Question

 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
4. Leading power factor

Answer 1

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 1^st wattmeter

W₂ = reading of 2^nd 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