An electrical bulb rated 220,V, 100,W is conn | vedclass

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(A) First,calculate the resistance of each bulb using the formula $R = \frac{V^2}{P}$.For the $100\,W$ bulb: $R_1 = \frac{220^2}{100} = 484\,\Omega$.F

Vedclass · Accepted Answer

$14$

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(A) First,calculate the resistance of each bulb using the formula $R = \frac{V^2}{P}$.For the $100\,W$ bulb: $R_1 = \frac{220^2}{100} = 484\,\Omega$.For the $60\,W$ bulb: $R_2 = \frac{220^2}{60} = \frac{4840}{6} = 806.67\,\Omega$.When connected in series,the total resistance $R_{eq} = R_1 + R_2 = 484 + 806.67 = 1290.67\,\Omega$.The current $I$ flowing through the series combination is $I = \frac{V_{total}}{R_{eq}} = \frac{220}{1290.67} \approx 0.17045\,A$.The power consumed by the $100\,W$ bulb is $P_1 = I^2 R_1 = (0.17045)^2 	imes 484 \approx 0.02905 	imes 484 \approx 14.06\,W$.Thus,the power consumed is approximately $14\,W$.

An electrical bulb rated $220\,V, 100\,W$ is connected in series with another bulb rated $220\,V, 60\,W$. If the voltage across the combination is $220\,V$,the power consumed by the $100\,W$ bulb will be about $........... W$.

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