Two bulbs of 100, W, 220, V and 200, W, 220, | vedclass

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(B) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb,$R_1 = \frac{220^2}{100} = 484\, \Omega$.For the second bulb,$R_2 = \f

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$66$

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(B) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb,$R_1 = \frac{220^2}{100} = 484\, \Omega$.For the second bulb,$R_2 = \frac{220^2}{200} = 242\, \Omega$.When connected in series,the total resistance is $R_{eq} = R_1 + R_2 = 484 + 242 = 726\, \Omega$.The total power consumed is $P_{total} = \frac{V^2}{R_{eq}} = \frac{220^2}{726} = \frac{48400}{726} \approx 66.67\, W$.Alternatively,for series connection,$P_{consumed} = \frac{P_1 P_2}{P_1 + P_2} = \frac{100 	imes 200}{100 + 200} = \frac{20000}{300} = 66.67\, W \approx 66\, W$.

Two bulbs of $100\, W, 220\, V$ and $200\, W, 220\, V$ are connected in series to a $220\, V$ supply. What is the total power consumed in $W$?

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