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Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1 and P2 of the two wattmeter’s are
- a)P1 = 3 kW, P2 = 6 kW
- b)P1 = 0 kW, P2 = 6 kW
- c)P1 = 2 kW, P2 = 6 kW
- d)P1 = 0 kW, P2 = 3 kW
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Power in a three-phase star connected balanced inductive load is measu...
Given that,
Phase voltage (Vph) = 200 V
Phase current (Iph) = 10 A
Power factor = cos ϕ = 0.5
⇒ ϕ = cos-1 (0.5) = 600
In two wattmeter method,
W1 = VLIL cos (30 + ϕ)
And
W2 = VLIL cos (30 – ϕ)
Phase voltage (Vph) = 200 V
Phase current (Iph) = 10 A
Power factor = cos ϕ = 0.5
⇒ ϕ = cos-1 (0.5) = 600
In two wattmeter method,
W1 = VLIL cos (30 + ϕ)
And
W2 = VLIL cos (30 – ϕ)
Most Upvoted Answer
Power in a three-phase star connected balanced inductive load is measu...
The power factor (PF) is given by the formula:
PF = (P1 - P2) / (P1 + P2)
We know that the power factor is 0.5, so we can substitute this value into the formula:
0.5 = (P1 - P2) / (P1 + P2)
Multiplying both sides of the equation by (P1 + P2), we get:
0.5(P1 + P2) = P1 - P2
Expanding the left side of the equation, we get:
0.5P1 + 0.5P2 = P1 - P2
Rearranging the equation, we get:
P1 + 2P2 = 2P1
Subtracting P1 from both sides, we get:
P2 = P1
Therefore, the readings of the two wattmeters (P1 and P2) are equal.
PF = (P1 - P2) / (P1 + P2)
We know that the power factor is 0.5, so we can substitute this value into the formula:
0.5 = (P1 - P2) / (P1 + P2)
Multiplying both sides of the equation by (P1 + P2), we get:
0.5(P1 + P2) = P1 - P2
Expanding the left side of the equation, we get:
0.5P1 + 0.5P2 = P1 - P2
Rearranging the equation, we get:
P1 + 2P2 = 2P1
Subtracting P1 from both sides, we get:
P2 = P1
Therefore, the readings of the two wattmeters (P1 and P2) are equal.
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Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer?
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Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer?.
Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer?.
Solutions for Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Power in a three-phase star connected balanced inductive load is measure by two wattmeter method. The phase voltage and phase current are 200 V and 10 A, respectively. The power-factor of load is 0.5. The readings P1and P2of the two wattmeter’s area)P1= 3 kW, P2= 6 kWb)P1= 0 kW, P2= 6 kWc)P1= 2 kW, P2= 6 kWd)P1= 0 kW, P2= 3 kWCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice GATE tests.
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