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In a three phase system power measured by using two wattmeter method and total power measured by both wattmeter is 50kW and power factor is 0.5 lagging. Reading of second wattmeter is –
- a)25 kW
- b)0 kW
- c)16.67 kW
- d)33.33 kW
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
In a three phase system power measured by using two wattmeter method a...
Total power,
P = 50kW
ϕ = 60 °
since at ϕ = 60 ° one watt meter reads 0 watts and entire power is reading by another watt meter
so at this condition the reading of another watt meter is zero.
P = 50kW
ϕ = 60 °
since at ϕ = 60 ° one watt meter reads 0 watts and entire power is reading by another watt meter
so at this condition the reading of another watt meter is zero.
Most Upvoted Answer
In a three phase system power measured by using two wattmeter method a...
To find the reading of the second wattmeter, we can use the formula:
Total power = √3 * V * I * power factor
Given that the total power is 50 kW and the power factor is 0.5 lagging, we can rearrange the formula to solve for the product of V * I:
V * I = Total power / (√3 * power factor)
V * I = 50 kW / (√3 * 0.5)
V * I = 50 kW / (1.732 * 0.5)
V * I = 50 kW / 0.866
V * I ≈ 57.74 kVA
Since the power factor is lagging, the power triangle will look like this:
|\
| \
Reactive Power | \ Apparent Power
| \
|____\
The true power (or real power) is the horizontal component of the power triangle, which is the power measured by the two wattmeters.
Given that the power factor is lagging (0.5 lagging), the angle between the apparent power and the true power is 60 degrees.
Therefore, the reading of the second wattmeter can be found by multiplying the apparent power (V * I) by the sine of the angle between the true power and the apparent power:
Reading of second wattmeter = (V * I) * sin(60 degrees)
Reading of second wattmeter = (57.74 kVA) * sin(60 degrees)
Reading of second wattmeter ≈ 49.99 kVA
Therefore, the reading of the second wattmeter is approximately 49.99 kVA.
Total power = √3 * V * I * power factor
Given that the total power is 50 kW and the power factor is 0.5 lagging, we can rearrange the formula to solve for the product of V * I:
V * I = Total power / (√3 * power factor)
V * I = 50 kW / (√3 * 0.5)
V * I = 50 kW / (1.732 * 0.5)
V * I = 50 kW / 0.866
V * I ≈ 57.74 kVA
Since the power factor is lagging, the power triangle will look like this:
|\
| \
Reactive Power | \ Apparent Power
| \
|____\
The true power (or real power) is the horizontal component of the power triangle, which is the power measured by the two wattmeters.
Given that the power factor is lagging (0.5 lagging), the angle between the apparent power and the true power is 60 degrees.
Therefore, the reading of the second wattmeter can be found by multiplying the apparent power (V * I) by the sine of the angle between the true power and the apparent power:
Reading of second wattmeter = (V * I) * sin(60 degrees)
Reading of second wattmeter = (57.74 kVA) * sin(60 degrees)
Reading of second wattmeter ≈ 49.99 kVA
Therefore, the reading of the second wattmeter is approximately 49.99 kVA.
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In a three phase system power measured by using two wattmeter method and total power measured by both wattmeter is 50kW and power factor is 0.5 lagging. Reading of second wattmeter is –a)25 kWb)0 kWc)16.67 kWd)33.33 kWCorrect answer is option 'B'. Can you explain this answer?
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