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(D) The power $P$ of an electric motor is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the motor.First,we calcul

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(D) The power $P$ of an electric motor is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the motor.First,we calculate the resistance $R$ using the initial conditions: $P_1 = 242 \ W$ and $V_1 = 220 \ V$.$R = \frac{V_1^2}{P_1} = \frac{220 	imes 220}{242} = \frac{48400}{242} = 200 \ \Omega$.Now,when the motor is operated at $V_2 = 200 \ V$,the current $I$ drawn by the motor is given by Ohm's law: $I = \frac{V_2}{R}$.$I = \frac{200 \ V}{200 \ \Omega} = 1 \ A$.Thus,the current drawn is $1 \ A$.

The power of an electric motor is $242 \ W$ when connected to a $220 \ V$ supply. When the motor is operated at $200 \ V$,the current drawn by it is (in $A$)

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