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Question

(C) The resistance of an appliance is given by $R = \frac{V^2}{P}$.For the heater,the resistance is $R_h = \frac{(220)^2}{500} = \frac{48400}{500} = 9

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$0.38$

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(C) The resistance of an appliance is given by $R = \frac{V^2}{P}$.For the heater,the resistance is $R_h = \frac{(220)^2}{500} = \frac{48400}{500} = 96.8\,\Omega$.For the bulb,the resistance is $R_b = \frac{(220)^2}{100} = \frac{48400}{100} = 484\,\Omega$.When connected in series,the total resistance of the circuit is $R_{eq} = R_h + R_b = 96.8 + 484 = 580.8\,\Omega$.The current in the series circuit is $I = \frac{V}{R_{eq}} = \frac{220}{580.8} \approx 0.3787\,A$,which rounds to $0.38\,A$.

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 current in the circuit when both are connected in series is ................ $A$.

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