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A 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz?
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A 50 Volt range spring controlled, electrodynamic voltmeter having a s...
Calculation of True Potential Difference across the Instrument
Given Parameters:
- Voltage Range (V) = 50V
- Full Scale Deflection (θ) = 900
- Current for Full Scale Deflection (I) = 0.05A
- Control Constant (K) = 0.5 x 10^-6 N-m/degree
- Initial Mutual Inductance (M) = 0.50H
- Frequency (f) = 50Hz
Calculation of Spring Constant (Ks):
The spring constant can be calculated using the formula:
Ks = (θ/2π) x (I^2 x M)/(V x K)
Substituting the given values, we get:
Ks = (900/2π) x (0.05^2 x 0.50)/(50 x 0.5 x 10^-6)
Ks = 318.3099 N/m
Calculation of Angular Frequency (ω):
Angular frequency can be calculated using the formula:
ω = 2πf
Substituting the given value of frequency, we get:
ω = 2 x 3.1416 x 50
ω = 314.16 rad/s
Calculation of Current (I1) at 50V:
At 50V, the deflection of the voltmeter is:
θ1 = (50/50) x 900 = 900
The current required for this deflection can be calculated using the formula:
I1 = (θ1/2π) x (V x K)/(M x Ks)
Substituting the given values, we get:
I1 = (900/2π) x (50 x 0.5 x 10^-6)/(0.50 x 318.3099)
I1 = 0.055A
Calculation of True Potential Difference:
The true potential difference can be calculated using the formula:
Vt = V x (I1/I)
Substituting the calculated values, we get:
Vt = 50 x (0.055/0.05)
Vt = 55V
Explanation:
A spring controlled, electrodynamic voltmeter operates on the principle of torque produced by the interaction of magnetic fields. The deflection of the voltmeter is proportional to the current flowing through it, and hence the potential difference can be measured. The spring provides the restoring torque, which is proportional to its spring constant. The control constant represents the torque produced by the interaction of magnetic fields, while the mutual inductance represents the coupling between the two
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A 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz?
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A 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz? 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 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz?.
A 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz? 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 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 Volt range spring controlled, electrodynamic voltmeter having a square low scale response takes 0.05A on D.C. for full scale deflection of 900 . The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the Instrument is 0.50 H. Find the true potential difference across the Instrument, when it reads 50 V at 50 Hz?.
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