Two electric bulbs marked 40,W, 220,V and 60, | vedclass

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(C) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb $(40\,W, 220\,V)$,$R_1 = \frac{220^2}{40} = 1210\,\Omega$.For the seco

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$0.24$

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(C) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb $(40\,W, 220\,V)$,$R_1 = \frac{220^2}{40} = 1210\,\Omega$.For the second bulb $(60\,W, 220\,V)$,$R_2 = \frac{220^2}{60} = 806.67\,\Omega$.When connected in series,the total resistance $R_s = R_1 + R_2 = \frac{220^2}{40} + \frac{220^2}{60} = 220^2 \left( \frac{3+2}{120} \right) = 220^2 \left( \frac{5}{120} \right) = \frac{220^2}{24}$.The effective power in series is $P_1 = \frac{V^2}{R_s} = \frac{220^2}{220^2/24} = 24\,W$.When connected in parallel,the effective power is $P_2 = P_1' + P_2' = 40\,W + 60\,W = 100\,W$.Therefore,the ratio $\frac{P_1}{P_2} = \frac{24}{100} = 0.24$.

Two electric bulbs marked $40\,W, 220\,V$ and $60\,W, 220\,V$ are connected in series across a $220\,V$ supply,resulting in an effective power $P_1$. When they are connected in parallel across the same $220\,V$ supply,the effective power is $P_2$. The ratio $\frac{P_1}{P_2}$ is:

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