Two bulbs rated (25, W - 220, V) and (100, W | vedclass

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(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.Resistance of the $25\, W$ bulb: $R_1 = \frac{220^2}{25} = 1936\, \Omega$.Resistance of t

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$25\, W$ bulb

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(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.Resistance of the $25\, W$ bulb: $R_1 = \frac{220^2}{25} = 1936\, \Omega$.Resistance of the $100\, W$ bulb: $R_2 = \frac{220^2}{100} = 484\, \Omega$.When connected in series to a $440\, V$ source,the total resistance is $R_{eq} = R_1 + R_2 = 1936 + 484 = 2420\, \Omega$.The current flowing through the circuit is $I = \frac{V_{total}}{R_{eq}} = \frac{440}{2420} = \frac{2}{11}\, A$.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 $25\, W$ bulb is designed to operate at $220\, V$ and the voltage across it is $352\, V$,it will fuse.

Two bulbs rated $(25\, W - 220\, V)$ and $(100\, W - 220\, V)$ are connected in series to a $440\, V$ line. Which one is likely to fuse?

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