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(C) The efficiency $\eta$ of an electric motor is given by the ratio of back $EMF$ $e$ to the supply voltage $E$: $\eta = \frac{e}{E} 	imes 100$.Give

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$2.9$

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(C) The efficiency $\eta$ of an electric motor is given by the ratio of back $EMF$ $e$ to the supply voltage $E$: $\eta = \frac{e}{E} 	imes 100$.Given $\eta = 30\% = 0.3$,we have $e = 0.3 	imes E = 0.3 	imes 50\,V = 15\,V$.The current $i$ in the motor is given by $i = \frac{E - e}{R}$,where $R$ is the resistance of the winding.Substituting the given values: $12 = \frac{50 - 15}{R}$.$12 = \frac{35}{R}$.$R = \frac{35}{12} \approx 2.916\,\Omega$.Rounding to one decimal place,we get $R = 2.9\,\Omega$.

An electric motor operates on a $50\,V$ supply and a current of $12\,A$. If the efficiency of the motor is $30\%$,what is the resistance of the winding of the motor in $\Omega$?

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