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(A) The input power to the motor is $P_{in} = V 	imes I = 60\, V 	imes 10\, A = 600\, W$.Given the efficiency $\eta = 50\%$,the output mechanical po

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(A) The input power to the motor is $P_{in} = V 	imes I = 60\, V 	imes 10\, A = 600\, W$.Given the efficiency $\eta = 50\%$,the output mechanical power is $P_{out} = \eta 	imes P_{in} = 0.5 	imes 600\, W = 300\, W$.The power lost as heat in the winding resistance $R$ is $P_{loss} = P_{in} - P_{out} = 600\, W - 300\, W = 300\, W$.Since $P_{loss} = I^2 R$,we have $300 = (10)^2 	imes R$.$300 = 100 	imes R$.Therefore,$R = \frac{300}{100} = 3\, \Omega $.

An electric motor operating on a $60\, V$ $dc$ supply draws a current of $10\, A$. If the efficiency of the motor is $50\%$,the resistance of its winding is.....$\Omega $

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