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In three-phase power measurement using the two-wattmeter method, one wattmeter reads zero. What is the load power factor?

Difficulty: Easy

Correct Answer: 0.5

Explanation:


Introduction / Context:
The two-wattmeter method measures total power in a 3-wire, three-phase system. For a balanced load with displacement angle φ, the wattmeter readings depend on φ, providing a convenient way to infer power factor from observed readings.


Given Data / Assumptions:

  • Balanced three-phase system.
  • Wattmeter readings: W1 = V_L I_L cos(30° + φ), W2 = V_L I_L cos(30° − φ).
  • Total power P = W1 + W2.


Concept / Approach:
When one wattmeter reads zero, its cosine term is zero. Setting cos(30° ± φ) = 0 yields the critical φ for which a reading vanishes. That angle translates directly to a power factor value cos φ.


Step-by-Step Solution:

Let W1 = 0 ⇒ cos(30° + φ) = 0 ⇒ 30° + φ = 90° ⇒ φ = 60°Thus pf = cos φ = cos 60° = 0.5


Verification / Alternative check:

At φ = 60°, the other wattmeter reads V_L I_L cos(30° − 60°) = V_L I_L cos(−30°) = V_L I_L (√3/2), nonzero, as expected.


Why Other Options Are Wrong:

pf = 1.0 gives equal positive readings; pf = 0 gives equal and opposite readings; pf = 0.8 does not null either meter.


Common Pitfalls:

Mixing up the ”equal and opposite” condition (pf = 0) with the ”one meter zero” condition (pf = 0.5).


Final Answer:

0.5
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