(B) Let each resistance be $R$ and the battery voltage be $V$. In series,the total resistance is $R_{eq} = 4R$. The power dissipated is $P_s = \frac{V

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(B) Let each resistance be $R$ and the battery voltage be $V$. In series,the total resistance is $R_{eq} = 4R$. The power dissipated is $P_s = \frac{V^2}{R_{eq}} = \frac{V^2}{4R} = 5 \, W$. From this,we get $\frac{V^2}{R} = 20 \, W$. In parallel,the total resistance is $R_{eq}' = \frac{R}{4}$. The total power dissipated is $P_p = \frac{V^2}{R_{eq}'} = \frac{V^2}{R/4} = 4 \left( \frac{V^2}{R} \right)$. Substituting the value,$P_p = 4 \times 20 = 80 \, W$.

Class 12 Physics Current Electricity Mix Examples-Current Electricity

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Four equal resistances dissipate $5 \, W$ of power together when connected in series to a battery of negligible internal resistance. The total power dissipated in these resistances when connected in parallel across the same battery would be ........... $W$.

A
$125$
B
$80$
C
$20$
D
$5$

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Four equal resistances dissipate $5 \, W$ of power together when connected in series to a battery of negligible internal resistance. The total power dissipated in these resistances when connected in parallel across the same battery would be ........... $W$.

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