Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A 50 V range spring controlled, electrodynami...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A 50 V range spring controlled, electrodynamic voltmeter having a squa...
Most Upvoted Answer
A 50 V range spring controlled, electrodynamic voltmeter having a squa...
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
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
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?
Question Description
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?.
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup