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(A) Let the resistance of each resistor be $R$ and the battery voltage be $V$.In a series connection,the equivalent resistance is $R_s = 4R$.The power

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$320$

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(A) Let the resistance of each resistor be $R$ and the battery voltage be $V$.In a series connection,the equivalent resistance is $R_s = 4R$.The power dissipated in series is $P_s = \frac{V^2}{R_s} = \frac{V^2}{4R} = 20 \ W$.From this,we find $\frac{V^2}{R} = 20 	imes 4 = 80 \ W$.In a parallel connection,the equivalent resistance is $R_p = \frac{R}{4}$.The power dissipated in parallel is $P_p = \frac{V^2}{R_p} = \frac{V^2}{R/4} = 4 	imes \frac{V^2}{R}$.Substituting the value of $\frac{V^2}{R}$,we get $P_p = 4 	imes 80 \ W = 320 \ W$.

Four identical resistors are connected in series with a battery,dissipating $20 \ W$ of power. What will be the power dissipated if the same resistors are connected in parallel to the same battery (in $W$)?

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