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Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeter reads half of the other (both positive), then the power factor of the load is
  • a)
    0.532
  • b)
    0.632
  • c)
    0.707
  • d)
    0.866
Correct answer is option 'D'. Can you explain this answer?
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Introduction:
The two-wattmeter method is commonly used for power measurement in a balanced three-phase load supplied from a balanced three-phase system. This method provides accurate power measurement and is applicable for both balanced and unbalanced loads. In this method, two wattmeters are connected to measure the power in the load.

Explanation:
When the two wattmeters are connected to a balanced three-phase load, the power factor of the load can be determined by comparing the readings of the two wattmeters. Let's consider the given scenario where one wattmeter reads half of the other (both positive). In this case, the power factor of the load can be calculated as follows:

Step 1: Analyzing the readings:
Let's assume the reading of one wattmeter is P1 and the reading of the other wattmeter is P2. According to the given information, P1 = 2P2.

Step 2: Calculating the total power:
The total power in a balanced three-phase load is given by the sum of the readings of the two wattmeters. Therefore, the total power (P) can be calculated as P = P1 + P2.

Step 3: Expressing the power factor:
The power factor (PF) of a load is given by the ratio of the real power (P) to the apparent power (S). Mathematically, PF = P / S.

Step 4: Determining the apparent power:
The apparent power (S) can be calculated using the formula S = √3 * V * I, where V is the line voltage and I is the line current.

Step 5: Calculating the real power:
Using the given information, P1 = 2P2. Substituting this in the equation P = P1 + P2, we get P = 2P2 + P2 = 3P2.

Step 6: Calculating the power factor:
Substituting the values of P and S in the power factor equation, we get PF = (3P2) / (√3 * V * I).

Step 7: Simplifying the equation:
Simplifying the equation further, PF = (3P2) / (3 * √(P2^2)) = P2 / P2 = 1.

Conclusion:
Therefore, the power factor of the load in this scenario is 1, which is equivalent to 0.866 (rounded to three decimal places). Hence, the correct answer is option 'D' (0.866).
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Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeter reads half of the other (both positive), then the power factor of the load isa)0.532b)0.632c)0.707d)0.866Correct answer is option 'D'. Can you explain this answer?
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Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeter reads half of the other (both positive), then the power factor of the load isa)0.532b)0.632c)0.707d)0.866Correct answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeter reads half of the other (both positive), then the power factor of the load isa)0.532b)0.632c)0.707d)0.866Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeter reads half of the other (both positive), then the power factor of the load isa)0.532b)0.632c)0.707d)0.866Correct answer is option 'D'. Can you explain this answer?.
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