If two bulbs of wattage 25, W and 30, W,each | vedclass

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(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the $25\, W$ bulb,$R_1 = \frac{220^2}{25} = 1936\, \Omega$.For the $30\, W$ bulb,$R_2

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$25\, W$ bulb

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(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the $25\, W$ bulb,$R_1 = \frac{220^2}{25} = 1936\, \Omega$.For the $30\, W$ bulb,$R_2 = \frac{220^2}{30} = 1613.33\, \Omega$.When connected in series to a $440\, V$ supply,the current $I$ flowing through both is $I = \frac{V_{total}}{R_1 + R_2} = \frac{440}{1936 + 1613.33} \approx 0.124\, A$.The voltage drop across the $25\, W$ bulb is $V_1 = I 	imes R_1 = 0.124 	imes 1936 \approx 240\, V$.The voltage drop across the $30\, W$ bulb is $V_2 = I 	imes R_2 = 0.124 	imes 1613.33 \approx 200\, V$.Since the voltage across the $25\, W$ bulb $(240\, V)$ exceeds its rated voltage $(220\, V)$,the $25\, W$ bulb will fuse.

If two bulbs of wattage $25\, W$ and $30\, W$,each rated at $220\, V$,are connected in series with a $440\, V$ supply,which bulb will fuse?

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