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Electronics Engineering QUESTION #11727
Question 1
A contactor rated 400 V AC, 50 A is used to switch a 400 V, 30 kW motor. The full-load current is approximately:
  • $43\,\text{A}$
  • $75\,\text{A}$
  • $43\,\text{A}$ at $\cos\phi = 1$
  • $54\,\text{A}$ at $\cos\phi = 0.85$, $\eta = 0.9$✔️
Correct Answer Explanation
$I_{FL} = \dfrac{P_{out}}{\sqrt{3} \times V_L \times \cos\phi \times \eta} = \dfrac{30000}{\sqrt{3} \times 400 \times 0.85 \times 0.9} = \dfrac{30000}{529} \approx \mathbf{56.7\,\text{A}}$. The contactor must be rated above this value. Approximately 54 A at typical efficiency/PF is the closest correct calculation here.