Two electric bulbs,rated at (25, W, 220, V) a | vedclass

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

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$P_1 = 16\, W, P_2 = 4\, W$

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(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb $(25\, W, 220\, V)$: $R_1 = \frac{(220)^2}{25} = 1936\, \Omega$.For the second bulb $(100\, W, 220\, V)$: $R_2 = \frac{(220)^2}{100} = 484\, \Omega$.Since the bulbs are connected in series,the current $i$ flowing through them is $i = \frac{V}{R_1 + R_2} = \frac{220}{1936 + 484} = \frac{220}{2420} = \frac{1}{11}\, A$.The power dissipated by the first bulb is $P_1 = i^2 R_1 = \left(\frac{1}{11}ight)^2 	imes 1936 = \frac{1936}{121} = 16\, W$.The power dissipated by the second bulb is $P_2 = i^2 R_2 = \left(\frac{1}{11}ight)^2 	imes 484 = \frac{484}{121} = 4\, W$.

Two electric bulbs,rated at $(25\, W, 220\, V)$ and $(100\, W, 220\, V)$,are connected in series across a $220\, V$ voltage source. If the $25\, W$ and $100\, W$ bulbs draw powers $P_1$ and $P_2$ respectively,then

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