A heater coil connected to a supply of 220 V | vedclass

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(C) The power dissipated by a resistor is given by $P = \frac{V^2}{R}$.Let the initial resistance of the coil be $R$. Then $P_1 = \frac{V^2}{R}$.When

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$1:4$

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(C) The power dissipated by a resistor is given by $P = \frac{V^2}{R}$.Let the initial resistance of the coil be $R$. Then $P_1 = \frac{V^2}{R}$.When the coil is cut into two equal halves,the resistance of each half becomes $\frac{R}{2}$.When these two halves are connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R/2} + \frac{1}{R/2} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}$,so $R_{eq} = \frac{R}{4}$.The new power dissipated is $P_2 = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = \frac{4V^2}{R}$.Therefore,the ratio $\frac{P_1}{P_2} = \frac{V^2/R}{4V^2/R} = \frac{1}{4}$.

$A$ heater coil connected to a supply of $220 \ V$ is dissipating some power $P_1$. The coil is cut into half and the two halves are connected in parallel. The heater now dissipates a power $P_2$. The ratio of power $P_1 : P_2$ is

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