Question

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

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

Answer 1

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

Let the resistance of each bulb be $R$.
when the switch $S$ is closed, the corresponding circuit is shown below:

The current gets distributed in inverse ratio of resistance in the parallel combination. Hence it will distribute equally i.e with $i/2$ value in each branch.

The two bulbs $B$ and $C$ are in parallel.
Equivalent resistance for the circuit,
$R_{\text{eq}}=R+\frac{R}{2}=\frac{3R}{2}$

Current supplied by the battery is,
$i=\frac{E}{R_{\text{eq}}}$
$\text{or},i=\frac{E}{\left(\frac{3R}{2}\right)}=\frac{2E}{3R}$
Hence the power consumed by bulb $B$ is,
$P_B=i_{\text{B}}^{2}R$

$\Rightarrow P_B=\left(\frac{i^2}{4}\right)R=\frac{1}{4}\times \frac{4E^2}{9R^2}\times R=\frac{E^2}{9R}$

Initially when swich $S$ is open:


Current supplied by the battery is,
$i^{\prime }=\frac{E}{2R}$
$(\because R_{\text{eq}}=R+R=2R)$
In this case the power consumed by bulb $B$ will be,
$P_B^{{}^{\prime }}=i^{\prime 2}R$
$\text{or},P_B^{{}^{\prime }}={\left(\frac{E}{2R}\right)}^2\times R=\frac{E^2}{4R}$

According to the question power consumed in bulb when switch $S$ is in open condition is $P$.
$P=P_B^{{}^{\prime }}=\frac{E^2}{4R}$
$\text{or},\frac{P_B}{P_B^{{}^{\prime }}}=\frac{\left(\frac{E^2}{9R}\right)}{\left(\frac{E^2}{4R}\right)}$
$\text{or},\frac{P_B}{P_B^{{}^{\prime }}}=\frac{4}{9}$
$\therefore P_B=\frac{4}{9}{P}_B^{{}^{\prime }}=\frac{4}{9}P$

Thus power consumed by same bulb $B$ when switch is closed will be $\frac{4P}{9}$
$$\begin{array}{c}\text{Why this question?}\\ \text{Tip: In such problems treat each bulb as a resistance}\\ \text{and draw the simplified circuit when switch is}\\ \text{open and closed respectively.The power dissipated}\\ \text{can be easily related when we find current through}\\ \text{the bulb in both cases.}\end{array}$$