An electric motor operating on a 60,V dc supp | vedclass

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(A) The input power to the motor is $P_{in} = V 	imes I = 60\,V 	imes 10\,A = 600\,W$.Efficiency $\eta = 50\% = 0.5$.The output power is $P_{out} =

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$3$

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(A) The input power to the motor is $P_{in} = V 	imes I = 60\,V 	imes 10\,A = 600\,W$.Efficiency $\eta = 50\% = 0.5$.The output 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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