Question

Three similar light bulbs are connected to a constant voltage d.c. supply as shown in the figure. 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. Both ammeter reading and bulb brightness increases.
• B. Ammeter reading increases and bulb brightness remains unchanged.
• C. Both ammeter reading and bulb brightness remains unchaanged.
• D. Ammeter reading decreases and bulb brightness remains unchanged.

Answer 1

The correct option is D Ammeter reading decreases and bulb brightness remains unchanged.

Suppose $V$ is the voltage of the supply and $R$ is the resistance of each bulb.

When all three bulbs are working properly, the equivalent resistance of the network will be

$R_{eq}=R/3$

$\therefore$ Current through ammeter will be

$I=V/R_{eq}=3V/R$

And power dissipated by each bulb will be

$P=\frac{V^2}{R}$

If the filament of one bulb breaks down, then equivalent resistance becomes

$R_{eq}^{\prime }=R/2$

Current in ammeter, $I^{\prime }=2V/R$

$\therefore I>I^{\prime }$

But power dissipation through each bulb remain same because potential difference across each bulb and resistance are same.

Therefore, current through ammeter decreases and since power dissipation through each bulb is $V^2/R$ the same as before, so brightness of bulbs is not effected.

So, option $\left(d\right)$ is correct.
$$\begin{array}{c}\text{Why this question?}\\ \text{Tip: In parallel if any one bulb gets fused,}\\ \text{then others will continue to glow.}\end{array}$$