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A 3 phase 20kVA load has power factor 0.866. If power is measured by two wattmeter method then find reading of each wattmeter when power factor is lagging –
- a)20/√3 W and 10/√3 W
- b)O W and 20/√3 W
- c)20/√3 W and OW
- d)10 W and 10 W
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A 3 phase 20kVA load has power factor 0.866. If power is measured by t...
P = √3 VlIL
VlIL= 20/√3
Pf is lagging,
So, P1 = VLILcos(30-ϕ)
= 20/√3×cos(30-30) = 20/√3
P1 = VLILcos(30-ϕ)
= 20/√3×cos(30-30)
= 10/√3
VlIL= 20/√3
Pf is lagging,
So, P1 = VLILcos(30-ϕ)
= 20/√3×cos(30-30) = 20/√3
P1 = VLILcos(30-ϕ)
= 20/√3×cos(30-30)
= 10/√3
Most Upvoted Answer
A 3 phase 20kVA load has power factor 0.866. If power is measured by t...
When the power factor is lagging, the power factor angle is positive. The power factor angle (θ) can be calculated using the power factor (pf) as follows:
pf = cos(θ)
Given that the power factor (pf) is 0.866, we can calculate the power factor angle (θ) as:
θ = arccos(0.866) ≈ 30.96 degrees
In a 3-phase load, the power is given by:
P = √3 * V * I * pf
Where P is the power in watts, V is the voltage in volts, I is the current in amps, and pf is the power factor.
We need to find the readings of each wattmeter, which can be calculated using the following equations:
Wattmeter 1 (W1) = P * cos(θ)
Wattmeter 2 (W2) = P * cos(θ + 120)
Since the power is given as 20 kVA, we can convert it to watts:
P = 20,000 watts
Now we can calculate the readings of each wattmeter:
W1 = 20,000 * cos(30.96) ≈ 17,321 watts
W2 = 20,000 * cos(30.96 + 120) ≈ -8,660.5 watts
Therefore, the reading of each wattmeter when the power factor is lagging is approximately:
Wattmeter 1: 17,321 watts
Wattmeter 2: -8,660.5 watts
pf = cos(θ)
Given that the power factor (pf) is 0.866, we can calculate the power factor angle (θ) as:
θ = arccos(0.866) ≈ 30.96 degrees
In a 3-phase load, the power is given by:
P = √3 * V * I * pf
Where P is the power in watts, V is the voltage in volts, I is the current in amps, and pf is the power factor.
We need to find the readings of each wattmeter, which can be calculated using the following equations:
Wattmeter 1 (W1) = P * cos(θ)
Wattmeter 2 (W2) = P * cos(θ + 120)
Since the power is given as 20 kVA, we can convert it to watts:
P = 20,000 watts
Now we can calculate the readings of each wattmeter:
W1 = 20,000 * cos(30.96) ≈ 17,321 watts
W2 = 20,000 * cos(30.96 + 120) ≈ -8,660.5 watts
Therefore, the reading of each wattmeter when the power factor is lagging is approximately:
Wattmeter 1: 17,321 watts
Wattmeter 2: -8,660.5 watts
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A 3 phase 20kVA load has power factor 0.866. If power is measured by two wattmeter method then find reading of each wattmeter when power factor is lagging –a)20/√3 W and 10/√3 Wb)O W and20/√3 Wc)20/√3 W and OWd)10 W and 10 WCorrect answer is option 'A'. Can you explain this answer?
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