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The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)
Correct answer is '17'. Can you explain this answer?
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The current coil of wattmeter is connected in series with ammeter and...
Given:
- Frequency of supply, f = 100 Hz
- Ammeter reading, I = 4.5 A
- Voltmeter reading, V = 240 V
- Wattmeter reading, P = 23 W
- Inductance of voltage coil circuit, L = 10 mH = 0.01 H
- Resistance of voltage coil circuit, R = 2000 Ω
To Find:
Percentage error in wattmeter reading
Solution:
Step 1: Calculate the apparent power, S
Apparent power is given by the product of voltage and current:
S = V x I
= 240 V x 4.5 A
= 1080 VA
Step 2: Calculate the power factor, PF
Power factor is the ratio of true power to apparent power:
PF = P / S
= 23 W / 1080 VA
= 0.0213
Step 3: Calculate the impedance, Z
Impedance is given by the ratio of voltage to current:
Z = V / I
= 240 V / 4.5 A
= 53.33 Ω
Step 4: Calculate the reactance, X
Reactance is given by the formula:
X = 2πfL
= 2 x 3.14 x 100 Hz x 0.01 H
= 6.28 Ω
Step 5: Calculate the resistance, R'
Resistance is given by the formula:
R' = √(Z^2 - X^2)
= √(53.33 Ω^2 - 6.28 Ω^2)
= √(2833.65 Ω^2)
= 53.24 Ω
Step 6: Calculate the corrected power factor, PF'
Corrected power factor is given by the formula:
PF' = R' / Z
= 53.24 Ω / 53.33 Ω
= 0.9983
Step 7: Calculate the percentage error in wattmeter reading, Error
Percentage error is given by the formula:
Error = (1 - PF') x 100
= (1 - 0.9983) x 100
= 0.17%
Therefore, the percentage error in wattmeter reading is approximately 0.2%, which is rounded to 0.2%.
- Frequency of supply, f = 100 Hz
- Ammeter reading, I = 4.5 A
- Voltmeter reading, V = 240 V
- Wattmeter reading, P = 23 W
- Inductance of voltage coil circuit, L = 10 mH = 0.01 H
- Resistance of voltage coil circuit, R = 2000 Ω
To Find:
Percentage error in wattmeter reading
Solution:
Step 1: Calculate the apparent power, S
Apparent power is given by the product of voltage and current:
S = V x I
= 240 V x 4.5 A
= 1080 VA
Step 2: Calculate the power factor, PF
Power factor is the ratio of true power to apparent power:
PF = P / S
= 23 W / 1080 VA
= 0.0213
Step 3: Calculate the impedance, Z
Impedance is given by the ratio of voltage to current:
Z = V / I
= 240 V / 4.5 A
= 53.33 Ω
Step 4: Calculate the reactance, X
Reactance is given by the formula:
X = 2πfL
= 2 x 3.14 x 100 Hz x 0.01 H
= 6.28 Ω
Step 5: Calculate the resistance, R'
Resistance is given by the formula:
R' = √(Z^2 - X^2)
= √(53.33 Ω^2 - 6.28 Ω^2)
= √(2833.65 Ω^2)
= 53.24 Ω
Step 6: Calculate the corrected power factor, PF'
Corrected power factor is given by the formula:
PF' = R' / Z
= 53.24 Ω / 53.33 Ω
= 0.9983
Step 7: Calculate the percentage error in wattmeter reading, Error
Percentage error is given by the formula:
Error = (1 - PF') x 100
= (1 - 0.9983) x 100
= 0.17%
Therefore, the percentage error in wattmeter reading is approximately 0.2%, which is rounded to 0.2%.
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Community Answer
The current coil of wattmeter is connected in series with ammeter and...
Given: LP = 10mH, Rp = 2000 Ω
Reading of ammeter = 4.5 A
Reading of voltmeter = 240 V
Reading of wattmeter = 23 W
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The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)Correct answer is '17'. Can you explain this answer?
Question Description
The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)Correct answer is '17'. 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 The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)Correct answer is '17'. 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 The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)Correct answer is '17'. Can you explain this answer?.
The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)Correct answer is '17'. 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 The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)Correct answer is '17'. 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 The current coil of wattmeter is connected in series with ammeter and inductive load. A voltmeter and voltage coil is connected across 100 Hz supply. The readings of ammeter, voltmeter and wattmeter are 4.5 A, 240 V, and 23 W respectively. The inductance of voltage coil circuit is 10 mH and its resistance is 2000 Ω. Neglect voltage drop across ammeter and current coil. The percentage error in wattmeter reading will be ___________. (Correct upto 1 decimal place)Correct answer is '17'. Can you explain this answer?.
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