A heater is rated 100 W, 200 V. It is cut int | vedclass

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(A) The power rating is $P = 100\ W$ and voltage is $V = 200\ V$.The resistance of the heater is $R = \frac{V^2}{P} = \frac{200^2}{100} = \frac{40000}

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$400$

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(A) The power rating is $P = 100\ W$ and voltage is $V = 200\ V$.The resistance of the heater is $R = \frac{V^2}{P} = \frac{200^2}{100} = \frac{40000}{100} = 400\ \Omega$.When the heater is cut into two equal parts,the resistance of each part becomes $R' = \frac{R}{2} = \frac{400}{2} = 200\ \Omega$.When these two parts are connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'} = \frac{2}{200} = \frac{1}{100}$.Thus,$R_{eq} = 100\ \Omega$.The energy released per second (power) in the new combination is $P' = \frac{V^2}{R_{eq}} = \frac{200^2}{100} = \frac{40000}{100} = 400\ W$.

$A$ heater is rated $100\ W, 200\ V$. It is cut into two equal parts. Both parts are connected in parallel to the same $200\ V$ source. The energy released per second in this new combination is ............. $W$.

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