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.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5.83\,\Omega \)
The motor takes electrical power \( VI \) from the supply. The efficiency is the fraction of it turned into useful mechanical power; the rest is taken as the Joule heating \( I^{2}R \) of the winding.
\( P_{\mathrm{in}} = VI = 100 \times 12 = 1200\,\mathrm{W} \)
\( P_{\mathrm{out}} = 30\%\ \text{of}\ 1200 = 360\,\mathrm{W} \)
\( P_{\mathrm{loss}} = 1200 - 360 = 840\,\mathrm{W} \)
The power lost is the heat produced in the winding:
\( \displaystyle I^{2}R = 840\,\mathrm{W} \)
\( \displaystyle R = \frac{840}{12^{2}} = \frac{840}{144} \approx 5.83\,\Omega \)