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A 50 V range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on d.c. for full scale deflection of 90°. The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the instrument is 200 mH. The total mutual inductance at90° deflection is
- a)200 mH
- b)220 mH
- c)228 mH
- d)254 mH
Correct answer is option 'C'. Can you explain this answer?
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A 50 V range spring controlled, electrodynamic voltmeter having a squa...
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Given:
- Range of the voltmeter = 50 V
- Full scale deflection current = 0.05 A
- Full scale deflection angle = 90 degrees
- Control constant = 0.5 x 10^-6 N-m/degree
- Initial mutual inductance = 200 mH
To find:
Total mutual inductance at 90 degrees deflection
Solution:
Step 1: Find the controlling torque
The controlling torque is given by the equation:
Controlling torque (Tc) = Control constant * Angle of deflection
Given that the control constant is 0.5 x 10^-6 N-m/degree and the angle of deflection is 90 degrees, we can calculate the controlling torque:
Tc = 0.5 x 10^-6 N-m/degree * 90 degrees
Tc = 45 x 10^-6 N-m
Step 2: Find the current required for full scale deflection
The current required for full scale deflection is given by the equation:
I = Tc / (M * B)
Where:
I = Current
Tc = Controlling torque
M = Mutual inductance
B = Magnetic field
We know the controlling torque (Tc) and the current (I) for full scale deflection. Rearranging the equation, we can solve for the mutual inductance (M):
M = Tc / (I * B)
Given that the current (I) for full scale deflection is 0.05 A, we can substitute the values to calculate the mutual inductance:
M = (45 x 10^-6 N-m) / (0.05 A * B)
Step 3: Find the total mutual inductance at 90 degrees deflection
The total mutual inductance at 90 degrees deflection is given by the equation:
M_total = M_initial + M_deflection
Where:
M_total = Total mutual inductance
M_initial = Initial mutual inductance
M_deflection = Change in mutual inductance due to deflection
Given that the initial mutual inductance is 200 mH, we need to calculate the change in mutual inductance due to deflection.
The change in mutual inductance due to deflection can be calculated using the equation:
M_deflection = Tc / (I * B)
Substituting the values, we get:
M_deflection = (45 x 10^-6 N-m) / (0.05 A * B)
Finally, we can calculate the total mutual inductance at 90 degrees deflection:
M_total = M_initial + M_deflection
M_total = 200 mH + (45 x 10^-6 N-m) / (0.05 A * B)
Since the options for total mutual inductance are given in mH, we need to convert the units from N-m/A to mH:
1 N-m/A = 1 V-s/A = 1 mH
Therefore, the total mutual inductance at 90 degrees deflection is:
M_total = 200 mH + (45 x 10^-6 mH)
M_total = 200 mH + 0.045 mH
- Range of the voltmeter = 50 V
- Full scale deflection current = 0.05 A
- Full scale deflection angle = 90 degrees
- Control constant = 0.5 x 10^-6 N-m/degree
- Initial mutual inductance = 200 mH
To find:
Total mutual inductance at 90 degrees deflection
Solution:
Step 1: Find the controlling torque
The controlling torque is given by the equation:
Controlling torque (Tc) = Control constant * Angle of deflection
Given that the control constant is 0.5 x 10^-6 N-m/degree and the angle of deflection is 90 degrees, we can calculate the controlling torque:
Tc = 0.5 x 10^-6 N-m/degree * 90 degrees
Tc = 45 x 10^-6 N-m
Step 2: Find the current required for full scale deflection
The current required for full scale deflection is given by the equation:
I = Tc / (M * B)
Where:
I = Current
Tc = Controlling torque
M = Mutual inductance
B = Magnetic field
We know the controlling torque (Tc) and the current (I) for full scale deflection. Rearranging the equation, we can solve for the mutual inductance (M):
M = Tc / (I * B)
Given that the current (I) for full scale deflection is 0.05 A, we can substitute the values to calculate the mutual inductance:
M = (45 x 10^-6 N-m) / (0.05 A * B)
Step 3: Find the total mutual inductance at 90 degrees deflection
The total mutual inductance at 90 degrees deflection is given by the equation:
M_total = M_initial + M_deflection
Where:
M_total = Total mutual inductance
M_initial = Initial mutual inductance
M_deflection = Change in mutual inductance due to deflection
Given that the initial mutual inductance is 200 mH, we need to calculate the change in mutual inductance due to deflection.
The change in mutual inductance due to deflection can be calculated using the equation:
M_deflection = Tc / (I * B)
Substituting the values, we get:
M_deflection = (45 x 10^-6 N-m) / (0.05 A * B)
Finally, we can calculate the total mutual inductance at 90 degrees deflection:
M_total = M_initial + M_deflection
M_total = 200 mH + (45 x 10^-6 N-m) / (0.05 A * B)
Since the options for total mutual inductance are given in mH, we need to convert the units from N-m/A to mH:
1 N-m/A = 1 V-s/A = 1 mH
Therefore, the total mutual inductance at 90 degrees deflection is:
M_total = 200 mH + (45 x 10^-6 mH)
M_total = 200 mH + 0.045 mH
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A 50 V range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on d.c. for full scale deflection of 90°. The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the instrument is 200 mH. The total mutual inductance at90° deflection is a)200 mH b)220 mHc)228 mH d)254 mHCorrect answer is option 'C'. Can you explain this answer?
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A 50 V range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on d.c. for full scale deflection of 90°. The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the instrument is 200 mH. The total mutual inductance at90° deflection is a)200 mH b)220 mHc)228 mH d)254 mHCorrect answer is option 'C'. 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 50 V range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on d.c. for full scale deflection of 90°. The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the instrument is 200 mH. The total mutual inductance at90° deflection is a)200 mH b)220 mHc)228 mH d)254 mHCorrect answer is option 'C'. 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 50 V range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on d.c. for full scale deflection of 90°. The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the instrument is 200 mH. The total mutual inductance at90° deflection is a)200 mH b)220 mHc)228 mH d)254 mHCorrect answer is option 'C'. Can you explain this answer?.
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