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While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are 100W and 250 W. The power factor of the load is ______________.
Correct answer is '0.802'. Can you explain this answer?
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While measuring power of a three-phase balanced load by the two-wattme...
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While measuring power of a three-phase balanced load by the two-wattme...
Given:
- Readings of the two-wattmeter method: 100W and 250W
To Find:
- Power factor of the load
Solution:
The two-wattmeter method is used to measure the power of a three-phase balanced load. In this method, two wattmeters are connected to measure the power in each phase, and the sum of the readings gives the total power.
Step 1: Calculating the Total Power
The total power (P) can be calculated by adding the readings of the two wattmeters.
P = P1 + P2
P = 100W + 250W
P = 350W
Step 2: Calculating the Reactive Power
The reactive power (Q) is given by:
Q = √(P^2 - S^2)
where P is the total power and S is the apparent power.
In a balanced load, the apparent power is given by:
S = √(P^2 + Q^2)
Substituting the values:
S = √(350^2 + Q^2)
Step 3: Calculating the Power Factor
The power factor (PF) is given by the ratio of the total power to the apparent power.
PF = P/S
PF = 350W / √(350^2 + Q^2)
Step 4: Solving for the Power Factor
To find the value of the power factor, we need to determine the value of Q.
Since the load is balanced, the reactive power in each phase is equal. Therefore, Q can be calculated as:
Q = √3 * |P1 - P2|
Q = √3 * |100W - 250W|
Q = √3 * |-150W|
Q = √3 * 150W
Q = 150√3 W
Substituting the value of Q:
PF = 350W / √(350^2 + (150√3)^2)
PF = 350W / √(350^2 + 150^2 * 3)
PF = 350W / √(122500 + 67500)
PF = 350W / √190000
PF = 350W / 435.89
PF ≈ 0.802
Therefore, the power factor of the load is approximately 0.802.
- Readings of the two-wattmeter method: 100W and 250W
To Find:
- Power factor of the load
Solution:
The two-wattmeter method is used to measure the power of a three-phase balanced load. In this method, two wattmeters are connected to measure the power in each phase, and the sum of the readings gives the total power.
Step 1: Calculating the Total Power
The total power (P) can be calculated by adding the readings of the two wattmeters.
P = P1 + P2
P = 100W + 250W
P = 350W
Step 2: Calculating the Reactive Power
The reactive power (Q) is given by:
Q = √(P^2 - S^2)
where P is the total power and S is the apparent power.
In a balanced load, the apparent power is given by:
S = √(P^2 + Q^2)
Substituting the values:
S = √(350^2 + Q^2)
Step 3: Calculating the Power Factor
The power factor (PF) is given by the ratio of the total power to the apparent power.
PF = P/S
PF = 350W / √(350^2 + Q^2)
Step 4: Solving for the Power Factor
To find the value of the power factor, we need to determine the value of Q.
Since the load is balanced, the reactive power in each phase is equal. Therefore, Q can be calculated as:
Q = √3 * |P1 - P2|
Q = √3 * |100W - 250W|
Q = √3 * |-150W|
Q = √3 * 150W
Q = 150√3 W
Substituting the value of Q:
PF = 350W / √(350^2 + (150√3)^2)
PF = 350W / √(350^2 + 150^2 * 3)
PF = 350W / √(122500 + 67500)
PF = 350W / √190000
PF = 350W / 435.89
PF ≈ 0.802
Therefore, the power factor of the load is approximately 0.802.
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While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are 100W and 250 W. The power factor of the load is ______________.Correct answer is '0.802'. Can you explain this answer?
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