Two-wattmeter method for three-phase power: identify the correct statement relating wattmeter readings and power factor in a 3-wire system (balanced or unbalanced loads).

Introduction / Context:
The two-wattmeter method is a standard technique for measuring total three-phase power in 3-wire systems. It works for both star and delta connections. Relationships between wattmeter readings and load power factor are well-known and provide diagnostic insight into system conditions.

Given Data / Assumptions:

• Three-phase, 3-wire system; method applies to star or delta loads.
• Wattmeter readings: W1 = V_L I_L cos(30° + φ), W2 = V_L I_L cos(30° − φ) for a balanced load with displacement angle φ.
• Total power P = W1 + W2.

Concept / Approach:
At unity power factor (φ = 0°), W1 = W2 = V_L I_L cos 30° (both positive and equal). At φ = 60° (power factor 0.5), one reading becomes zero: cos(30° + 60°) = cos 90° = 0. At φ = 90° (purely reactive; power factor zero), readings are equal in magnitude and opposite in sign because cos(120°) = −cos(60°) = −0.5 and cos(−60°) = 0.5. Hence the only correct statement among the choices is that equal and opposite readings indicate pf = 0.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

Common Pitfalls:

Final Answer: