A wire when connected to 220,V mains supply h | vedclass

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(B) Let the resistance of the original wire be $R$. The power dissipated is given by $P_1 = \frac{V^2}{R}$.When the wire is cut into two equal pieces,

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$4$

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(B) Let the resistance of the original wire be $R$. The power dissipated is given by $P_1 = \frac{V^2}{R}$.When the wire is cut into two equal pieces,the resistance of each piece becomes $R' = \frac{R}{2}$.When these two pieces are connected in parallel to the same supply $V$,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}$.Thus,$R_{eq} = \frac{R}{4}$.The power dissipated in the parallel combination is $P_2 = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = 4 \left( \frac{V^2}{R} \right) = 4P_1$.Therefore,the ratio $\frac{P_2}{P_1} = 4$.

$A$ wire when connected to $220\,V$ mains supply has power dissipation $P_1$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_2$. Then $P_2:P_1$ is

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