# Homework Help: Finding inductor value for same current/voltage phase AC analysis

1. Apr 7, 2014

### Maylis

1. The problem statement, all variables and given/known data
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)?

2. Relevant equations

3. 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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2. Apr 7, 2014

### Maylis

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

3. Apr 8, 2014

### Staff: Mentor

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.

4. Apr 8, 2014

### Maylis

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.

5. Apr 8, 2014

### Staff: Mentor

Shouldn't you be setting 1/(j(10^4)(4x10^-6)) + j(10^4)L = 0

6. Apr 8, 2014

### Maylis

I'm not seeing why. Arent they in parallel? That appears to be an addition in series.

7. Apr 8, 2014

### Staff: Mentor

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

8. Apr 9, 2014

### rude man

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?

9. Apr 9, 2014

### Maylis

I got 2.5 mH, thank you!