Inductor (and Capacitor) Discharge

  • Thread starter Willa
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  • #1
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Hello folks, just having a few problems getting my head round LRC circuits and so have gone back to basics, first starting with Capacitors and Inductors discharging through resistors. At this stage I ask of you kind sirs two things: 1. Confirm that my understanding of the capacitor discharge (below) is correct. 2. Give a similar style explanation for Inductor discharge (I'm struggling with what causes the current to start to decrease in that one).

Capacitor Discharge: Charge starts to flow from one plate to the other through the resistor, as it does, the pd across the capacitor drops due to V = Q/C. By kirchoff's 2nd law, going round the circuit the pd across the resistor is equal and opposite to the pd across the capacitor, so the pd across the resistor drops. By I=V/R, the current drops, this means the rate at which charge is transferred between the two plates of the capacitor decreases...resulting in an exponential decay of charge on capacitor.

Is that all correct?

And so if someone can do a similar (qualitative) explanation of the inductor discharge that would be brill!

Cheers
 

Answers and Replies

  • #2
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Charge isnt being transferred between the two plates, charge is being stored on the plates and the potential is increasing between the two plates, thats what makes it a capacitor. This is while charging. Capacitors inhibit the flow of electrons, resulting in exponential growth of potential to the peak value, at the expense of exponential decay of current.

While discharging, the opposite occurs, the charge just decreases gradually, goes through the resistor, and out the ground.

I dont know much about inductors.
 
  • #3
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when i started talking about capacitor discharge, I implicitly assumed it started charged. And a capacitor doesnt need to be grounded to discharge it, just connect the two ends through a resistor and it discharges....that's the whole point! That's what i'm trying to describe. I'm trying to describe the discharge more thoroughly than "the charge gradually decreases"....I have given a reason for it's exponential discharge.
thanks for trying neway
 
  • #4
chroot
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whozum,

Please do not attempt to answer questions when your own knowledge is so sketchy.

Willa,

Your explanation of capacitor discharging appears to be accurate.

As for an inductor, there's a basic problem with your question. There's no way to "charge" an inductor, and then disconnect it from the circuit, leaving it "charged." The inductor stores energy in a magnetic field, and the inductor has to have current flowing through it to maintain that magnetic field. When you disconnect it from the circuit that was providing that current, the inductor will generate large voltages in attempt to keep the current continuous, and will (one way or another) dissipate its energy.

You cannot walk around with a charged inductor in your hand the way you can with a capacitor.

- Warren
 
  • #5
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Willa, did you edit your message? This isnt the same question I remember answering.
 
  • #6
chroot
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Also, for a mathematical explanation of capacitor discharge, consider that for a capacitor,

[tex]I = C \frac{dV}{dt}[/tex]

Also, consider Ohm's law:

[tex]V = R I[/tex]

Combining,

[tex]V = R C \frac{dV}{dt}[/tex]

which is a differential equation in V. A differential equation of this form has solutions that are exponentials.

- Warren
 
  • #7
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Lol I certainly havent edited my original question, I dont know what drugs you're on atm :p

And When I say "discharge an inductor" I guess I mean "Describe what happens in a circuit with an inductor and a resistor and there is an initial current in the circuit (god knows how it got there, it's just there). The decay is then a current decay, but I cant seem to describe it qualitively
 
  • #8
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I must've misread your question. I probably confused it with another one I answered just before. Sorry about that confusion, although E&M is definitely not my strongpoint, I do know a few things about it but I'll try to be safe.

Sorry for the confusion.
 
  • #9
chroot
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It's the same story, Willa:

[tex]
\begin{equation*}
\begin{split}
V &= L \frac{dI}{dt}\\
V &= I \cdot R\\
I &= \frac{L}{R} \frac{dI}{dt}
\end{split}
\end{equation*}
[/tex]

- Warren
 
  • #10
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No no, I said "qualitative". I fully understand how to push around equations to get a nice exponential decay, I just can't explain it QUALITATIVELY! Well at least I can't with confidence....that's what I'm after!
 
  • #11
chroot
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The qualitative explanation: The current and voltage drop exponentially to their final values of zero (volts and amps).

- Warren
 
  • #12
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No no no....that's the overall effect, which I know! But I want to know qualitatively WHY the current drop is exponential...without any equations. Something along the lines of "The current drops a little, the indcutor reacts by enforcing the current....this causes something, which means the rate at which current decreases gets smaller...hence exponential decay"....can you fill in the gaps in my understanding there please!?
 
  • #13
chroot
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Current flowing through the resistor dissipates power. The power dissipated by the resistor is equal to I^2R. As the current decays, the power dissipated decays even more quickly, causing the current to decay more slowly at later times.

- Warren
 
  • #14
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but hold on, that implies the inductor has pretty much nothing to do with the circuit....i.e. just a resistor on it's own with current flowing through it would decay. Surely there's something missing!?
 
  • #15
chroot
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Willa said:
but hold on, that implies the inductor has pretty much nothing to do with the circuit....i.e. just a resistor on it's own with current flowing through it would decay.
Not so. A resistor on its own would have no current flowing through it at all.

- Warren
 
  • #16
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once again, i propose the idea that a resistor on it's own, with an instanteous current flowing through it...what happeneds to the current. are you suggesting it would behave exactly like a resistor and inductor?
 
  • #17
chroot
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Willa,

Once again, I specify that even your instananeous-current model would not work. The instant a voltage source is removed from the circuit, the current through the resistor goes instantly to zero. This is quite unlike the case with an inductor present.

- Warren
 
  • #18
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but then your previous explanation doesnt seem complete. You said that the power dissipitated is I^2R, and that this causes the current to decay exponentially. That implies that a resistor on it's own with some current through it has exponential decay. Unless of course, you have been lacking in your explanation of current decay.

I guess what I'm after is something that directly involves the induced EMF of the inductor...i.e. a description along the lines of "The EMF induced is in such and such a direction...this causes this to happen to the current, which means the EMF changes like this, and that causes the current to drop by less". That would be ideal, because that's the type of understanding i'm after.

Sorry to be a complete pain
 
  • #19
chroot
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Willa said:
but then your previous explanation doesnt seem complete. You said that the power dissipitated is I^2R, and that this causes the current to decay exponentially. That implies that a resistor on it's own with some current through it has exponential decay. Unless of course, you have been lacking in your explanation of current decay.
No, I didn't say that. Try "reading."

- Warren
 
  • #20
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chroot said:
Current flowing through the resistor dissipates power. The power dissipated by the resistor is equal to I^2R. As the current decays, the power dissipated decays even more quickly, causing the current to decay more slowly at later times.

- Warren
you just state the part in bold without explaining

Cant you see that your explanation is completely independant of the inductor - you describe a situation with current flowing through a resistor, which then dissipitates power. That is all you've described....I cant see any connection in the above statement to an idcutor or it's induced emf behaviour!
 
  • #21
chroot
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The inductor is not an infinite energy source, so its current must eventually decrease. When it decreases, for any reason, the power dissipated by the resistor decreases, causing the current to decrease more slowly.

You are correct that the behavior has nothing specifically to do with the inductor; this is a common physical situation of which the inductor is just one example. For the sake of this explanation, the inductor is just a current source with a finite energy supply. The exact same behavior (governed by the differential equation) would be observed with a battery, for example.

- Warren
 
  • #22
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The inductor is not an infinite energy source, so its current must eventually decrease. When it decreases, for any reason, the power dissipated by the resistor decreases, causing the current to decrease more slowly.

You are correct that the behavior has nothing specifically to do with the inductor; this is a common physical situation of which the inductor is just one example. For the sake of this explanation, the inductor is just a current source with a finite energy supply. The exact same behavior (governed by the differential equation) would be observed with a battery, for example.

- Warren
tat reason is what willa's after...
 
  • #23
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Note on Inductor "discharge"

I prefer to call this de-energizing. The inductor doesn't store charge, it stores energy in it's magnetic field.

There us a marked qualitative difference between the energizing and de-energizing of an inductor in a DC circuit.

Energizing: CLOSE a switch to connect an inductor to a battery (6V) and the current builds up inverse exponentially, peaking at Imax = V/R, while the voltage starts at 12V and decays exponentially as described in previous posts. Put a neon lamp (90V) in parallel with the battery and the lamp won't light up since the voltage won't reach anywhere near 90 V, it won't get above 6V.

De-energising: Already have current flowing in the inductor, then suddenly OPEN the switch and an initial large transient voltage is induced across the inductor. Enough to light up the neon lamp! How come?

Shouldn't the current reduce gradually as before (when the switch was closed)? Turns out no, it's not symmetric w.r.t to opening and closing the switch, not necessarily anyway.

The induced voltage across the inductor is proportional to dI/dt, and since this is very large as the current drops to zero extremely rapidly when the switch is opened, the inductor back EMF in reaction against the rapid drop in current is large, large enough to light up the lamp usually (with a good quality switch).

This is an easy experiment effect to demo. Nevertheless, this effect is hard to explain to students. I myself still do not quite understand WHY the current drops more rapidly with time when the switch is opened, more rapidly than when the current increases when the switch is closed. Is it due to the property of switches? Or the fact that in one case the battery is being introduced into the circuit and the other case it is being dropped from the circuit? Why this asymmetry in the initial rate of change of current?
 
  • #24
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I think you might be looking for something like I=(V/R)*(e^((-Rt)/L)). But then again, maybe not.
 

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