Two 220; V, 100; W bulbs are connected first | vedclass

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(B) First, calculate the resistance of each bulb using the formula $R = \frac{V^2}{P}$.$R = \frac{220 	imes 220}{100} = 484\; \Omega$.In series, the

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$50\; W, 200\; W$

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(B) First, calculate the resistance of each bulb using the formula $R = \frac{V^2}{P}$.$R = \frac{220 	imes 220}{100} = 484\; \Omega$.In series, the equivalent resistance is $R_{eq} = R + R = 484 + 484 = 968\; \Omega$.The power drawn is $P_{series} = \frac{V^2}{R_{eq}} = \frac{220 	imes 220}{968} = 50\; W$.In parallel, the equivalent resistance is $R_{eq} = \frac{R}{2} = \frac{484}{2} = 242\; \Omega$.The power drawn is $P_{parallel} = \frac{V^2}{R_{eq}} = \frac{220 	imes 220}{242} = 200\; W$.Thus, the power drawn in series and parallel combinations is $50\; W$ and $200\; W$ respectively.

Two $220\; V, 100\; W$ bulbs are connected first in series and then in parallel. Each time the combination is connected to a $220\; V\; AC$ supply line. The power drawn by the combination in each case respectively will be

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