Question

A 200 V, 50 Hz, 1-phase energy meter makes 960 revolutions to supply a load of 0.8 pf for 2 hrs. If motor constant is 160 rev/kWh, then load current is:

• A. 1.875 A
• B. 15 A
• C. 18.75 A
• D. 37.5 A

Hint:

Use energy meter formula: Energy = \( \frac{\text{Revolutions}}{\text{rev/kWh}} \). Then apply: \( P = VI\cos\phi \).

Answer 1

The Correct Option is C

Energy recorded = \( \frac{960}{160} = 6 \text{ kWh} \) Time = 2 hrs → Power = \( \frac{6}{2} = 3 \text{ kW} \) Using \( P = VI\cos\phi \Rightarrow I = \frac{3000}{200 \times 0.8} = 18.75 \, \text{A} \)