Question
Download Solution PDFTwo wattmeters are used to measure the power in a 3-phase balanced system. What is the power factor of the load when one wattmeter reads twice the other?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In a two-wattmeter method, for lagging load
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
Power factor = cos ϕ
Calculation:
Given that, W1 = 2 W2
⇒ ϕ = 30°
Power factor = cos 30° = 0.866
Important Point:
|
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 |
|
|
√3 VLIL |
W1 = W2 |
|
30° |
0.866 |
|
VLIL |
1.5 VL IL |
W2 = 2W1 |
|
60° |
0.5 |
0 |
|
|
W1 = 0 |
|
90° |
0 |
|
|
0 |
W1 = -ve W2 = +ve |
Last updated on Jun 23, 2025
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