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(D) First,calculate the resistance of the heater:$R_{	ext{heater}} = \frac{V^2}{P} = \frac{(100)^2}{1000} = 10 \ \Omega$.When the heater operates at

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$5$

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(D) First,calculate the resistance of the heater:$R_{	ext{heater}} = \frac{V^2}{P} = \frac{(100)^2}{1000} = 10 \ \Omega$.When the heater operates at $P' = 62.5 \ W$,the voltage across it $(V')$ is:$P' = \frac{(V')^2}{R_{	ext{heater}}} \Rightarrow V' = \sqrt{P' \cdot R_{	ext{heater}}} = \sqrt{62.5 	imes 10} = \sqrt{625} = 25 \ V$.The voltage across the series resistor $(10 \ \Omega)$ is $V_s = 100 \ V - 25 \ V = 75 \ V$.The total current in the circuit is $I = \frac{V_s}{10 \ \Omega} = \frac{75}{10} = 7.5 \ A$.The current through the heater is $I_H = \frac{V'}{R_{	ext{heater}}} = \frac{25}{10} = 2.5 \ A$.The current through the resistor $R$ is $I_R = I - I_H = 7.5 \ A - 2.5 \ A = 5 \ A$.Since $R$ is in parallel with the heater,the voltage across $R$ is also $25 \ V$.Therefore,$R = \frac{V'}{I_R} = \frac{25}{5} = 5 \ \Omega$.

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