Non-ideal inductor voltage

1. Nov 14, 2012

DrOnline

1. The problem statement, all variables and given/known data
An inductor is subjected to the current in the graph:

L = 100mH = 0.1H

I) Calculate and graph the voltage over the ideal inductor.

II) Repeat for non ideal inductor with a resistance of:
RL = 50 Ω

2. Relevant equations

$$U_{L}=L*\frac{di}{dt}$$
$$U_{av}=L*\frac{Δi}{Δt}$$

The voltage over an inductor is equal to the rate of change of the current through the coil.

3. The attempt at a solution

I)

This is what I came up with. As an example, from 2 ms to 4 ms:

$$U_{av}=0.1H*\frac{-2A}{2*10^-3 s} = -100V$$

Ok, some question, I am pretty sure the answer is yes to these two, but I'd love to have them confirmed, so I know I am on the right track:

*Is my square wave form correct? Should it be square, like this?

*Am I doing this correctly? It seems to me $$U_{av} = U$$. So I could write just U =...

II)

This is when I become confused, and I have spent hours and hours trying to solve this... sadly heh. I don't know where to start!

The way I understand it:

*Regardless of the internal resistance, the current will still be the same, as the coil and the internal resistance will be in series.

*So for charging up during 0-1ms, the voltage over the coil will be the same.

*Am I still going to have a square wave form?

I've been given as a clue that $$U_{coil}(3ms) = -50 V$$

Please, just a nudge in the right direction, I'm totally stuck here..

Last edited: Nov 14, 2012
2. Nov 14, 2012

aralbrec

Yes, all good assuming your algebra is right.

Yes you are but I don't even look at the Uav equation. What is important is the slope of i(t) at any particular time and in your graphs the slope di/dt is constant over periods of time.

That's right, you are shoving the same current through the non-ideal inductor as before. The outside current source will have to adjust its terminal voltage to accomplish this.

The non-ideal inductor is now modelled as a resistance in series with an ideal inductor. The voltage you measure across the non-ideal inductor is equal to the voltage across the ideal part plus ?

3. Nov 14, 2012

DrOnline

$$v_{real} = v_{ideal} + v_{R_{i}}$$

Matching the variables to my drawing's labels:
$$v_{non-ideal} = v_{ideal} + v_{R}$$

Alright, so the voltage from 2ms to 4ms is -100 over the ideal.

And the average current is 1 A. That makes the voltage over the resistor: R * I = 50 ohm * 1A = 50V.

$$v_{real} = v_{ideal} + v_{R_{i}} = -100V + 50V = -50V$$

Is this sound reasoning? I think it is. What troubled me was the linear drop in the current, and I was having problems understanding how that influenced the voltage of the resistor... I think I get it now.

Last edited: Nov 14, 2012
4. Nov 14, 2012

aralbrec

Yes :). But let's not look at the average, let's look at a specific instant of time.

$$U_{coil}(3ms) = -50 V$$

At t=3ms, the voltage across the ideal inductor is -100. The current through the inductor is (from your graph) 1A. So the voltage measured across the nonideal inductor is: Videal + iR = -100 + 50 = -50V, in agreement with the hint.

I just wanted to make sure you were looking at the current at a particular instant of time t=3ms instead of some average over an interval.

You can see that the voltage measured across the non-ideal inductor will be a scaled version of your i(t) graph (iR) plus the square wave voltage you found for the ideal inductor. So your squares are going to disappear except where the current is constant.

5. Nov 14, 2012

DrOnline

Thank you so much.

I kinda knew that when I finally get this task solved, I would look back at it and wonder what took me so long.

The problem was I kept feeling the drop in current coming through the inductor would change the voltage over the conductor all the time, causing some weird differential equation task, where the voltage rose abruptly, but dropped exponentially. And then I was constantly wondering about whether I had built my work on faulty logic from the start, so it's a huge help just that you confirmed for me the first steps, that I had done it right.

Thanks again! ;)