An electric bulb of 60 , W, 120 , V is to be | vedclass

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(C) First, calculate the resistance of the bulb using the formula $R = \frac{V^2}{P}$.$R = \frac{120 	imes 120}{60} = 240 \, \Omega$.Next, calculate

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$200 \, \Omega$

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(C) First, calculate the resistance of the bulb using the formula $R = \frac{V^2}{P}$.$R = \frac{120 	imes 120}{60} = 240 \, \Omega$.Next, calculate the rated current of the bulb using $I = \frac{P}{V}$.$I = \frac{60}{120} = 0.5 \, A$.For the bulb to glow properly, it must draw the same current $I = 0.5 \, A$ when connected to the $220 \, V$ source.The total resistance $R_{total}$ required in the circuit is $R_{total} = \frac{V_{source}}{I} = \frac{220}{0.5} = 440 \, \Omega$.Since the bulb and the additional resistor are in series, $R_{total} = R + R_{series}$.Therefore, $R_{series} = R_{total} - R = 440 \, \Omega - 240 \, \Omega = 200 \, \Omega$.

An electric bulb of $60 \, W, 120 \, V$ is to be connected to a $220 \, V$ source. What resistance should be connected in series with the bulb so that the bulb glows properly?

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