Current through short-circuit inductor

[Fairouz84]

Hi,
If we short circuit a simple basic inductor in an AC system, what happen to the inductor and the current at the branch?

Inductor do not has resistance, so do the short circuit line.
Does the current will divide equally through each branch and only some lagging at the inductor branch ?

Or is it that the current will only go through the short circuit line?

Thanks for your interest

 

[vk6kro]

If it is an AC circuit, the inductor will have some reactance but a perfect short circuit will have no resistance or inductance.

So, in this case all the current would go through the short circuit.

 

[the_emi_guy]

An inductor with its leads shorted is analogous to a capacitor with its leads open. Whatever current was flowing through the inductor when the leads were shorted (initial condition) will continue to flow around this loop forever. This is analogous to opening the leads of a capacitor, it will simply hold, indefinitely, whatever voltage was across its plates when the leads were opened. Both the shorted inductor and the open capacitor are used as energy storage in high energy experiments (using superconductors in the inductor case).

 

[yungman]

Yes, the lower the resistance the shorting line, the longer it takes for the current to die down. But there is no perfect conductor in normal situation, so the current will die down and all current will eventually goes through the wire not the inductor.

 

[sophiecentaur]

Yes, the lower the resistance the shorting line, the longer it takes for the current to die down. But there is no perfect conductor in normal situation, so the current will die down and all current will eventually goes through the wire not the inductor.

For this setup I think you have to assume that the Inductor also has zero resistance - so the current ( from the external circuit) will be shared by both.
But, yet again, the results get crazy when we start to use truly ideal components.

 

[yungman]

For this setup I think you have to assume that the Inductor also has zero resistance - so the current ( from the external circuit) will be shared by both.
But, yet again, the results get crazy when we start to use truly ideal components.

Yes, I stand corrected if the inductor is consider perfect conductor. But as I said, it real life that is hard to get. A simple short is going to be much lower resistance than a coil and current dominates in the short circuit side.

 

[vk6kro]

For this setup I think you have to assume that the Inductor also has zero resistance - so the current ( from the external circuit) will be shared by both.
But, yet again, the results get crazy when we start to use truly ideal components.

The inductor at least has inductance, so some inductance has infinitely greater impedance than a perfect short circuit in an AC circuit.

 

[yungman]

Getting too theoretical here, in any normal sense, the resistance of the inductor is higher than the dead short and most current goes through the dead short. Super conductor is not practical in common cases.

call me ignorant, have we achieve true super conductor yet? I mean that the resistance of a 1mm diameter wire has the same resistance of a 1m diameter wire per long length where both are 0.000000000000000000000000000000000...ohms?

 

[vk6kro]

Just using the assumptions in the question, the inductor has no resistance but it does have inductance and therefore impedance in an AC circuit.

The short circuit has no resistance

Just on those assumptions, then ALL the AC current will go through the short circuit. Some impedance is infinitely greater than no impedance.

If you change the assumptions, then the result will be different.

 

[yungman]

Just using the assumptions in the question, the inductor has no resistance but it does have inductance and therefore impedance in an AC circuit.

The short circuit has no resistance

Just on those assumptions, then ALL the AC current will go through the short circuit. Some impedance is infinitely greater than no impedance.

If you change the assumptions, then the result will be different.

If short out the signal, you have no AC, you have no impedance in the inductor!!

Sorry, I am just pulling your leg! Have a nice weekend.

 

[the_emi_guy]

Getting too theoretical here, in any normal sense, the resistance of the inductor is higher than the dead short and most current goes through the dead short. Super conductor is not practical in common cases.

call me ignorant, have we achieve true super conductor yet?

Are we talking about a textbook problem here or a real application. If it is a real application you may need to describe the components, voltages, currents, and frequency in more detail.
I mentioned superconductor because I assumed this was a textbook problem with ideal components.
This is off-topic but, in fact, superconductors have *exactly* zero resistance. Check out the Wikipedia page.

 

[yungman]

Are we talking about a textbook problem here or a real application. If it is a real application you may need to describe the components, voltages, currents, and frequency in more detail.
I mentioned superconductor because I assumed this was a textbook problem with ideal components.
This is off-topic but, in fact, superconductors have *exactly* zero resistance. Check out the Wikipedia page.

That's a good question, I don't know. I tend to think in real circuit, not theoretical zero resistance, current goes on forever when you short the inductor and all. I don't even take this post seriously anymore with the super conductor came in.

 

[sophiecentaur]

Are we talking about a textbook problem here or a real application. If it is a real application you may need to describe the components, voltages, currents, and frequency in more detail.
I mentioned superconductor because I assumed this was a textbook problem with ideal components.
This is off-topic but, in fact, superconductors have *exactly* zero resistance. Check out the Wikipedia page.

The "zero resistance" of a superconductor is the same as the "zero resistance" of the lines on a schematic circuit diagram. Any 'real' superconducting conductor will be joined into a circuit with 'real' wires.

These concepts don't have a place when we're trying to talk about taking practical things as far as we can. There's always an R in there.