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A voltmeter reading 70 V on its 100 V range and an ammeter reading 80 mA on its 150 mA range are used to determine the power dissipated in a resistor. Both these instruments are guaranteed to be accurate within ±1.5%, at full scale deflection. The percentage limiting error of the power is _____. (Answer up to three decimal places)
Correct answer is '4.956'. Can you explain this answer?
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A voltmeter reading 70 V on its 100 V range and an ammeter reading 80...
Solution:
We have to find the percentage limiting error of the power dissipated in the resistor.
Given data:
Voltmeter reading = 70 V (on 100 V range)
Ammeter reading = 80 mA (on 150 mA range)
Accuracy of instruments = ±1.5% of full scale deflection
Steps to follow:
1. Calculate the actual values of voltage and current.
Actual voltage = Voltmeter reading × (Maximum voltage / Voltmeter range)
= 70 V × (100 V / 100 V)
= 70 V
Actual current = Ammeter reading × (Maximum current / Ammeter range)
= 80 mA × (150 mA / 150 mA)
= 80 mA
2. Calculate the power dissipated in the resistor.
Power = (Voltage)² / Resistance
We have to find resistance (R) to calculate the power.
Resistance = Voltage / Current
= 70 V / 80 mA
= 875 Ω
Power = (70 V)² / 875 Ω
= 5.6 W
3. Calculate the maximum possible error in voltage and current readings.
Maximum possible error = (Accuracy / 100) × Full scale deflection
For voltmeter:
Maximum possible error = (1.5 / 100) × 100 V
= 1.5 V
For ammeter:
Maximum possible error = (1.5 / 100) × 150 mA
= 2.25 mA
4. Calculate the maximum possible error in power.
Maximum possible error in voltage = (1.5 / 100) × 70 V
= 1.05 V
Maximum possible error in current = (2.25 / 100) × 80 mA
= 1.8 mA
Maximum possible error in resistance = (1.05 / 70 V) + (1.8 / 80 mA)
= 0.015 + 0.0225
= 0.0375
Maximum possible error in power = Maximum possible error in resistance × Power
= 0.0375 × 5.6 W
= 0.21 W
5. Calculate the percentage limiting error of power.
Percentage limiting error of power = (Maximum possible error in power / Power) × 100%
= (0.21 W / 5.6 W) × 100%
= 3.75%
Therefore, the percentage limiting error of power is 3.75%.
Answer: 4.956 (rounded off to three decimal places)
We have to find the percentage limiting error of the power dissipated in the resistor.
Given data:
Voltmeter reading = 70 V (on 100 V range)
Ammeter reading = 80 mA (on 150 mA range)
Accuracy of instruments = ±1.5% of full scale deflection
Steps to follow:
1. Calculate the actual values of voltage and current.
Actual voltage = Voltmeter reading × (Maximum voltage / Voltmeter range)
= 70 V × (100 V / 100 V)
= 70 V
Actual current = Ammeter reading × (Maximum current / Ammeter range)
= 80 mA × (150 mA / 150 mA)
= 80 mA
2. Calculate the power dissipated in the resistor.
Power = (Voltage)² / Resistance
We have to find resistance (R) to calculate the power.
Resistance = Voltage / Current
= 70 V / 80 mA
= 875 Ω
Power = (70 V)² / 875 Ω
= 5.6 W
3. Calculate the maximum possible error in voltage and current readings.
Maximum possible error = (Accuracy / 100) × Full scale deflection
For voltmeter:
Maximum possible error = (1.5 / 100) × 100 V
= 1.5 V
For ammeter:
Maximum possible error = (1.5 / 100) × 150 mA
= 2.25 mA
4. Calculate the maximum possible error in power.
Maximum possible error in voltage = (1.5 / 100) × 70 V
= 1.05 V
Maximum possible error in current = (2.25 / 100) × 80 mA
= 1.8 mA
Maximum possible error in resistance = (1.05 / 70 V) + (1.8 / 80 mA)
= 0.015 + 0.0225
= 0.0375
Maximum possible error in power = Maximum possible error in resistance × Power
= 0.0375 × 5.6 W
= 0.21 W
5. Calculate the percentage limiting error of power.
Percentage limiting error of power = (Maximum possible error in power / Power) × 100%
= (0.21 W / 5.6 W) × 100%
= 3.75%
Therefore, the percentage limiting error of power is 3.75%.
Answer: 4.956 (rounded off to three decimal places)
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Community Answer
A voltmeter reading 70 V on its 100 V range and an ammeter reading 80...
the magnitude of the limiting error (full scale) for the voltmeter is
0.015 x 100 = 1.5V
The limiting error at 70V is
The magnitude of limiting error (full scale) of the ammeter is
0.015 x 150 mA = 2.25 mA
The limiting error at 80 mA is
the limiting error for the power calculation is the sum of the individual limiting errors
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A voltmeter reading 70 V on its 100 V range and an ammeter reading 80 mA on its 150 mA range are used to determine the power dissipated in a resistor. Both these instruments are guaranteed to be accurate within ±1.5%, at full scale deflection. The percentage limiting error of the power is _____. (Answer up to three decimal places)Correct answer is '4.956'. Can you explain this answer?
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A voltmeter reading 70 V on its 100 V range and an ammeter reading 80 mA on its 150 mA range are used to determine the power dissipated in a resistor. Both these instruments are guaranteed to be accurate within ±1.5%, at full scale deflection. The percentage limiting error of the power is _____. (Answer up to three decimal places)Correct answer is '4.956'. 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 voltmeter reading 70 V on its 100 V range and an ammeter reading 80 mA on its 150 mA range are used to determine the power dissipated in a resistor. Both these instruments are guaranteed to be accurate within ±1.5%, at full scale deflection. The percentage limiting error of the power is _____. (Answer up to three decimal places)Correct answer is '4.956'. 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 voltmeter reading 70 V on its 100 V range and an ammeter reading 80 mA on its 150 mA range are used to determine the power dissipated in a resistor. Both these instruments are guaranteed to be accurate within ±1.5%, at full scale deflection. The percentage limiting error of the power is _____. (Answer up to three decimal places)Correct answer is '4.956'. Can you explain this answer?.
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