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
|