Value on ammeter display

  • Thread starter sol59
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  • #1
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Homework Statement


What will the pointer of electrodynamic ammeter show (its natural frequency of oscillation is 5 Hz), if its input is energized by current:
a) i (t) = 1.5 + 4 sin (3146 t)
b) i (t) = 1.5 + 4 sin (2 t)
Ammeter has a range of 6 A and the scale has 100 divisions.

Homework Equations


The Attempt at a Solution

[/B]
My solution is 25 but I don't understand the difference between a) (where the frequency is 500 Hz) and b) (0.318 Hz).
 

Answers and Replies

  • #2
NascentOxygen
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Science Advisor
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Homework Statement


What will the pointer of electrodynamic ammeter show (its natural frequency of oscillation is 5 Hz), if its input is energized by current:
a) i (t) = 1.5 + 4 sin (3146 t)
b) i (t) = 1.5 + 4 sin (2 t)
Ammeter has a range of 6 A and the scale has 100 divisions.

Homework Equations


The Attempt at a Solution

[/B]
My solution is 25 but I don't understand the difference between a) (where the frequency is 500 Hz) and b) (0.318 Hz).
Hi sol59.
smiley_sign_welcome.gif


Describe what you understand by an electrodynamic ammeter.

What do you understand is meant by a natural frequency of oscillation of 5 Hz in the context of this problem.

Until you show you have done some work by yourself, we are not able to help you further.
 
  • #3
21
0
Hi sol59.
smiley_sign_welcome.gif


Describe what you understand by an electrodynamic ammeter.

What do you understand is meant by a natural frequency of oscillation of 5 Hz in the context of this problem.

Until you show you have done some work by yourself, we are not able to help you further.

Hi, thanks for the reply:) It's ammeter with electrodynamic measuring mechanism (with one movable coil and one fixed coil). The natural frequency is the frequency at which the measuring mechanism tends to oscillate before the current is led into its input. I calculated the deflection as: (1.5*100)/6 = 25 If somebody could give me advice how to solve this problem I would be very glad.
 

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