Forty electric bulbs are connected in series | vedclass

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(B) Let the resistance of each bulb be $R$. When $40$ bulbs are connected in series,the total resistance is $R_{eq1} = 40R$. The current in the circui

Vedclass · Accepted Answer

More with $39$ bulbs than with $40$

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(B) Let the resistance of each bulb be $R$. When $40$ bulbs are connected in series,the total resistance is $R_{eq1} = 40R$. The current in the circuit is $I_1 = V / (40R)$. The power consumed by each bulb is $P_1 = I_1^2 R = (V / 40R)^2 R = V^2 / (1600R)$. When $39$ bulbs are connected in series,the total resistance is $R_{eq2} = 39R$. The current in the circuit is $I_2 = V / (39R)$. The power consumed by each bulb is $P_2 = I_2^2 R = (V / 39R)^2 R = V^2 / (1521R)$. Since $1521R  P_1$. Therefore,the illumination is more with $39$ bulbs than with $40$ bulbs.

Forty electric bulbs are connected in series across a $220\, V$ supply. After one bulb is fused,the remaining $39$ are connected again in series across the same supply. The illumination will be

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