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 bulbs 2
B. only bulbs 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) = 3 R //2`
Equivalent resistance decreases, so current increases in circuit and in '1' also. Hence brightness of '1' increases. It means p.d. across '1' increases, so across '2' p.d. decreases, hence bightness of 2 decreases.
Initially : p.d. across '4'
`V_(4i) = (1)/(2) [((2 R //3) epsilon)/(2 R // 3 + R)] = (epsilon)/(5)`
Finally : `V_(4f) = ((R//2) epsilon)/(R//2 + R) = (epsilon)/(3)`
`V_(4f) gt V_(4i)`, hence brightness of 4 increases.