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A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively. The power in the system is measured using the two wattmeter method. The first wattmeter connected between R-line and Y-line reads zero. The reading of the second wattmeter (connected between B-line and Y-line) in watt is _________.
Correct answer is '-3464.10'. Can you explain this answer?
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Explanation:
Given data:
- Symmetrical three-phase three-wire RYB system
- Balanced delta-connected load
- RMS values of line current = 10 A
- RMS value of line-to-line voltage = 400 V
- First wattmeter reading = 0
To find:
- Reading of the second wattmeter (connected between B-line and Y-line) in watt
Three-Phase Power Measurement:
In a three-phase power system, power can be measured using the two wattmeter method. This method is used to measure the total power in a three-phase system. The two wattmeters are connected in the system in such a way that they measure the power in two individual phases and the sum of the two readings gives the total power.
Two Wattmeter Method:
In the two wattmeter method, the wattmeters are connected between the line voltages and the line currents. The power measured by each wattmeter is given by the formula:
W1 = VL * IL * cos(θ1)
W2 = VL * IL * cos(θ2)
Where:
- W1 and W2 are the readings of the first and second wattmeters
- VL is the line-to-line voltage
- IL is the line current
- θ1 and θ2 are the phase angles between the line voltage and line current for the first and second wattmeters, respectively
Balanced Delta-Connected Load:
In a balanced delta-connected load, the line current is equal to the phase current. Therefore, in this case, the line current (IL) is equal to 10 A.
First Wattmeter Reading:
The first wattmeter connected between R-line and Y-line reads zero. This indicates that the power in the R-phase is zero. Therefore, θ1 is equal to 90 degrees.
Second Wattmeter Reading:
To find the reading of the second wattmeter (W2), we need to calculate the phase angle θ2.
Using the power formula for the second wattmeter, we have:
W2 = VL * IL * cos(θ2)
Since the load is delta-connected, the line voltage (VL) is equal to the phase voltage. Therefore, VL is equal to 400 V.
To calculate θ2, we can use the power formula for the total power in a three-phase system:
W_total = W1 + W2
Since the first wattmeter reading is zero, we have:
W_total = W2
The total power can also be calculated using the apparent power formula:
S = √3 * VL * IL
Where S is the apparent power.
Using the given values, we can calculate S:
S = √3 * 400 V * 10 A
S = √3 * 4000 VA
The power factor (PF) is given by the formula:
PF = W_total / S
Since W_total = W2, we have:
PF = W2 / S
We can rearrange this formula to solve for W2:
W2 = PF * S
The power factor can be calculated using the formula:
PF = cos(θ2)
Since cos(θ2) is equal to the power factor, we have:
W2 = cos(θ2) * S
By substituting the given values, we can calculate W2:
W2 = cos(θ2) * √3 * 4000 VA
To calculate
Given data:
- Symmetrical three-phase three-wire RYB system
- Balanced delta-connected load
- RMS values of line current = 10 A
- RMS value of line-to-line voltage = 400 V
- First wattmeter reading = 0
To find:
- Reading of the second wattmeter (connected between B-line and Y-line) in watt
Three-Phase Power Measurement:
In a three-phase power system, power can be measured using the two wattmeter method. This method is used to measure the total power in a three-phase system. The two wattmeters are connected in the system in such a way that they measure the power in two individual phases and the sum of the two readings gives the total power.
Two Wattmeter Method:
In the two wattmeter method, the wattmeters are connected between the line voltages and the line currents. The power measured by each wattmeter is given by the formula:
W1 = VL * IL * cos(θ1)
W2 = VL * IL * cos(θ2)
Where:
- W1 and W2 are the readings of the first and second wattmeters
- VL is the line-to-line voltage
- IL is the line current
- θ1 and θ2 are the phase angles between the line voltage and line current for the first and second wattmeters, respectively
Balanced Delta-Connected Load:
In a balanced delta-connected load, the line current is equal to the phase current. Therefore, in this case, the line current (IL) is equal to 10 A.
First Wattmeter Reading:
The first wattmeter connected between R-line and Y-line reads zero. This indicates that the power in the R-phase is zero. Therefore, θ1 is equal to 90 degrees.
Second Wattmeter Reading:
To find the reading of the second wattmeter (W2), we need to calculate the phase angle θ2.
Using the power formula for the second wattmeter, we have:
W2 = VL * IL * cos(θ2)
Since the load is delta-connected, the line voltage (VL) is equal to the phase voltage. Therefore, VL is equal to 400 V.
To calculate θ2, we can use the power formula for the total power in a three-phase system:
W_total = W1 + W2
Since the first wattmeter reading is zero, we have:
W_total = W2
The total power can also be calculated using the apparent power formula:
S = √3 * VL * IL
Where S is the apparent power.
Using the given values, we can calculate S:
S = √3 * 400 V * 10 A
S = √3 * 4000 VA
The power factor (PF) is given by the formula:
PF = W_total / S
Since W_total = W2, we have:
PF = W2 / S
We can rearrange this formula to solve for W2:
W2 = PF * S
The power factor can be calculated using the formula:
PF = cos(θ2)
Since cos(θ2) is equal to the power factor, we have:
W2 = cos(θ2) * S
By substituting the given values, we can calculate W2:
W2 = cos(θ2) * √3 * 4000 VA
To calculate
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A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. Can you explain this answer?
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A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. 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 A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. Can you explain this answer?.
A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. 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 A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. Can you explain this answer?.
Solutions for A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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ample number of questions to practice A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively.The power in the system is measured using the two wattmeter method. The first wattmeterconnected between R-line and Y-line reads zero. The reading of the second wattmeter (connectedbetween B-line and Y-line) in watt is _________.Correct answer is '-3464.10'. Can you explain this answer? tests, examples and also practice GATE tests.
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