Finding R,L,C at resonant state

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    Resonant State
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Discussion Overview

The discussion revolves around finding the values of resistance (R), inductance (L), and capacitance (C) in a resonant circuit based on given conditions of current and angular frequency. Participants explore the relationships between these variables and the implications of their calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the magnitude of the current in the resonant state is 3[A] and provides conditions for current at different angular frequencies.
  • Another participant challenges the results, indicating that they calculated a negative capacitance value, which raises concerns about the validity of the equations used.
  • Multiple participants express uncertainty about the correctness of their equations and calculations, with one suggesting that there are multiple solutions but only one set where L and C are positive.
  • There is a discussion about the implications of taking square roots in equations, with participants noting the potential for multiple solutions and the importance of considering both polarities.
  • One participant acknowledges a misunderstanding regarding "losing solutions" and presents new values for C and L, which are then affirmed by another participant.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the equations or the values derived from them. There are multiple competing views regarding the calculations and the implications of the results.

Contextual Notes

Participants express uncertainty about the elimination of squares in their equations and the handling of negative values for capacitance. There is a noted dependence on the definitions and assumptions made in the calculations.

MissP.25_5
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Hello again. I hope someone could check if I got the answers correct.

The magnitude of the current I in resonant state was 3[A]. When the angular frequency of the power supply was 1 [rad/s] and 2[rad/s], the magnitude of current I became 1/√2 of the value when resonance occurred. Find R, L, C.
 

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When I solve your equations as you've given them I'm seeing -3/8 for the capacitance... which shouldn't be.

You have to be a bit careful about 'losing' solutions when you eliminate squares.
 
MissP.25_5 said:
Hello again. I hope someone could check if I got the answers correct.

The magnitude of the current I in resonant state was 3[A]. When the angular frequency of the power supply was 1 [rad/s] and 2[rad/s], the magnitude of current I became 1/√2 of the value when resonance occurred. Find R, L, C.

Your approach looks right, but I got a different result. Check your equations against mine & recheck your math:

(L - 1/C)^2 = 16,
(2L - 1/2C)^2 = 16
 
gneill said:
When I solve your equations as you've given them I'm seeing -3/8 for the capacitance... which shouldn't be.

You have to be a bit careful about 'losing' solutions when you eliminate squares.

Yes, at first I got -3/8 for the Capacitance but since it's an absolute value, I made it positive. Well, looks like I have to redo my calculation again :(
 
gneill said:
When I solve your equations as you've given them I'm seeing -3/8 for the capacitance... which shouldn't be.

You have to be a bit careful about 'losing' solutions when you eliminate squares.

I redid it and still got the same simultaneous equations. Can you tell me where I am wrong?
 

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MissP.25_5 said:
I redid it and still got the same simultaneous equations. Can you tell me where I am wrong?

Do you agree with my equations or not? If so, what solutions do you get?

Hint: there are multiple solutions but only one where L and C are positive, as they must be.
 
rude man said:
Do you agree with my equations or not? If so, what solutions do you get?

Hint: there are multiple solutions but only one where L and C are positive, as they must be.

Yes, yours seems to be the same as mine. But can we cancel out the squares by just adding a root to both sides?
 
MissP.25_5 said:
Yes, yours seems to be the same as mine. But can we cancel out the squares by just adding a root to both sides?

When you take a square root you have to realize there are two polarities involved. Here we are taking two roots so be prepared to get four sets of answers for L and C.

But only one of them has L > 0 and C > 0.
 
rude man said:
When you take a square root you have to realize there are two polarities involved. Here we are taking two roots so be prepared to get four sets of answers for L and C.

But only one of them has L > 0 and C > 0.

Oh, I totally forgot about that. So that's what "losing solutions" means. So C=1/8 and L=4.
Thank you soooo much! Could you take a look at my new thread? I need help.
 
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MissP.25_5 said:
Oh, I totally forgot about that. So that's what "losing solutions" means. So C=1/8 and L=4.
Thank you soooo much! Could you take a look at my new thread? I need help.

Yes, you have got it!
Will check your new thread.
 
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