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.

Options

A. only bulb 2
B. only bulb 4
C. only bulbs 1 and 4
D. only bulbs 2 and 4

Answer

C

Explanation

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.