Question

Three identical bulbs are connected as shown in figure. When switch $S$ is closed, the power consumed in bulb $B$ is $P$. What will be the power consumed by the same bulb when switch $S$ is opened?

• A. $\frac{9P}{4}$
• B. $\frac{16P}{9}$
• C. $\frac{9P}{16}$
• D. $\frac{4P}{9}$

Answer 1

The correct option is D $\frac{4P}{9}$

As all the bulbs are identical So,

According to this question in first case

Voltage across bulbs $=\frac{E}{2}$ (By voltage divider rule)

So, $R=\frac{V^2}{P}=\frac{E^2}{4P}$

When $S$ is closed effective resistance of $B$ and $C$

$\begin{array}{c}=\frac{R\times R}{R+R}\left(\parallel com\right)\\ =\frac{R}{2}\end{array}$

So,

the voltage across $B$

$\begin{array}{c}B=\frac{\frac{R}{2}}{\frac{3R}{2}}E\\ =\frac{E}{3}\left(Byvoltagedriverrule\right)\end{array}$

Now the power consumed by $B$

$\begin{array}{c}B=\frac{{\left(\frac{E}{3}\right)}^2}{R}\\ =\frac{E^2\times 4P}{9\times E^2}\\ =\frac{4P}{9}\end{array}$

Hence, Option $D$ is the correct answer.