Two bulbs,one of 200 V, 60 W and the other | vedclass

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(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb $(P_1 = 60 \ W, V = 200 \ V)$:$R_1 = \frac{200^2}{60} = \frac{40000}{6

Vedclass · Accepted Answer

$37.5$

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(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb $(P_1 = 60 \ W, V = 200 \ V)$:$R_1 = \frac{200^2}{60} = \frac{40000}{60} = \frac{2000}{3} \ \Omega$.For the second bulb $(P_2 = 100 \ W, V = 200 \ V)$:$R_2 = \frac{200^2}{100} = \frac{40000}{100} = 400 \ \Omega$.When connected in series,the total resistance is $R_{eq} = R_1 + R_2 = \frac{2000}{3} + 400 = \frac{2000 + 1200}{3} = \frac{3200}{3} \ \Omega$.The power consumed by the series combination connected to a $200 \ V$ supply is:$P_{total} = \frac{V^2}{R_{eq}} = \frac{200^2}{3200/3} = \frac{40000 	imes 3}{3200} = \frac{400 	imes 3}{32} = \frac{1200}{32} = 37.5 \ W$.

Two bulbs,one of $200 \ V, 60 \ W$ and the other of $200 \ V, 100 \ W$,are connected in series to a $200 \ V$ supply. The power consumed will be ................ $W$.

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