Question
Download Solution PDFThe reading of wattmeter and ammeter is 1 kW and 10 A respectively in the three phase circuit given below. What is the value of power factor of the circuit, if the circuit is balanced?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
3-phase power measurement: for a balanced load 3-phase power measurement one wattmeter required
|
3-phase Load |
Wattmeter method |
|
Balanced load |
1 wattmeter |
|
Balanced + unbalanced load |
2 wattmeter |
|
Unbalanced load |
3 wattmeter |
3-phase power measurement of a balanced load by 1 wattmeter method
P3ϕ = 3 Vph Iph cosϕ
Vph = phase voltage
Iph = phase current
Calculation:
As per the given circuit diagram, a balanced load wattmeter reading will be
Iph = 10 A, Vph = Vl /√ 3 = 200/ √3 = 115.47
\(W = {V_{ph}}{I_L}\cos ϕ \)
\(1000 = 115.47 \times 10 \times \cos ϕ \)
\( \Rightarrow \cos ϕ = 0.866\)
Power factor \(\cos ϕ = 0.866\)
Important Point:
Power is measured according to Blondel's theorem
|
Phase/conductor |
Wattmeter required |
|
n |
n-1 |
|
n with neutral wire |
n |
Last updated on Jun 16, 2025
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