Self-Inductance

cscott

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

When the current in a circuit containing a inductor in parallel with a galvanometer and battery rises (after closing a switch), why do I see a maximum voltage across the inductor that quickly diminishes to a steady voltage?

I thought during a current rise the emf is less due to self-inductance?

Or does the voltage jump then fall to steady when the current stops rising and there is no self-inductance?

Please help me understand what's going on.

George Jones

There a couple of parts to this.

First, what's an expression for the voltage across a pure inductance?

cscott

I'm not sure I understand what you're asking me

cscott

Are you referring to Faraday's law?

George Jones

Yes, which gives the magnitude of the voltage across and inductor as

[tex]L \frac{dI}{dt}.[/tex]

cscott

I'm confused as to exactly which voltage this represents...

Would this be the measured voltage across the inductor?

George Jones

Yes, assuming the inductor had zero resistance.

cscott

Alright thanks I understand that bit now.

Is there a circumstance where the voltage would drop to a value > 0 when the current became steady?

George Jones

Yes. A real inductor is made of wire that doesn't have zero resistance. A real inductor is modeled by a resistor in series with an ideal inductor. What happens when the current becomes steady?

cscott

V = IR for a real inductor?

I'm trying to interpret some dad for a lab:

<conditions>: <maximum>/<steady> (voltage)

laminated bar on E-core, close switch: 18/15
laminated bar on E-core, open switch: -6/0

so there is a toroidal coil on the middle post of the "E" core and a laminated core goes across the top to complete the magnetic circuit

Does this make any sense?

George Jones

Yes, this the voltage across a coil of resistance R when a steady DC current I goes through it.

I didn't actually seeing what went on in the lab, but I think this makes sense. When the switch is opened, dI/dt is negative, and the steady-state current is zero since the switch is open.

cscott

Ahh that make sense.

Thanks a lot for your help.

cscott

Could you verify one more thing for me?

If a iron bar is stuck to a core of a coil and a current is put the coil how does a piece of paper affect the amount of hysteresis of magnetism in the iron bar?

I think the paper would be a material of lower permeability so it should lower the hysteresis and the strength of the magnetization. Am I correct?

George Jones

I haven't run into the term hysteresis since I was an undergrad, which was more years ago than I care to admit.

Fair warning.

I, too, think so.

I'm not at all sure, but I think this increases hysteresis ("lagging" between B and H), which increases flux loss.

Someone who knows this stuff better than do I should comment.

cscott

Thanks for your help.

I just noticed you're in Saint John. I lived in Fredericton but I'm in Ontario going to Waterloo right now. Small world. :)