First slide

Electrical power

Question

Three identical bulbs $B_{1}, B_{2} and B_{3}$ are connected as shown in the figure. Total power consumed by the circuit is ‘p’. If the bulb $B_{2}$ is fused, total power consumed by the system will be $P^{1}$ . Then

Moderate

A

2P/3
B

3P/4
C

P/4
D

P/3

Solution

If resistance of each bulb is R, then initial equivalent resistance of the system is,
$R_{C} = \frac{R \times R}{R + R} + R = 3 R / 2$ .
Write $B_{2}$ gets fused, $R_{C}^{1} = R + R = 2 R$
Since power consumed by the system $= \frac{V^{2}}{R_{C}}$
We can write, $\frac{P^{1}}{P} = \frac{R_{C}}{R_{C}^{1}} = \frac{3 R / 2}{2 R} = \frac{3}{4}$
$\Rightarrow P^{1} = \frac{3 P}{4}$

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