Reading an Electrodynamic Ammeter with 6 A Range and 100 Divisions

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SUMMARY

The discussion focuses on calculating the deflection of an electrodynamic ammeter with a 6 A range and 100 divisions when energized by two different current inputs: i(t) = 1.5 + 4 sin(3146t) and i(t) = 1.5 + 4 sin(2t). The natural frequency of oscillation for the ammeter is 5 Hz. The calculated deflection for the first input is 25 divisions, but the user seeks clarification on the impact of the differing frequencies of the inputs, specifically 500 Hz for the first and 0.318 Hz for the second.

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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).
 
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sol59 said:

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.
 
NascentOxygen said:
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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