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 \(ϕ = {\tan ^{  1}}\left( {\frac{{\sqrt 3 \left( {{W_1}  {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
W_{1} = reading of 1^{st} wattmeter
W_{2} = 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
W_{1} = 300 Watt and W_{2} = 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
