Question

Power consumed by a three-phase balanced load is measured using two-wattmeter method. The per-phase average power drawn by the load is 30 kW at \( \frac{\sqrt{3}}{2} \) lagging power factor. The readings of the wattmeters will be

• A. 15 kW and 15 kW
• B. 22.5 kW and 7.5 kW
• C. 60 kW and 30 kW
• D. 45 kW and 45 kW

Hint:

In the two-wattmeter method, for a balanced 3-phase system: - \( W_1 + W_2 = {Total power} \) - \( \phi = \cos^{-1}({Power factor}) \) - Use angle addition/subtraction formulas for phase angles.

Answer 1

The Correct Option is C

1. Total 3-phase power:
\[ P_{total} = 3 \times 30\,{kW} = 90\,{kW} \]
2. Using two-wattmeter method:
\[ W_1 + W_2 = 90\,{kW} \]
\[ W_1 - W_2 = \sqrt{3}V_LI_L\sin\phi \]
3. Calculate \(V_LI_L\):
\[ \sqrt{3}V_LI_L\cos\phi = 90\,{kW} \]
\[ V_LI_L = \frac{90}{\sqrt{3}\times\frac{\sqrt{3}}{2}} = 60\,{kVA} \]
4. Wattmeter readings:
\[ W_1 = V_LI_L\cos(30^\circ - \phi) = 60\cos(0^\circ) = 60\,{kW} \]
\[ W_2 = V_LI_L\cos(30^\circ + \phi) = 60\cos(60^\circ) = 30\,{kW} \]
Verification:
\[ 60\,{kW} + 30\,{kW} = 90\,{kW} \quad {(matches total power)} \]