Question

Three similar light bulbs are connected to a constant voltage dc supply as shown in Fig. 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?

• 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 1

The correct option is D Ammeter reading decreases, Bulb brightness unchanged

Suppose V is the voltage of the supply and R is the resistance of each bulb.
As they are in parallel so equivalent resistance $R_{eq}=(1/R+1/R+1/R{)}^{-1}=R/3$
Now current reading in ammeter is $I=\frac{V}{R_{eq}}=3V/R$
Current through each bulb $=V/R$.
If one of them is broken, equivalent resistance become $R_{eq}^{\prime }=(1/R+1/R{)}^{-1}=R/2$
now current in ammeter $I^{\prime }=V/R_{eq}^{\prime }=2V/R$
but current in remaining bulb is same as before. So brightness remain same.