Two equal resistances when connected in serie | vedclass

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Question

(B) Let the value of each resistance be $R$ and the voltage of the battery be $V$. In series combination,the equivalent resistance is $R_s = R + R = 2

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$240$

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(B) Let the value of each resistance be $R$ and the voltage of the battery be $V$. In series combination,the equivalent resistance is $R_s = R + R = 2R$. The power consumed in series is $P_s = \frac{V^2}{R_s} = \frac{V^2}{2R} = 60 \ W$. From this,we get $\frac{V^2}{R} = 120 \ W$. In parallel combination,the equivalent resistance is $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$. The power consumed in parallel is $P_p = \frac{V^2}{R_p} = \frac{V^2}{R/2} = 2 	imes \frac{V^2}{R}$. Substituting the value of $\frac{V^2}{R}$,we get $P_p = 2 	imes 120 = 240 \ W$.

Two equal resistances when connected in series to a battery,consume electric power of $60 \ W$. If these resistances are now connected in parallel combination to the same battery,the electric power consumed will be............. $W$

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