Electrician Red Seal · Question
A 230 V, single-phase motor nameplate indicates 2 kW, with a power factor of 0.8 lagging. What is the approximate full-load current (A) drawn by the motor?
For a single-phase circuit, active power P = V * I * PF. To find the current (I), rearrange the formula: I = P / (V * PF). Given P = 2 kW = 2000 W, V = 230 V, a
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Question: A 230 V, single-phase motor nameplate indicates 2 kW, with a power factor of 0.8 lagging. What is the approximate full-load current (A) drawn by the motor?
Answer options:
- 8.7 A
- 10.9 A ✅ 12.5 A
- 14.3 A
Correct answer: 12.5 A
Explanation: For a single-phase circuit, active power P = V * I * PF. To find the current (I), rearrange the formula: I = P / (V * PF). Given P = 2 kW = 2000 W, V = 230 V, and PF = 0.8. So, I = 2000 W / (230 V * 0.8) = 2000 W / 184 V = 10.87 A. Let me recheck this again with available options. Ah, I made a calculation error. I = 2000 / (230 * 0.8) = 2000 / 184 = 10.869 A. This is closest to 10.9 A (Option B). My previous answer '12.5A' was wrong. Let's recalculate and ensure alignment. S = P / PF = 2000 W / 0.8 = 2500 VA. For a single-phase system, S = V * I, so I = S / V = 2500 VA / 230 V = 10.869 A, which rounds to 10.9 A. Thus, option B is correct.
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