# In 3-phase power measurement by two-wattmeter method, the two wattmeters read as W1 = 300 W and W2 = 300 W. Then the load is said to be operating at _

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In 3-phase power measurement by two-wattmeter method, the two wattmeters read as W1 = 300 W and W2 = 300 W. Then the load is said to be operating at _________.

1. Unity power factor
2. Lagging power factor
3. Zero power factor

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Correct Answer - Option 1 : Unity power factor

Concept:

In two wattmeter method, the power factor is given by

$cosϕ = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]$

Here $ϕ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)$

W1 = reading of 1st wattmeter

W2 = reading of 2nd wattmeter

When both the wattmeter’s show the same positive reading, ϕ becomes zero and hence power factor is unity.

Calculation:

Given that

W1 = 300 Watt and W2 = 300 Watt

Now power factor

$\cosϕ = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]$

cos ϕ = 1 (Unity power factor)

 p.f. angle (ϕ) p.f.(cos ϕ) W1 [VLIL cos (30 + ϕ)] W2 [VLIL cos (30 - ϕ)] W = W1 + W2 [W = √3VLIL cos ϕ] Observations 0° 1 $\frac{{\sqrt 3 }}{2}{V_L}{I_L}$ $\frac{{\sqrt 3 }}{2}{V_L}{I_L}$ √3 VLIL W1 = W2 30° 0.866 $\frac{{{V_L}I}}{2}$ VLIL 1.5 VL IL W2 = 2W1 60° 0.5 0 $\frac{{\sqrt 3 }}{2}{V_L}{I_L}$ $\frac{{\sqrt 3 }}{2}{V_L}{I_L}$ W1 = 0 90° 0 $\frac{{ - {V_L}{I_L}}}{2}$ $\frac{{{V_L}{I_L}}}{2}$ 0 W1 = -ve W2 = +ve