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(B) For series connection:The equivalent resistance of each bulb is $R = \frac{V^2}{P} = \frac{250^2}{100} = 625 \ \Omega$.When two identical bulbs ar

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$50 \ W, 200 \ W$

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(B) For series connection:The equivalent resistance of each bulb is $R = \frac{V^2}{P} = \frac{250^2}{100} = 625 \ \Omega$.When two identical bulbs are connected in series,the total resistance $R_s = R + R = 2R = 1250 \ \Omega$.The total power consumed in series is $P_s = \frac{V^2}{R_s} = \frac{250^2}{1250} = \frac{62500}{1250} = 50 \ W$.Alternatively,$P_s = \frac{P_1 P_2}{P_1 + P_2} = \frac{100 	imes 100}{100 + 100} = 50 \ W$.For parallel connection:When two identical bulbs are connected in parallel,the total resistance $R_p = \frac{R}{2} = \frac{625}{2} = 312.5 \ \Omega$.The total power consumed in parallel is $P_p = \frac{V^2}{R_p} = \frac{250^2}{312.5} = \frac{62500}{312.5} = 200 \ W$.Alternatively,$P_p = P_1 + P_2 = 100 + 100 = 200 \ W$.Thus,the total power in series and parallel cases is $50 \ W$ and $200 \ W$ respectively.

Two bulbs of $250 \ V$ and $100 \ W$ are first connected in series and then in parallel with a supply of $250 \ V$. Total power in each of the cases will be . . . . . . respectively.

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