In a two-wattmeter method of measuring power in a balance 3-phase circuit, the ratio of the two wattmeter reading is 1 : 2. The circuit power factor is

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ESE Electrical 2015 Paper 1: Official Paper
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  1. 0.707
  2. 0.5
  3. 0.866
  4. indeterminate

Answer (Detailed Solution Below)

Option 3 : 0.866
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Detailed Solution

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Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Calculation:

Given that, W2 = 2W1

\(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - 2{W_1}} \right)}}{{\left( {{W_1} + 2{W_2}} \right)}}} \right) = - 30^\circ \)

Power factor = cos ϕ = 0.866

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