What happens when you short circuit a source?

[Bassalisk]

I attached a relevant image.

I was doing some circuit solving methods, and I came across this as a riddle. If I short circuit a resistor, logically, all current will go through a wire on top.

But will it be the same if you short circuit a source? I mean can it be that source goes out of existence?

I mean from a REAL situation, a wire has some internal resistance r, and this wouldn't be a problem, but in a circuit analysis, where wire is ideal, and there are no losses, what will happen?

 

[Quinzio]

Yes, just like you say, the source goes out of existence, because it's easy to short circuit with a wire of rather low resistance. And probably, in real world, you would see some smoke coming out of the source.

 

[Bassalisk]

Hmm but its really unintuitive to think that the source will be wiped out...

 

[phinds]

Hmm but its really unintuitive to think that the source will be wiped out...

Why. It DOES seem intuitive to me to think that the source will be wiped out, and if you've ever short-circuited a power supply with a big enough wire (or whatever) you'll know that "wiped out" can be very literal.

 

[DragonPetter]

What does the R represent?

You have not short circuited the voltage source that you show in your schematic, unless that is supposed to be the internal series resistance of the voltage source.

 

[DragonPetter]

OP, think of short circuiting a voltage source as drawing infinite current out of it, this pushes whatever real source to a physical limit of trying to generate infinite power.

Also think about it this way, the voltage source has a finite voltage across it, but by shorting this voltage with a 0 ohm wire, you make the voltage across it 0. This is a contradiction between the voltage source voltage and the voltage drop across the wire, and so you're creating a physical impossibility, since there is no such thing as infinite current or infinite energy for that matter. This is an example where you can see a difference between an ideal mathematical circuit and the real world physics of circuits. Of course, the mathematical circuit is still valid, but it shows you where the limits are.

The physical process of any particular voltage source (chemical, electromechanical, etc.) will have a finite maximum amount of charge that it can move from one potential to another per a second, and so there is an internal resistance to the source that limits the current even when you attach it to a circuit that will not limit it. This is a way of looking at how real world physics avoids this violation of the laws of thermodynamics that a simple mathematical model might allow for.

 

[Bassalisk]

Yea I think I understand now. Thanks all.