a t i o n, t h e c u r r e n t t h r o u g h B_{1} a n d B_{2} i s i = \frac{250}{R_{1} + R_{2}} = \frac{250}{\frac{V^{2}}{100} + \frac{V^{2}}{60}} = \frac{250}{V^{2}} (\frac{100 \times 60}{100 + 60}) = \frac{250}{V^{2}} (\frac{75}{2}) S u b s t i t u t e t h e v a l u e s o f i, R_{1}, a n d R_{2} t o g e t t h e o u t p u t p o w e r s W_{1} = i^{2} R_{1} \approx {(\frac{250}{V^{2}} \times \frac{75}{2})}^{2} \frac{V^{2}}{100} = \frac{250^{2}}{V^{4}} {(\frac{75}{2})}^{2} \frac{V^{2}}{100} = 14 {(\frac{250}{V})}^{2}, W_{2} = i^{2} R_{2} \approx {(\frac{250}{V^{2}} \times \frac{75}{2})}^{2} \frac{V^{2}}{60} = \frac{250^{2}}{V^{4}} {(\frac{75}{2})}^{2} \frac{V^{2}}{60} = 23 {(\frac{250}{V})}^{2}, a n d W_{3} = \frac{250^{2}}{R_{3}} = 60 {(\frac{250}{V})}^{2}$

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