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A dynamometer wattmeter with its voltage coil connected across the load side of the instrument reads 250 W, If the load voltage be 200 V, what power is being taken by the load?
(The voltage coil has a resistance of 2000 Ω)
(The voltage coil has a resistance of 2000 Ω)
- a)250 W
- b)270 W
- c)230 W
- d)can’t be determined
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A dynamometer wattmeter with its voltage coil connected across the loa...
Power loss in the voltage coil =
instrument reading = 250 watt
∴ Actual load = 250 - 20
= 230 watt
Most Upvoted Answer
A dynamometer wattmeter with its voltage coil connected across the loa...
To find the power being taken by the load, we need to consider the power factor of the load.
The power measured by the dynamometer wattmeter is given by:
P = VIcosθ
Where P is the power, V is the voltage, and θ is the phase angle between the voltage and current.
In this case, the voltage coil of the wattmeter is connected across the load, so the voltage measured by the wattmeter is the same as the load voltage, which is 200 V.
The power measured by the wattmeter is 250 W. Let's assume the load has a power factor of cosθ.
So, we have:
250 W = 200 V * I * cosθ
We are given that the voltage coil has a resistance of 2000 ohms.
The current passing through the voltage coil can be calculated using Ohm's law:
I = V/R
I = 200 V / 2000 ohms
I = 0.1 A
Substituting this value of current into the power equation:
250 W = 200 V * 0.1 A * cosθ
cosθ = 250 W / (200 V * 0.1 A)
cosθ = 1.25
Since the power factor is a value between 0 and 1, it is not possible for cosθ to be greater than 1. Therefore, there is an error in the given data or calculation.
Please check your values and provide the correct information.
The power measured by the dynamometer wattmeter is given by:
P = VIcosθ
Where P is the power, V is the voltage, and θ is the phase angle between the voltage and current.
In this case, the voltage coil of the wattmeter is connected across the load, so the voltage measured by the wattmeter is the same as the load voltage, which is 200 V.
The power measured by the wattmeter is 250 W. Let's assume the load has a power factor of cosθ.
So, we have:
250 W = 200 V * I * cosθ
We are given that the voltage coil has a resistance of 2000 ohms.
The current passing through the voltage coil can be calculated using Ohm's law:
I = V/R
I = 200 V / 2000 ohms
I = 0.1 A
Substituting this value of current into the power equation:
250 W = 200 V * 0.1 A * cosθ
cosθ = 250 W / (200 V * 0.1 A)
cosθ = 1.25
Since the power factor is a value between 0 and 1, it is not possible for cosθ to be greater than 1. Therefore, there is an error in the given data or calculation.
Please check your values and provide the correct information.
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A dynamometer wattmeter with its voltage coil connected across the load side of the instrument reads 250 W, If the load voltage be 200 V, what power is being taken by the load?(The voltage coil has a resistance of 2000 Ω)a)250 Wb)270 Wc)230 Wd)can’t be determinedCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 dynamometer wattmeter with its voltage coil connected across the load side of the instrument reads 250 W, If the load voltage be 200 V, what power is being taken by the load?(The voltage coil has a resistance of 2000 Ω)a)250 Wb)270 Wc)230 Wd)can’t be determinedCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dynamometer wattmeter with its voltage coil connected across the load side of the instrument reads 250 W, If the load voltage be 200 V, what power is being taken by the load?(The voltage coil has a resistance of 2000 Ω)a)250 Wb)270 Wc)230 Wd)can’t be determinedCorrect answer is option 'C'. Can you explain this answer?.
A dynamometer wattmeter with its voltage coil connected across the load side of the instrument reads 250 W, If the load voltage be 200 V, what power is being taken by the load?(The voltage coil has a resistance of 2000 Ω)a)250 Wb)270 Wc)230 Wd)can’t be determinedCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 dynamometer wattmeter with its voltage coil connected across the load side of the instrument reads 250 W, If the load voltage be 200 V, what power is being taken by the load?(The voltage coil has a resistance of 2000 Ω)a)250 Wb)270 Wc)230 Wd)can’t be determinedCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dynamometer wattmeter with its voltage coil connected across the load side of the instrument reads 250 W, If the load voltage be 200 V, what power is being taken by the load?(The voltage coil has a resistance of 2000 Ω)a)250 Wb)270 Wc)230 Wd)can’t be determinedCorrect answer is option 'C'. Can you explain this answer?.
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