I was recently asked this question to calculate the voltage across a capacitor. The problem is described as follows :-

Consider a capacitor "C" connected across a voltage source "V". Once the capacitor is charged fully, it is disconnected from the source. Now this fully charged capacitor is connected across another uncharged capacitor of capacitance "C". Assuming ideal conditions, what are the various parameters of the new circuit. Like the voltage, Equivalent Capacitance, Energy etc. at the equilibrium point?

Note :- Guys this is not a homework or assignment question. This was one of the question that was asked to test my ability in which I failed.

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    \$\begingroup\$ I am sorry I do not have time for an answer now, but what you ask is a well known "paradox". It is true, some energy will be lost even if the wires are perfect conductors. The simplifications engineers use are not suitable for this problem ( think about EM radiation losses). In many physics books there is a treatment of this problem, but you can find it in the EE book by Anant Agartwal in chapter 9, if I remember correctly \$\endgroup\$ – Fiat Lux Feb 17 '13 at 23:32
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    \$\begingroup\$ Moreover, you can also think that even if the resistance can be zero, the inductance of a circuit cannot, so you can model your system with inductors in addition to the capacitors. You will find energy will switch back and forth the components of your circuit, but the total will be conserved. You can even try with a circuit simulator to see what happens. \$\endgroup\$ – Fiat Lux Feb 17 '13 at 23:44
  • \$\begingroup\$ Thanks buddy for letting me know about the text book. I had previously tried modelling the differential equation in Simulink. But wasn't able to do it. It gave me solutions that were unstable. I think the answer listed in the book is what I was looking for. \$\endgroup\$ – LJanardhan Feb 18 '13 at 0:15

Solve the problem by first assuming that the capacitors are connected through a resistor with some resistance, say \$R\$ ohms. As long as there is any difference between the voltages across the capacitors, current will flow through that resistor and thus transfer charge from the first capacitor into the second capacitor. So the first capacitor discharges slowly (exponential decay!) into the second capacitor and the current ceases at \$t = \infty\$ when the two capacitors each have charge \$Q/2\$, and thus voltage \$V/2\$. Half the energy stored in the first capacitor is lost since when the discharge of one into the other is complete, they each have one-fourth of the initial energy. Where did this energy go? Will the answer change if we use a different resistor in this experiment?

After this, think about what happens if \$R\$ is infinitesimally small.


Since this may be a homework question, I'll just give a few suggestions:

What is conserved in this circuit? You mention energy in the title, you need to think about charge.

In all solutions of this sort you have governing equations and basic assumptions that you make about quantities that are conserved.

\$ Q=CV\$ is a conserved quantity. That means that once charged up, when you attach a new Capacitor you are changing the C value in the equation. But \$CV = \$ constant.

In other words, Energy is NOT conserved.

A more physical view may help. Lets make the capacitors with an air gap and the ability to change the plate spacing (ignoring fringe fields etc. - it's a thought experiment). To change the value of C (increase it) we need to decrease the plate spacing. Since the energy is decreasing that means that the forces present will drag the plates towards each other and do positive work against the system that controls the plate spacing. That means the system in order to decrease the energy must be attractive, i.e. the forces are generated that bring the plates together. We know this is the case because one plate will have +'ve charge and the other plate will have -'ve charge (charge balance again). SO the system is consistent.

For your system of two two caps, work IS done when the displacement current flows during charge equilibration.

  • \$\begingroup\$ Well this is not an Homework or assignment problem. It was a question that someone asked me when they were trying to challenge. But I failed miserably to solve at that moment of time & still trying to figure out. The problem looks simple. But when you try to use equations then it looks kinda difficult. When you say Q = CV = const. For the new ckt, C = 2C as they are in parallel. So Q = 2C*Vnew = CV. So Vnew = V/2. E = 1/2*2C*(V/2)^2 = 1/4 CV. which means that half of previous energy is lost. Which is impossible by using law of conservation of energy. So there should be something else. \$\endgroup\$ – LJanardhan Feb 17 '13 at 22:56
  • \$\begingroup\$ @LJanardhan then you have a fundamental problem that you have to resolve. Either you are creating/destroying charge OR you are expecting that there is no work done in a system where you are moving charge through a potential. \$\endgroup\$ – placeholder Feb 17 '13 at 23:13

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