Correct Answer - D
Suppose `V` is the voltage of the supply and `R` is the resistance of each bulb. Now , `R_(P) = R//3` and current in ammeter is `I = V//R_(p) = 3 V//R`, provided all three bulbs are working properly. If one bulb has broken down , then
`R_(P) = ( R )/( 2) and I = 2(V)/( R )`
Therefore , current decreases and since current through each bulb is `V// R`, the same as before , brightness of bulbs is not affected.