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(B) The heat energy produced is given by $H = \frac{V^2}{R} t$. Since the heat produced $H$ and the voltage $V$ are the same for both cases,we have $H

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$15$

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(B) The heat energy produced is given by $H = \frac{V^2}{R} t$. Since the heat produced $H$ and the voltage $V$ are the same for both cases,we have $H = \frac{V^2}{R_1} t_1 = \frac{V^2}{R_2} t_2$.Given $t_1 = 20 \ min$ and $t_2 = 60 \ min$,we get $\frac{20}{R_1} = \frac{60}{R_2}$,which implies $R_2 = 3R_1$.When connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_1} + \frac{1}{3R_1} = \frac{4}{3R_1}$.Thus,$R_{eq} = \frac{3R_1}{4}$.For the same heat $H$ to be produced by the parallel combination in time $t$,we have $H = \frac{V^2}{R_{eq}} t$.Equating this to the heat produced by the first coil: $\frac{V^2}{R_1} 	imes 20 = \frac{V^2}{(3R_1/4)} 	imes t$.$20 = \frac{4}{3} t \Rightarrow t = \frac{20 	imes 3}{4} = 15 \ min$.

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