Two capacitors are parallel connected with an open switch. Both have a different capacity in which: $$c_1>c_2$$ and both charged with a different voltage $$v_1\neq v_2$$ and now we close the switch.

What will the voltage be on the capacitors and will it hold Tellegen's theorem?

I believe it won't, but I couldn't write a proper proof or to find the common voltage.

  • \$\begingroup\$ yes, they are not connected until we close the switch \$\endgroup\$
    – user107761
    Jan 26, 2014 at 17:33
  • \$\begingroup\$ well, of course voltages will be equalize, but to what voltage, and what will happen to the energy within the circuit? \$\endgroup\$
    – user107761
    Jan 26, 2014 at 17:37
  • \$\begingroup\$ I think you can't really treat this as a lumped network because you get a singularity (infinite current). You'd have to add resistance or inductance. In the latter case, energy would slosh back and forth between the two capacitors until something else (like EM radiation or eddy currents) dissipated it. So, Tellegen's theorem can't be applied. \$\endgroup\$ Jan 26, 2014 at 18:01

3 Answers 3


Consider this circuit:


simulate this circuit – Schematic created using CircuitLab

I know you didn't specify a resistor in the circuit. Its purpose will become clear later.

Let's say that initially \$V_{C1} = 1V\$ and \$V_{C2} = 0V\$. The charge in C1 is:

$$ Q_{C1} = CV = 1F \cdot 1V = 1C $$

The total energy in the circuit is the same as the energy in C1, because there is no other stored energy elsewhere in the circuit:

$$ E_{C1} = \frac{Q^2}{2C} = \frac{(1C)^2}{2F} = 0.5J $$

When the switch is closed, some current flows. The total charge in the circuit must remain the same, and we can see that the voltage across the capacitors must be equal once the circuit reaches equilibrium.

$$ Q_{C1} = Q_{C2} = 0.5C $$

$$ V_{C1} = V_{C2} = \frac{Q}{C} = \frac{0.5C}{1F} = 0.5V $$

The energy in the capacitors is:

$$ E_{C1} = E_{C2} = \frac{(0.5C)^2}{2F} = 0.125J $$

We have two of these capacitors so the total energy is twice that, 0.25J. Initially we had 0.5J. Where did we lose half the energy?

Consider that in the instant the switch was closed, there is 1V across R. The current is thus 1V/R. The power is thus:

$$ P = EI = 1V \cdot \frac{1V}{R} = \frac{(1V)^2}{R} $$

As you decrease R, the power goes up, approaching infinity:

$$ \lim_{R \searrow 0} \frac{(1V)^2}{R} = \infty $$

Thus, the lost energy was lost as heat in R. The energy lost is the same for any value of R. R can't be made equal to 0Ω without resulting in infinite power, which is impossible.

Incidentally, this is why charge pumps can't be 100% efficient.


The problem here is that connecting two capacitors with different charges will result in an infinite amount of current and this is the basic problem in analysing the circuit. If you introduced a small resistor (call it the switch contact resistance), you can derive a formula that predicts the final voltage across the capacitors.

But energy will be lost in the resistor so, using the formula you can assume R gets progressively smaller and smaller and for each reduction in R you will find that the initial current gets bigger and bigger and you should be able to notice that the \$i^2\$R.t loss does not actually get any smaller - it approaches a constant value and the smaller R gets you'll find that the energy loss remains the same. This indicates the final energy loss.

Please consider that you can't short the two capacitors together and hope to get sensible results by just assuming that the initial individual energies stored in each capacitor will equal the final energy once they are in parallel. This doesn't happen in the real world and it won't happen in the theoretical world either.

  • \$\begingroup\$ Even if we assume that the current is infinite for some period of time, won't its integral be finite and equal to the charge that moves through the switch? \$\endgroup\$
    – Joe Hass
    Jan 26, 2014 at 17:55
  • \$\begingroup\$ I think your claim, " it won't happen in the theoretical world either" needs to be explained. Why should this be the case? By the way, it is not only a theoretical question. What happens in the real word when there is a superconducting switch and superconducting wires. Where is the energy going, you claim that is lost? That would violate the conservation of energy principle, wouldnt it? \$\endgroup\$
    – Andreas H.
    Jan 26, 2014 at 17:56
  • \$\begingroup\$ @AndreasH. in the real world, with superconductors there will still be inductance and you will find that the voltage oscillates between one cap and the other. \$\endgroup\$
    – Andy aka
    Jan 26, 2014 at 18:00
  • \$\begingroup\$ i think there will be a spark. that will be the energy loss. \$\endgroup\$
    – user107761
    Jan 26, 2014 at 18:02
  • \$\begingroup\$ @AndreasH. With superconductors, you can eliminate the resistance, but you still have inductance. What you have then is an LC tank circuit. It won't oscillate forever: you can also think of it as a small loop antenna, and the "resistance" in the circuit is the radiation resistance of the antenna. That should make it obvious where the energy goes: it's lost to EM radiation and near-field coupling with non-superconducting surroundings. \$\endgroup\$
    – Phil Frost
    Jan 27, 2014 at 2:29

What has been forgotten is that as the resistance R decreases the effect of the inductance L of the circuit becomes more significant - you have a loop of wire.
So in fact you have a series LCR circuit with the condition for critical damping

$$R^2 = \dfrac{4L}{C}$$

When you do your analysis of a CR circuit you are in fact analysing an LCR circuit but with the assumption that the resistance of the circuit is much greater than that for critical damping and so the circuit is over-damped and you get nice exponential curves.

If the resistance is very low then the circuit is under-damped and the charges/currents/voltages undergo damped simple harmonic motion.
Under such conditions the accelerating charges because they are unbound (free electrons in a conductor) emit electromagnetic radiation which is the basis of a radio transmitter.

So some of the energy dissipated in the circuit ends up as heat (ohmic heating) and some as electromagnetic radiation and at no time does the power dissipated become infinite.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.