If two bulbs of wattage 25, W and 100, W resp | vedclass

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(B) Resistance of $25\, W$ bulb is $R_1 = \frac{V^2}{P_1} = \frac{220^2}{25} = 1936\,\Omega$. Its maximum safe current is $I_1 = \frac{P_1}{V} = \frac

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$25\, W$ bulb

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(B) Resistance of $25\, W$ bulb is $R_1 = \frac{V^2}{P_1} = \frac{220^2}{25} = 1936\,\Omega$. Its maximum safe current is $I_1 = \frac{P_1}{V} = \frac{25}{220} \approx 0.1136\, A$. Resistance of $100\, W$ bulb is $R_2 = \frac{V^2}{P_2} = \frac{220^2}{100} = 484\,\Omega$. Its maximum safe current is $I_2 = \frac{P_2}{V} = \frac{100}{220} \approx 0.4545\, A$. When connected in series to a $440\, V$ supply,the total resistance is $R_{eq} = R_1 + R_2 = 1936 + 484 = 2420\,\Omega$. The current flowing through the series circuit is $I = \frac{V_{total}}{R_{eq}} = \frac{440}{2420} = \frac{2}{11} \approx 0.1818\, A$. Since the current $I \approx 0.1818\, A$ is greater than the safe current $I_1 \approx 0.1136\, A$ of the $25\, W$ bulb,the $25\, W$ bulb will fuse.

If two bulbs of wattage $25\, W$ and $100\, W$ respectively,each rated at $220\, V$,are connected in series with a supply of $440\, V$,then which bulb will fuse?

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