# Voltage across an ideal inductor

Discussion in 'Homework Help' started by LightAce, Sep 23, 2010.

1. ### LightAce Thread Starter New Member

Sep 19, 2010
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The question is: Assuming that both the switch and the inductor are ideal, what voltage appears across the inductor when the switch is opened?

I've spent a while with other people trying to figure it out but we couldn't get an answer or any formula to get an answer, so we all agreed on infinity which is not very likely. Any ideas on answering this question is much appreciated.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Infinity is the only option you have - unless the voltmeter is non ideal.

3. ### Georacer Moderator

Nov 25, 2009
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In an inductor, the current that flows through it is a continuous function. That is, it can't make leaps. If in a moment $t_-$ it has a value of I, the exact next moment $t_+$ it will be I too, no matter the voltage applied to it. The inductor will develop the necessary voltage on its terminals to make sure this continuity will hold.

Consequently, when the inductor of your exercise is penetrated by 2 Amps, and the switch is opened, suddenly the loop is broken the current stops flowing through the inductor. But the inductor needs to maintain the 2 Amps through it and through the switch.
Since now the switch is opened, its effective resistance is infinite. We can replace it with an infinite resistor R. So, if the current through the inductor is 2 Amps, we can write the equation
$I=\frac{V}{R}\\
\Leftrightarrow 2=V/\infty\\
\Leftrightarrow V=\infty$

since 2 is finite.

When the switch is not ideal, it has great but not infinite resistance at the moment it opens. The great but not infinite voltage created at the inductor is enough to penetrate the air gap between the poles of the switch and create an electric arc.

Note: Can someone please revise the logic path for explaining the infinite voltage? I searched on books and the web, but couldn't find anything other than RL circuits.

4. ### Ghar Active Member

Mar 8, 2010
655
73

$V = L \frac{di}{dt}$

When you open the switch the current goes from some value to 0 in 0 time. That's an infinite slope:
$V = L \frac{di}{dt} \to L \infty \\
V \to \infty$

You can plug in a ton of identities in that equation and work back to Faraday's law or whatever you want...

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I've been wondering if the inductor has a notional physical shape - whilst maintaining an ideal electrical 'form' - whether some of the stored energy would possibly be lost as a radiating electromagnetic wave. Would the voltage still be infinite?

Last edited: Sep 24, 2010
6. ### Georacer Moderator

Nov 25, 2009
5,150
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Does that actually happen, or current continues to flow through the open branch, polarising it?

7. ### budo New Member

Sep 23, 2010
5
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a: The current will continue to flow. Voltage will go negative until something breaks or a spark occurs across the switch.

b: negative infinity

c: It could be damaged by overvoltage

8. ### Ghar Active Member

Mar 8, 2010
655
73
Well I guess the logic is that the current flows through the infinite resistance open which relates to the infinite voltage which drops the current in zero time.

Since in reality the infinite resistance doesn't exist you get current through a finite resistance with finite voltage and it drops to 0 in non-zero time.

@T_n_k:

I'd need to confirm some things but I think you're right, arcing does radiate energy which is why some power tools and such can interfere with radio. The rapidly rising and collapsing voltage from arcing creates a broadband electric field. If you limit the arcing voltage you should reduce that radiated loss.

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The main thing that would limit the magnitude of the voltage that would appear across the switch if no arcing occurs would be the distributed capacitance of the inductor. If there were no resistance in the wire of the inductor, the distributed capacitance would form a tank and the inductor would then sustain an oscillation forever if there were no radiation. The peak voltage would be less than infinite, but it might be large. Even without radiation any nearby lossy dielectrics or conductors would eventually absorb enough energy to damp out the oscillations.