Finding inductor value for same current/voltage phase AC analysis

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Discussion Overview

The discussion revolves around determining the value of an inductor (L) in an AC circuit such that the current (i(t)) is in phase with the voltage (v(t)). The focus is on the theoretical and mathematical aspects of AC analysis, particularly involving impedance and reactance in the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about what it means for i(t) to be in phase with v(t), questioning whether both can be assumed to be cosine functions with the same phase angle.
  • Another participant suggests that if the current is in phase with the applied voltage, the circuit appears purely resistive, implying that the net reactance equals zero ohms.
  • One participant attempts to calculate L using the formula 1/j(10^4)(4x10^-6) = j(10^4)L, arriving at a negative value for L and questioning how to resolve this issue.
  • Another participant challenges the approach, suggesting that the equation should involve setting the total impedance to zero, indicating a need to equate the imaginary components of the equivalent impedance.
  • There is a discussion about whether the components are in parallel or series, with one participant initially misunderstanding the configuration but later correcting themselves.
  • Ultimately, one participant confirms that they arrived at a positive value for L (2.5 mH) after further consideration.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to solving for L, with some suggesting different equations and interpretations of the circuit configuration. The discussion does not reach a consensus on the method, but one participant does confirm a successful calculation of L.

Contextual Notes

There are unresolved assumptions regarding the definitions of phase relationships in AC circuits and the implications of using different equations for impedance and admittance. The discussion also reflects confusion about the configuration of circuit components (parallel vs. series).

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Homework Statement


In the circuit below, what should the value of L be at ω = 10^4 rad/s so that i(t) is in-phase with v(t)?

Homework Equations


The Attempt at a Solution


I am a little uncertain exactly what is meant by i(t) being in phase with v(t). Do I assume that both are cosine functions, and that means that the phase angle for both of them are the same?

I am not sure if I should just divide the phase voltage (unknown phase) by phase current to get the equivalent impedance? I don't know L, so it is a little tough with the expression I have for L to get something. There are no numerical values for the voltage nor current, so I mean how would I even be able to calculate an equivalent impedance with an unknown inductor value.

For one I just said that L is zero for a DC current, but there must be an AC solution as well.
 

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Here is my attempt, it looks like you cannot edit a thread using a phone?

ImageUploadedByPhysics Forums1396932452.217375.jpg
 
With current being in phase with applied voltage, you can say the circuit seems (to that source) to be purely resistive. That is, there is no imaginary component of the impedance, or the nett reactance equals zero ohms.
 
Great thanks.

I set 1/j(10^4)(4x10^-6) = j(10^4)L and solve for L, getting -2.5 mH. I know the answer is 2.5 mH, so I am wondering how to get rid of this negative term.
 
Maylis said:
Great thanks.

I set 1/j(10^4)(4x10^-6) = j(10^4)L and solve for L, getting -2.5 mH. I know the answer is 2.5 mH, so I am wondering how to get rid of this negative term.
Shouldn't you be setting 1/(j(10^4)(4x10^-6)) + j(10^4)L = 0
 
I'm not seeing why. Arent they in parallel? That appears to be an addition in series.
 
Maylis said:
I'm not seeing why. Arent they in parallel? That appears to be an addition in series.

Oos, I wasn't paying close attention. :redface: I meant shouldn't you be equating the imaginary term in your Zeq to zero?
 
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Maylis said:
Great thanks.

I set 1/j(10^4)(4x10^-6) = j(10^4)L and solve for L, getting -2.5 mH. I know the answer is 2.5 mH, so I am wondering how to get rid of this negative term.

L and C are in parallel so you need to add admittances, not impedances.
So solve jwC + 1/jwL = 0 for L, what do you get?
 
I got 2.5 mH, thank you!
 

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