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A wattmeter reads 5.54 kW when its current coil is connected in the red phase and its voltage coil is connected between the neutral and red phase of a symmetrical three phase system supplying a balanced load of 30 A at 400 V. What will be the reading of the instrument if the current coil remains unchanged and voltage coil be connected between blue and yellow phases. The phase sequence is RYB.
- a)7.2 KVAR
- b)12.47 KVAR
- c)16.63 KW
- d)9.6 KW
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A wattmeter reads 5.54 kW when its current coil is connected in the re...
Let V and I be the phase voltage and phase current respectively.
Phase current = 30 A
In the first case, the wattmeter measures the power in one phase.
VI cos ϕ = 5.54 × 103
When the current coil is connected in the red phase and the pressure coil circuit is connected across the yellow and blue phase, the reading of wattmeter is,
Phase current = 30 A
In the first case, the wattmeter measures the power in one phase.
VI cos ϕ = 5.54 × 103
When the current coil is connected in the red phase and the pressure coil circuit is connected across the yellow and blue phase, the reading of wattmeter is,
Most Upvoted Answer
A wattmeter reads 5.54 kW when its current coil is connected in the re...
Given data:
Current, I = 30 A
Voltage, V = 400 V
Power, P = 5.54 kW
To find: Reading of the instrument when voltage coil is connected between blue and yellow phases.
Solution:
Let's first calculate the power factor of the load using the given data.
Apparent power, S = VI = 30 x 400 = 12000 VA
Power factor, cosφ = P/S = 5.54/12 = 0.46
Sinφ = √(1 - cos²φ) = √(1 - 0.46²) = 0.89
Now let's calculate the phase angle between the voltage and current using the power factor.
Tanφ = sinφ/cosφ = 0.89/0.46 = 1.93
φ = tan⁻¹(1.93) = 63.65°
We know that the power measured by a wattmeter is given by P = VIcosφ. Since the current coil remains unchanged, the reading of the instrument will only depend on the voltage and power factor.
When the voltage coil is connected between the neutral and red phase, the phase angle between the voltage and current is zero, so the power factor is cosφ = 1. In this case, P = VIcosφ = 30 x 400 x 1 = 12000 W.
When the voltage coil is connected between the blue and yellow phases, the phase angle between the voltage and current is φ = 63.65°, so the power factor is cosφ = 0.46. In this case, P = VIcosφ = 30 x 400 x 0.46 = 5520 W.
The difference between the two readings is the reactive power of the load, which is given by Q = √(S² - P²) = √(12000² - 5520²) = 10368 VAR.
Since the load is balanced and the phase sequence is RYB, the reactive power is the same for all three phases. Therefore, the reading of the instrument when the voltage coil is connected between the blue and yellow phases is Q/3 = 10368/3 = 3456 VAR = 3.456 kVAR.
Therefore, the correct answer is option (A) 7.2 kVAR. (Note that this answer is incorrect and the correct answer should be 3.456 kVAR).
Current, I = 30 A
Voltage, V = 400 V
Power, P = 5.54 kW
To find: Reading of the instrument when voltage coil is connected between blue and yellow phases.
Solution:
Let's first calculate the power factor of the load using the given data.
Apparent power, S = VI = 30 x 400 = 12000 VA
Power factor, cosφ = P/S = 5.54/12 = 0.46
Sinφ = √(1 - cos²φ) = √(1 - 0.46²) = 0.89
Now let's calculate the phase angle between the voltage and current using the power factor.
Tanφ = sinφ/cosφ = 0.89/0.46 = 1.93
φ = tan⁻¹(1.93) = 63.65°
We know that the power measured by a wattmeter is given by P = VIcosφ. Since the current coil remains unchanged, the reading of the instrument will only depend on the voltage and power factor.
When the voltage coil is connected between the neutral and red phase, the phase angle between the voltage and current is zero, so the power factor is cosφ = 1. In this case, P = VIcosφ = 30 x 400 x 1 = 12000 W.
When the voltage coil is connected between the blue and yellow phases, the phase angle between the voltage and current is φ = 63.65°, so the power factor is cosφ = 0.46. In this case, P = VIcosφ = 30 x 400 x 0.46 = 5520 W.
The difference between the two readings is the reactive power of the load, which is given by Q = √(S² - P²) = √(12000² - 5520²) = 10368 VAR.
Since the load is balanced and the phase sequence is RYB, the reactive power is the same for all three phases. Therefore, the reading of the instrument when the voltage coil is connected between the blue and yellow phases is Q/3 = 10368/3 = 3456 VAR = 3.456 kVAR.
Therefore, the correct answer is option (A) 7.2 kVAR. (Note that this answer is incorrect and the correct answer should be 3.456 kVAR).
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A wattmeter reads 5.54 kW when its current coil is connected in the re...
12.46
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A wattmeter reads 5.54 kW when its current coil is connected in the red phase and its voltage coil is connected between the neutral and red phase of a symmetrical three phase system supplying a balanced load of 30 A at 400 V. What will be the reading of the instrument if the current coil remains unchanged and voltage coil be connected between blue and yellow phases. The phase sequence is RYB.a)7.2 KVARb)12.47 KVARc)16.63 KWd)9.6 KWCorrect answer is option 'A'. Can you explain this answer?
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A wattmeter reads 5.54 kW when its current coil is connected in the red phase and its voltage coil is connected between the neutral and red phase of a symmetrical three phase system supplying a balanced load of 30 A at 400 V. What will be the reading of the instrument if the current coil remains unchanged and voltage coil be connected between blue and yellow phases. The phase sequence is RYB.a)7.2 KVARb)12.47 KVARc)16.63 KWd)9.6 KWCorrect answer is option 'A'. 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 A wattmeter reads 5.54 kW when its current coil is connected in the red phase and its voltage coil is connected between the neutral and red phase of a symmetrical three phase system supplying a balanced load of 30 A at 400 V. What will be the reading of the instrument if the current coil remains unchanged and voltage coil be connected between blue and yellow phases. The phase sequence is RYB.a)7.2 KVARb)12.47 KVARc)16.63 KWd)9.6 KWCorrect answer is option 'A'. 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 A wattmeter reads 5.54 kW when its current coil is connected in the red phase and its voltage coil is connected between the neutral and red phase of a symmetrical three phase system supplying a balanced load of 30 A at 400 V. What will be the reading of the instrument if the current coil remains unchanged and voltage coil be connected between blue and yellow phases. The phase sequence is RYB.a)7.2 KVARb)12.47 KVARc)16.63 KWd)9.6 KWCorrect answer is option 'A'. Can you explain this answer?.
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