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The two-wattmeter method is sued to measure power in 3-phase, 3-wire balanced inductive circuit. The line voltage and line current are 400 V and 10 A respectively. If the load power factor is 0.866 lagging, then readings of the two wattmeter’s are
- a)6000 W and 0 W
- b)5000 W and 1000 W
- c)4500 W and 1500 W
- d)4000 W and 2000 W
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The two-wattmeter method is sued to measure power in 3-phase, 3-wire b...
Given that, line voltage (VL) = 400 V
Line current (IL) = 10 A
Cos ϕ = 0.866
⇒ ϕ = cos-1(0.866) = 30°
W1 = VLIL cos (30 – ϕ)
= 400 × 10 × cos (30 – 10) = 4000 W
W2 = VLIL cos (30 + ϕ)
= 400 × 10 × cos (30 + 30) = 2000 W
Most Upvoted Answer
The two-wattmeter method is sued to measure power in 3-phase, 3-wire b...
To find the readings of the two wattmeters, we first need to calculate the total power in the circuit.
Total Power = √3 * Line Voltage * Line Current * Power Factor
Total Power = √3 * 400 V * 10 A * 0.866
Total Power = 6928.2 W
Since the load power factor is lagging, the two wattmeters will give different readings. Let's assume the readings of the two wattmeters are W1 and W2.
The power measured by the first wattmeter (W1) is given by:
W1 = √3 * Line Voltage * Line Current * cos(θ)
W1 = √3 * 400 V * 10 A * cos(arccos(0.866))
W1 = 6928.2 W * 0.866
W1 = 6000 W
The power measured by the second wattmeter (W2) is given by:
W2 = √3 * Line Voltage * Line Current * cos(θ + 120°)
W2 = √3 * 400 V * 10 A * cos(arccos(0.866) + 120°)
W2 = 6928.2 W * 0.866
W2 = 6000 W
So, the readings of the two wattmeters (W1 and W2) will be 6000 W each.
Total Power = √3 * Line Voltage * Line Current * Power Factor
Total Power = √3 * 400 V * 10 A * 0.866
Total Power = 6928.2 W
Since the load power factor is lagging, the two wattmeters will give different readings. Let's assume the readings of the two wattmeters are W1 and W2.
The power measured by the first wattmeter (W1) is given by:
W1 = √3 * Line Voltage * Line Current * cos(θ)
W1 = √3 * 400 V * 10 A * cos(arccos(0.866))
W1 = 6928.2 W * 0.866
W1 = 6000 W
The power measured by the second wattmeter (W2) is given by:
W2 = √3 * Line Voltage * Line Current * cos(θ + 120°)
W2 = √3 * 400 V * 10 A * cos(arccos(0.866) + 120°)
W2 = 6928.2 W * 0.866
W2 = 6000 W
So, the readings of the two wattmeters (W1 and W2) will be 6000 W each.
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The two-wattmeter method is sued to measure power in 3-phase, 3-wire balanced inductive circuit. The line voltage and line current are 400 V and 10 A respectively. If the load power factor is 0.866 lagging, then readings of the two wattmeter’s area)6000 W and 0 Wb)5000 W and 1000 Wc)4500 W and 1500 Wd)4000 W and 2000 WCorrect answer is option 'D'. Can you explain this answer?
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