Two electrical bulbs marked 40, W, 220, V and | vedclass

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(C) The resistance of a bulb is given by $R = V^2 / P$. For the first bulb, $R_1 = 220^2 / 40$. For the second bulb, $R_2 = 220^2 / 60$. When connecte

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$0.24$

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(C) The resistance of a bulb is given by $R = V^2 / P$. For the first bulb, $R_1 = 220^2 / 40$. For the second bulb, $R_2 = 220^2 / 60$. When connected in series, the effective resistance is $R_s = R_1 + R_2$. The power consumed is $P_1 = V^2 / (R_1 + R_2)$. When connected in parallel, the effective resistance is $R_p = (R_1 R_2) / (R_1 + R_2)$. The power consumed is $P_2 = V^2 / R_p = V^2 (R_1 + R_2) / (R_1 R_2)$. The ratio is $P_1 / P_2 = [V^2 / (R_1 + R_2)] / [V^2 (R_1 + R_2) / (R_1 R_2)] = (R_1 R_2) / (R_1 + R_2)^2$. Substituting the values: $R_1 = 1210\, \Omega$ and $R_2 = 806.67\, \Omega$. $P_1 / P_2 = (40 	imes 60) / (40 + 60)^2 = 2400 / 10000 = 0.24$.

Two electrical bulbs marked $40\, W, 220\, V$ and $60\, W, 220\, V$ are connected in series across a $220\, V$ supply. The effective power is $P_1$. When they are connected in parallel across the same supply, the effective power is $P_2$. The ratio $(P_1/P_2)$ is:

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