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.

Let the resistance of each bulb be R. Initially: Req = 5R/3,    Finally : Req = 3R/2.Equivalent resistance decreases, so current increases in circuit and in '1' also. Hence brightness of '1' increases. It means potential difference across '1' increases, so across '2' potential difference decreases, hence brightness of 2 decreases. Initially: Potential difference across '4'V4i =  12  2R/3ε2R/3+R=ε5Finally:  V4f  = R/2εR/2+R =ε3Since V4f > V4i, hence brightness of 4 increases.