Why Does Voltage Across an Inductor Peak and Then Diminish?

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Homework Statement



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.
 
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There a couple of parts to this.

First, what's an expression for the voltage across a pure inductance?
 
I'm not sure I understand what you're asking me
 
Are you referring to Faraday's law?
 
cscott said:
Are you referring to Faraday's law?

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

L \frac{dI}{dt}.
 
I'm confused as to exactly which voltage this represents...

Would this be the measured voltage across the inductor?
 
cscott said:
I'm confused as to exactly which voltage this represents...

Would this be the measured voltage across the inductor?

Yes, assuming the inductor had zero resistance.
 
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?
 
cscott said:
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?

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?
 
  • #10
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?
 
  • #11
cscott said:
V = IR for a real inductor?

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

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?

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.
 
  • #12
Ahh that make sense.

Thanks a lot for your help.
 
  • #13
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?
 
  • #14
cscott said:
Could you verify one more thing for me?

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.

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

I, too, think so.

so it should lower the hysteresis

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.
 
Last edited:
  • #15
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. :)
 
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