An electric motor operating at a 100 V D.C. supply draws a current of 12 A. If the efficiency of the motor is 30%, calculate the resistance of its winding.

5.83 Ω

The correct answer is Option (4) → 5.83 Ω

The efficiency ($η$) of an electric motor is,

$η=\frac{Output\,Power}{Input\,Power}×100$

$P_{in}=V×I=100V×12A=1200W$

$∴P_{out}=η×P_{in}=\frac{30}{100}×1200=360W$

$∴P_{loss}=P_{in}-P_{out}$

$=1200W-360W=840W$

Also,

$P_{loss}=I^2R$

$⇒R=\frac{P_{loss}}{I^2}=\frac{840W}{(12A)^2}=5.83 Ω$