Three identical bulbs B_1, B_2 and B_3 are co | vedclass

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(B) Let the resistance of each bulb be $R$. Initially,when switch $S$ is closed,bulbs $B_2$ and $B_3$ are in parallel,and their equivalent resistance

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(B) Let the resistance of each bulb be $R$. Initially,when switch $S$ is closed,bulbs $B_2$ and $B_3$ are in parallel,and their equivalent resistance is $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$. This parallel combination is in series with bulb $B_1$. So,the total equivalent resistance of the circuit is $R_{eq} = R + \frac{R}{2} = \frac{3R}{2}$. The total current drawn from the source is $I = \frac{V}{R_{eq}} = \frac{V}{3R/2} = \frac{2V}{3R}$. Since bulb $B_1$ is in series with the source,the current through $B_1$ is $I_1 = I = \frac{2V}{3R}$. When switch $S$ is opened,bulb $B_3$ is disconnected. Now,only bulbs $B_1$ and $B_2$ remain in series. The new equivalent resistance is $R'_{eq} = R + R = 2R$. The new total current is $I' = \frac{V}{2R}$. Comparing the currents,$I_1 = \frac{2V}{3R} \approx 0.66 \frac{V}{R}$ and $I'_1 = I' = \frac{V}{2R} = 0.5 \frac{V}{R}$. Since $I'_1 As the power dissipated in the bulb is $P = I^2 R$,the decrease in current leads to a decrease in the incandescence (brightness) of bulb $B_1$.

Three identical bulbs $B_1, B_2$ and $B_3$ are connected to the mains as shown in the figure. If $B_3$ is disconnected from the circuit by opening switch $S$,then the incandescence of bulb $B_1$ will

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