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(A) The resistance of an appliance is given by $R = \frac{V^2}{P}$.For the heater,$R_h = \frac{220^2}{500} = 96.8\, \Omega$.For the bulb,$R_b = \frac{

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$13.98\, W, 69.89\, W$

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(A) The resistance of an appliance is given by $R = \frac{V^2}{P}$.For the heater,$R_h = \frac{220^2}{500} = 96.8\, \Omega$.For the bulb,$R_b = \frac{220^2}{100} = 484\, \Omega$.When connected in series,the total resistance $R_{eq} = R_h + R_b = 96.8 + 484 = 580.8\, \Omega$.The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{220}{580.8} \approx 0.3788\, A \approx 0.38\, A$.The power consumed by the heater is $P_h = I^2 R_h = (0.3788)^2 	imes 96.8 \approx 13.88\, W \approx 13.98\, W$ (using $0.38\, A$ approximation).The power consumed by the bulb is $P_b = I^2 R_b = (0.3788)^2 	imes 484 \approx 69.44\, W \approx 69.89\, W$ (using $0.38\, A$ approximation).

An electric heater and an electric bulb are rated $500\, W, 220\, V$ and $100\, W, 220\, V$ respectively. Both are connected in series to a $220\, V$ $a.c.$ mains. The power consumed by the heater and the bulb respectively is:

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