Two bulbs rated 25 W - 220 V and 100 W - 2 | vedclass

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(B) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb: $R_1 = \frac{220^2}{25} = 1936 \ \Omega$.For the second bulb: $R_2 =

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$25 \ W$ bulb fuses

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(B) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb: $R_1 = \frac{220^2}{25} = 1936 \ \Omega$.For the second bulb: $R_2 = \frac{220^2}{100} = 484 \ \Omega$.Since the bulbs are connected in series,the total resistance is $R_{net} = R_1 + R_2 = 1936 + 484 = 2420 \ \Omega$.The current flowing through the circuit is $I = \frac{V_{supply}}{R_{net}} = \frac{440}{2420} = \frac{2}{11} \ A$.The potential difference across the $25 \ W$ bulb is $V_1 = I 	imes R_1 = \frac{2}{11} 	imes 1936 = 352 \ V$.The potential difference across the $100 \ W$ bulb is $V_2 = I 	imes R_2 = \frac{2}{11} 	imes 484 = 88 \ V$.Since the potential difference across the $25 \ W$ bulb $(352 \ V)$ exceeds its rated voltage $(220 \ V)$,the $25 \ W$ bulb will fuse.

Two bulbs rated $25 \ W - 220 \ V$ and $100 \ W - 220 \ V$ are connected in series to a $440 \ V$ supply. Which of the following happens?

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