# LR circuit confusion

Discussion in 'General Electronics Chat' started by Vorador, Mar 3, 2013.

Oct 5, 2012
87
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Hi,

I was going through my circuit theory text-book and saw an example where the voltage induced in the inductor after the removal of the original source using a switch that includes larger resistors in series with the same inductor than the resistors before throwing the switch. The example demonstrated the property of inductors to effect no instantaneous change in the current flowing through it. As a result, more voltage is necessary to drive the same amount of current through larger resistances so the inductor self-induces a higher voltage across it than the source!

The example made no mention about it but something doesn't feel right to me about that, because it seems to me that the inductor would be liberating more energy than it stored, violating the law of conservation of energy. I used the equation P=V*I to calculate the power before the source was removed using the switch and after and I got a higher value for when the source is not present.

I don't understand why this must be.
Any help would be very much appreciated!

2. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,435
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On the contrary, this is precisely how inductors work. One application of this principle is the spark coil in an automobile: the distributor opens a switch driving the coil, which kicks back a large voltage across the spark plug causing ignition.

Energy is always conserved. However, energy stored in an inductor is not related to V*I but instead equals (1/2) * L*I^2 L (half the inductance times current squared ).

3. ### #12 Expert

Nov 30, 2010
16,664
7,311
The inductor is not producing more power, it is producing more voltage.

Oct 5, 2012
87
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Ah right. I totally overlooked the fact that V*I applies to resistors..

So I'm guessing that, assuming a constant current, the inductance must decrease to keep the energy conserved?

5. ### crutschow Expert

Mar 14, 2008
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3,361
V*I is the power dissipated in the series resistors not the inductor.

6. ### WBahn Moderator

Mar 31, 2012
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But more voltage at the same current IS more power!

The inductor is, indeed, producing more power. Assuming the current was constant before the switch was thrown, the inductor was neither producing nor absorbing any power at all.

But don't confuse "conservation of energy" with "conservation of power". The first is a law, the second is a fiction -- there's no such principle.

If you draw energy out of the inductor at a higher power than it was put in, all that means is that it will take less time to extract the energy than it took to put it in. Nothing more, nothing less.

7. ### #12 Expert

Nov 30, 2010
16,664
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OK. Time is a factor in this one.

8. ### tgil New Member

May 18, 2011
19
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V*I applies to inductors but it refers to the power dissipated in the inductor rather than the energy stored in it. The voltage drop across the inductor is proportional to the change in current (V=L*di/dt). When the current is constant through an inductor, it does not dissipate any power (L*di/dt=0=V) but it does store energy equal to 1/2LI^2 in the form of magnetic fields.

When you flip the switch, the current tries to go to zero immediately because the potential driving the current is removed. However, the magnetic fields around the inductor induce a voltage that counteracts the dropping current preventing the current from going to zero instanateously. While this is happening, the inductor is consuming power equal to V*I. However, most of the energy is discharged when the induced voltage gets high enough to cause a spark.

So you are seeing a power spike rather than an energy spike.

Oct 5, 2012
87
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That's interesting. I've always thought about how conservation of energy implies conservation of power (all the books I've read seem to take it for granted that it does).

Thanks for this very informative post and everyone else for your kind and useful response!

Last edited: Mar 3, 2013
10. ### WBahn Moderator

Mar 31, 2012
18,084
4,917
Hopefully it's a matter of having to read the books more carefully.

Consider almost any explosion -- imagine the the power into the explosive had to equal the power out of the explosive. Or the rate at which plant matter is converted to fossil fuels had to match the rate at which it is released in a power plant or car. Or if the trees had to take carbon dioxide out of the air and grow at the same rate they are converted back to gas when burned in a fireplace (or forest fire!).

11. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
Ok, I skipped may responses after reading the first couple, so please forgive me. It all depends on what the series resistance was while the voltage was applied. In certain circuits one can charge an inductor using one resistance, and discharge it using a different one. Ohm's law will always apply, and so V=IR. If R increases, then V increases. The total energy, however, should stay the same.

12. ### crutschow Expert

Mar 14, 2008
13,475
3,361
Don't confuse power with energy. Power is the using, transferring, or dissipation of energy over time, so conservation of power really doesn't occur (or have meaning), expect as to how it relates to the conservation of energy. Thus, for a given amount of energy the power generated will change with (and is inversely proportional to) the time over which it is occurring.

For example, the energy into generating an explosion can be provided at a small power level over a large period of time (such as blowing up a balloon) but generates a much larger power level over a short period of time when it is burst.

13. ### crutschow Expert

Mar 14, 2008
13,475
3,361
As long as you understand the difference between the energy dissipated in the resistors and the energy stored in the inductor. When charging an inductor, any energy dissipated in the series resistor has no direct relation to the energy stored in the inductor.

14. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998

We were discussing the energy stored vs the energy released in an inductor.

15. ### WBahn Moderator

Mar 31, 2012
18,084
4,917
Maybe. The OP was actually rather imprecise. He talked about using P=V*I but didn't say which V was being used. I rather suspect that he was using the voltage across the resistors. If that's the case, then he was comparing apples and oranges to begin with unless he was clear that he was talking about the power being dumped into the resistors by the source compared to the power being dumped into the resistors by the inductor and that the comparison says nothing about the energy stored in the inductor or the power applied to the inductor as it was being stored.

16. ### Brownout Well-Known Member

Jan 10, 2012
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He stated energy stored in the inductor, but used the wrong equation. The very next post corrected the equation.

Oct 5, 2012
87
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Yeah, probably. My bad.

Alright so I think I understand it now, but I have just one more question concerning a source-free parallel RLC circuit problem. The initial data given in the problem is inductor current IL(0)=6A and capacitor voltage Vc(0)=0V.
Since it's a parallel circuit, the R,L and C all should have the same voltage across them which is 0 here. Yet there's a 6A current flowing through L and C. How is the current flowing without a voltage? I think this will only happen for a very small instant of time since the inductor resists instantaneous changes in current. Am I correct on this one?

You all have been a great help. Thank you so much!

18. ### WBahn Moderator

Mar 31, 2012
18,084
4,917
Current can flow without a voltage and it doesn't have to be for just an instant. In a superconducing magnet, currents of hundreds of amperes can be sustained for months without a power source. As long as the current isn't changing, there is no voltage.

This situation in your circuit would be as follows: Imagine a 6V votlage source with a 1 ohm internal resistance. Connect this source in parallel with your RLC circuit and wait for everything to settle down. What will the inductor current and capacitor voltage be? Now disconnect the source.

19. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
You're still thinking you need a voltage for current, ie ohm's law, I=E/R. But when you have inductances and capacitances, then you're not constrained by ohm's law. For example, for a capacitance, I=C*dv/dt. Even if V=0, dv/dt can still be nonzero, and current can flow through the capacitor for more than just an instance.

Another way to think about this is to realize that in an inductor or capacitor, curent and voltage are out of phase, and so current can be flowing even as voltage passes through zero.