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For the circuit shown, a shorting wire of negligible resistance is added to the circuit between points `A` and `B`. When this shorting wire is added, bulb 3 goes out. Which bulb (s) in the circuit brighten ? All bulbs are identical. A. only bulb 2B. only bulb 4C. only bulbs 1 and 4D. only bulbs 2 and 4 |
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Answer» Correct Answer - C c. Initially, `R_(eq) = 5R//3`. Finally, `R_(eq) = 3R//2` . Equivalent resistance decreases, so current increases in circuit and in 1 also. Hence, brightness of 1 increases. It means pd across 1 increases, so across 2 pd decreases, hence brightness of 2 decreases. Initially, pd across 4 is `V_(4i) = 1/2 [((2R//3)epsilon)/(2R//3+R)] = epsilon/5` Finally, `V_(4f) = ((R//2)epsilon)/(R//2+R) = epsilon/3` Since `V_(4f) gt V_(4i)`, brightness of 4 increases. |
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