Two equal resistances when connected in serie | vedclass

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

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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 $V$.When connected in series,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 $\frac{V^2}{R} = 120 \, W$.When connected in parallel,the equivalent resistance is $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$.The power consumed in parallel is $P_p = \frac{V^2}{R_p} = \frac{V^2}{R/2} = 2 \left( \frac{V^2}{R} ight)$.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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