Electronics Question - Inductor and Current Rate of Increase

danza · 11 Feb 2012 at 11:15

This question is confusing me. I can't work it from the supporting information I've been given, so I'm asking here. The question is:

A 10V, 250A power supply feeds to an inductor or 120H, and operates as a voltage source set to 10V. The leads to the inductor have a resistance 20m ohms.

- What is the initial rate at which the current increases, just after the power supply is turned on?

- When the current is approaching 250A, what is the rate at which the current increases?

All I've been given that seems relevant is V=L.(dI/dt) What would dt be? I guess all I need is the figure for dt just after the point the power is turned on, then I can transpose for dI? But I can't figure out what dt would be? Or am I being totally retarded here and getting the wrong end of the stick completely?

What do I use as a value for t (time)? :confused:

It's not homework as such, just an exercise in understanding. AND I WANT TO UNDERSTAND!!

danza · 11 Feb 2012 at 13:16

Anyone?

What do I need to do to account for the resistance of the leads?

I'm not the best in the world at electronics (I'm more mechanical), but remember all the stuff I did to do with multiple loop circuits/Kirchoff, RL, RC and RLC circuits and resonant frequencies, just didn't touch on this!

danza · 11 Feb 2012 at 13:50

I'm getting:

V=IR > 10 = 250 x 20x10^-3 = 5V, so 5V through the inductor.

V=L.(dI/dt) > 5 = 120.(dI/dt)

Which becomes:

5/120 = (dI/dt) = 0.0417 Amp/sec.

This is totally wrong isn't it?

danza · 11 Feb 2012 at 17:26

Anybody?

Amirite?

danza · 12 Feb 2012 at 11:02

(Dust) anyone?

mattbrown91 · 12 Feb 2012 at 11:33

First thing I'll say is that it's a series circuit and the current must be the same all the way around the circuit.

Next thing.

For that relationship between potential difference and rate of change of current, you can see that it is not possible to have an instantaneous change in current as it would require an infinite voltage.

Does this answer part 1?

...actually, I'm not sure if it does.. I'll get a pencil and paper.

mattbrown91 · 12 Feb 2012 at 11:46

Well. The way I see it is that you model the resistance in the wires as resistors which are connected to the other components by wires with no resistance.

So you'll go +10V ---- 20mR-----120H-----20mR-----0V
Ok?

So for question 1, with the current being zero at all points around the circuit...because the current can't change instantaneously... this means there is no voltage drop over the resistors (from V=IR) which means that there is the full 10V across the inductor.
So, if you plug that into your V = L(dI/dt) then that gives you dI/dt = 1/12 amperes/sec

For question 2, let the current be 250A all around the circuit. This means you've have voltage drops of 5V over each of the wires (from V=IR) which leaves no potential difference over the inductor. so 0=L(dI/dt) therefore dI/dt = 0 amperes/sec.

Anything you want more detail on?

danza · 12 Feb 2012 at 12:48

That makes sense, thanks

mattbrown91 · 12 Feb 2012 at 12:52

There's just a couple of golden rules:
The current can't instantaneously change through an inductor
The potential difference can't instantaneously change across a capacitor.

Is this A-Level or are you in a degree?

Tute · 12 Feb 2012 at 12:56

Are you talking about a DC source? As grounding DC through an inductor would cause one hell of a mess, it looks like a short to DC.

danza · 12 Feb 2012 at 13:09

Not A level or degree. It's an internally run work course. I'm doing a HND in Mechanical/Manufacturing engineering, and all that makes perfect sense. Electrical is something I've barely touched on. I'm employed as a mechanical design engineer, but the company is running this course so that the designers (and eventually the project team) have a better understanding of the electromagnetics involved in what we manufacture.

danza · 13 Feb 2012 at 19:12

Still having trouble as I progress.

- If the PSU was exchanged for one with a 1000A rating, what is the maximum current it could drive the inductor to and why?

danza · 13 Feb 2012 at 20:05

Am I right in saying:

1000A x 40x10^-3R = 40V, which basically means that the volt drop over the resisitors at 1000A is 40V, and seeing as the PSU is only 10V, it can only drive the inductor to 250A. So a 40V 1000A PSU would be required?