Question

Three identical bulbs are connected in parallel with a battery.The current drawn from the battery is $6\text{A}$. If one of the bulbs gets fused, what will be the total current drawn from the battery ?

• A. $1\text{A}$
• B. $2\text{A}$
• C. $3\text{A}$
• D. $4\text{A}$

Answer 1

The correct option is D $4\text{A}$

Let the potential difference maintained by the battery be $V$ and let the resistance of each bulb be $R$ (Figure a).
If the equilvalent resistance of the circuit is $r$.

$\frac{1}{r}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}$
$r=\frac{R}{3}$
The current is $i=\frac{V}{r}=\frac{3V}{R}$.
It is given that this current is $6\text{A}$.
So, $6=\frac{3V}{R}\frac{V}{R}=2\text{A}$
If one of the bulbs gets fused (Figure b), only two bulbs remain connected in parallel. The equivalent resistance $r^{\prime }$ in that case is given by
$\frac{1}{r^{\prime }}=\frac{1}{R}+\frac{1}{R}{r}^{\prime }=\frac{R}{2}$
The current in the battery will be
$i^{\prime }=\frac{v}{r^{\prime }}=\frac{2V}{R}=2\times \left(2\right)=4\text{A}$