# Capacitor short circuited !

1. May 28, 2014

### Hardik Batra

I have attached one circuit. I don't know why we consider the rightmost capacitor gets short circuited.
And what do you mean by short circuit ?

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2. May 28, 2014

### Drakkith

Staff Emeritus
The rightmost vertical wire is connecting both plates of the capacitor. There is no way for the capacitor to maintain a voltage difference between its plates, as any buildup of charge on one plate is take away through the wire to the other plate. So the wire "short circuits" the capacitor.

Another way to look at it:
Since the capacitor and the wire are parallel with each other they must have an equal voltage drop across them. If we assume the wire is a perfect conductor with zero resistance, the voltage drop across it is zero, leading to zero voltage drop across the capacitor as well.

Last edited: May 28, 2014
3. May 28, 2014

### Hardik Batra

voltage drop across it is zero then how the capacitor short circuit?

4. May 28, 2014

### ehild

The "short circuit" is that short piece of wire that connects the plates of the capacitor. We say: "the capacitor is short circuited".
If you have short circuit in some electrical appliance, it means that parts which should be separated, get into contact. There will be smoke and burning, unless the fuse blows ....

ehild

5. May 28, 2014

### CWatters

A "short circuit" exists between two points when something (eg a wire) having relatively low resistance is connected between those two points. That's what you have in your circuit...

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6. May 28, 2014

### Hardik Batra

What will happened when we connected low resistance wire?

7. May 28, 2014

### Hardik Batra

Short piece of wire that connect the capacitor. But from where short wire comes to connects the capacitor.

8. May 28, 2014

### sophiecentaur

It's on the diagram.

9. May 29, 2014

### CWatters

What do you think?

What is the voltage drop across a wire with low (or no) resistance?

V = I*R
R -> 0
V -> ???

10. May 29, 2014

### CWatters

See red wires on the drawing I posted.

11. May 29, 2014

### Delta²

Hm, viewing this problem from the point of view of classical physics i sense a contradiction... IF we accept that the rightmost loop has zero resistance, then by kirchoff's law the current will become infinite there and the capacitor will be discharged in zero time. BUT we know that the electromagnetic field will propagate with limited speed even through a zero-resistance wire, the induced current wave will propagate also with a limited speed, hence there can be no discharge in zero time. What do u say?

12. May 29, 2014

### Staff: Mentor

No contradiction, even in classical physics. This is just the difference between considering the steady state the system settles into when the voltages stabilizes (capacitor fully discharged, and stays that way) and the short-lived period during which the voltages are still changing. The latter case cannot be correctly analyzed in terms of ideal wires with zero resistance - no matter how small the resistance of the wire is, you have to take it into account when calculating the discharge time of the capacitor.

13. May 29, 2014

### Delta²

Lets just say that there is no contradiction because in that short-lived period during which the voltage and current changes rapidly, Kirchoff's laws do not hold. (They hold if we take into account the self-inductance of the closed loop and assume that the wave length of the current is much bigger than the dimension of the closed loop).

14. Jun 2, 2014

### CWatters

This thread is in danger of wandering from the OPs original question.