Question
Download Solution PDFIn a 3-phase system, two-wattmeter method is used to measure the power. If one of the wattmeters shows a negative reading and the other shows a positive reading, and the magnitude of the readings are not the same, then what will be the power factor (p.f.) of the load?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In a two-wattmeter method,
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
\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
|
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 |
\(\frac{{{V_L}I_L}}{2}\) |
VLIL |
1.5 VL IL |
W1= 2W2 |
|
60° |
0.5 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W2 = 0 |
|
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W2 = -ve W1 = +ve |
Explanation:
W1 = -ve, W2 = +ve
|W1| ≠ |W2|, then the powere factor range is given as
The range of ϕ is, 60 < ϕ < 90
Now the range of power factor is, 0.5 > cos ϕ > 0.
Last updated on Jun 16, 2025
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