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(D) The resistance of the bulb is given by $R = \frac{V^2}{P} = \frac{60^2}{10} = 360 \, \Omega$.The voltage across the bulb is $V_R = 60 \, V$ and th

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$480 \, \Omega$

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(D) The resistance of the bulb is given by $R = \frac{V^2}{P} = \frac{60^2}{10} = 360 \, \Omega$.The voltage across the bulb is $V_R = 60 \, V$ and the source voltage is $V_S = 100 \, V$.In an $RC$ series circuit,the relationship between voltages is $V_S^2 = V_R^2 + V_C^2$,where $V_C$ is the voltage across the capacitor.$100^2 = 60^2 + V_C^2 \Rightarrow V_C^2 = 10000 - 3600 = 6400 \Rightarrow V_C = 80 \, V$.The current flowing through the circuit is $I = \frac{V_R}{R} = \frac{60}{360} = \frac{1}{6} \, A$.Using Ohm's law for the capacitor,$V_C = I X_C$,where $X_C$ is the capacitive reactance.$X_C = \frac{V_C}{I} = \frac{80}{1/6} = 480 \, \Omega$.

Calculate the capacitive reactance of a capacitor in order to run a bulb rated at $10 \, W, 60 \, V$ when connected in series to an $a.c.$ source of $100 \, V$.

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