Three similar light bulbs are connected to a constant to a constant voltage dc supply as shown in `Fig . 7.32`. Each bulb operates at normal brightness and the ammeter ( of negligible resistance) registers a steady current. The filament of one of the bulbs breaks. What happens to the ammeter reading and to the brightness of the remaining bulbs?

Options

A. Ammeter reading - increases, Bulb brightness - increases
B. Ammeter reading - increases, Bulb brightness - unchanged
C. Ammeter reading - unchanged , Bulb brightness - unchanged
D. Ammeter reading - decreases, Bulb brightness - unchanged

Answer

D

Explanation

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.