Question

For the circuit shown, a wire of negligible resistance is connected between points $A$ and $B$. When this wire is connected, bulb $3$ goes out. All bulbs are identical. The brightness of which bulbs will increase?

• A. only bulb $1$ and $4$
• B. only bulb $2$ and $4$
• C. only bulb $2$
• D. only bulb $4$

Answer 1

The correct option is A only bulb $1$ and $4$

When bulb $3$ is still there

$R_{eq}=R+\frac{\left(2R\right)\left(R\right)}{2R+R}=R+\frac{2R}{3}=\frac{5R}{3}$

When bulb $3$ goes out

$R_{eq}=R+\frac{R\times R}{R+R}=R+\frac{R}{2}=\frac{3R}{2}$

$\Rightarrow$ Equivalent resistance will decrease.

Since $R_{eq}$ decreasess $\Rightarrow$ current in the circuit will increase.

Therefore brightness of bulb $1$ will increase Final current $\left(i\right)$

$=\frac{E}{\frac{3R}{2}}=\frac{2E}{3R}$

Initial current $\left(i_0\right)$

$=\frac{E}{\frac{5R}{2}}=\frac{3E}{5R}$

Current across bulb $2$

$=\frac{i}{2}=\frac{E}{2R}$

Initial current across bulb $2$

$=\frac{2i_0}{3}=\frac{2E}{5R}$

$\Rightarrow$ Current in bulb $2$ has decreased.

In other words, its brightness will also decrease.

Also, initial current through bulb $4$

$=\frac{i_0}{3}=\frac{E}{5R}$

Final current through bulb $4$

$\frac{i}{2}=\frac{2E}{3R}\times \frac{1}{2}=\frac{E}{3R}$

$\Rightarrow$ Current in bulb $4$ has increased.

In other words, its brightness will also increase.

Hence, option (c) is correct.