Energy loss when two capacitors are placed in parallel

[usernam]

Suppose an uncharged capacitor is placed in parallel with another already charged capacitor of the same capacitance then the charge is split in half between the two capacitors and the total energy is reduced by half.
Where is the energy lost?
Is it lost during the transfer of charge between the two capacitors or in some other manner?

 

[Borber]

In the school we used to say that when you do this experiment the half of the energy disapeared in spark.

 

[samyer]

This is a simplified answer:

If a capacitor is charged to a given level, it 'holds' an amount of energy that is POTENTIAL energy. That energy becomes KINETIC during the discharge phase. Assuming that the second capacitor has the SAME electrical characteristics as the first, i.e., capacitance, ESR, working voltage, etc., then the only additional external component(s) that essentially come into play are the resistance of the conductors and the connections, i.e., wiring, etc. between the two capacitors. Using your example, what will happen is the following:

The first capacitor will discharge into the second capacitor at a rate determined by the time constant established by the capacitance of the circuit and the circuit (including the capacitors themselves) resistance and some capacitive reactance (we will ignore inductive reactance, although there will be some amount). If the first capacitor is charged to Vx, then the second capacitor will attempt to charge to the same level. However, there will be losses in the circuit due to resistance, etc. The second capacitor can only charge based on the current supplied by the first capacitor which is decreasing at the time constant rate established above. The charge curves for the two capacitors will be the inverse of each other. The total energy supplied to the first capacitor will be evenly distributed (minus the losses) between the two capacitors. The only energy 'lost' is that which is lost due to resistance heating, etc., based on the parameters of the given circuit. An analogy can be drawn with two buckets - - one empty and one full of water. The two buckets are at the same level and a tube is connected at the bottom of each bucket. When the water begins to flow, it will eventually reach a point where the levels are essentially the same. You might ask why can't I just move all of the water from the full bucket to the empty bucket? To draw a correlation to the capacitor circuit - that implies that the first capacitor's potential energy (e.g., the height of the full bucket) is greater than what it is at rest and fully charged. In other words, that is why you can completely move all of the charge from the one capacitor to the other without any external influence - - just the circuit you described.

Hope this helps.

I hope this helps.

 

[Borber]

OK your explanation is realistic. But in your explanation, it is obvious, some energy disappeared for heating and remaining energy is less than a half. The problem is raised without mentioning any losses, it has to be understood as abstract one.
E1=0
E2=C*V²/2

E3=2*C*(V/2)²/2=C*V²/4 ratio of E2 to E3 is 2.

 

[flatulent]

This is a classical problem in circuits class. What you do is start with the calculations of what happens when a resistor is used to connect the two capacitors. Then take the limit when the resistance goes to zero value. You will always get half the energy going into heating the resistor even at zero resistance.

This is very much like sinx/x is 1 for x being zero. You have to take the limiting case.

 

[usernam]

Yes I suppose it's not a question of whether the resistance is present or not.
Ultimately it reduces to the fact that there is a potential difference and charge is flowing across this potential difference.
So some energy is lost in this process.
After all potential difference is defined as the work required to move a charge from one point to the other.

 

[aryajur]

I think the best analogy I can think of in this case would be the potential energy of a body in the gravitational field. At a higher height it is more when it comes down it is less. Its more of a matter of state rather than losses.

 

[samyer]

I need to correct one thing in my response - - I need to change "...can completely move all of the charge" to "....CANNOT completely move all of the charge".
For the person asking this question, I've read the answers from others and I just wanted to point out how much you learn from various points of view.
Apologies,
Sam

 

[VVV]

Energy is lost primarily in the connection wires and internal resistances of the capacitors, while the charge is being transferred. A transfer of charge is a current and a current produces heating when passing through a conductor.
Then, of course, there is energy being dissipated in the spark itself and the electomagnetic field created by the changing current during the discharge.

Interestingly, if the two capacitors are connected by a resistor rather than directly, the same amount of energy is dissipated, this time mostly in this resistor.