The figure shows three similar lamps L_1, L_2 | vedclass

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(B) Let the resistance of each lamp be $R$. Initially,$L_2$ and $L_3$ are in parallel,and their equivalent resistance is $R_p = \frac{R 	imes R}{R +

Vedclass · Accepted Answer

Brilliance of $L_1$ decreases and that of $L_2$ increases

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(B) Let the resistance of each lamp be $R$. Initially,$L_2$ and $L_3$ are in parallel,and their equivalent resistance is $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$.This combination is in series with $L_1$. The total resistance of the circuit is $R_{eq} = R + \frac{R}{2} = \frac{3R}{2}$.The total current from the source is $I = \frac{E}{R_{eq}} = \frac{2E}{3R}$.This current $I$ flows through $L_1$. The voltage across $L_1$ is $V_1 = I 	imes R = \frac{2E}{3}$. The voltage across the parallel combination ($L_2$ and $L_3$) is $V_p = I 	imes \frac{R}{2} = \frac{E}{3}$.When $L_3$ fuses,it acts as an open circuit. The circuit now consists of $L_1$ and $L_2$ in series.The new total resistance is $R'_{eq} = R + R = 2R$.The new current in the circuit is $I' = \frac{E}{2R}$.For $L_1$: The new current $I' = \frac{E}{2R}$ is less than the initial current $I = \frac{2E}{3R} = \frac{0.66E}{R}$. Thus,the brilliance of $L_1$ decreases.For $L_2$: The new voltage across $L_2$ is $V'_2 = I' 	imes R = \frac{E}{2} = 0.5E$. The initial voltage across $L_2$ was $V_p = \frac{E}{3} \approx 0.33E$. Since $0.5E > 0.33E$,the voltage across $L_2$ increases,so the brilliance of $L_2$ increases.

The figure shows three similar lamps $L_1, L_2$,and $L_3$ connected across a power supply. If the lamp $L_3$ fuses,how will the light emitted by $L_1$ and $L_2$ change?

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