A Curious Question
What happens if the switch is closed?
What happens if the switch is closed? It should not come as a big surprise, the one capacitor will charge the other, and some energy gets lost in the resistor. We know the electrons which have charged the Capacitor C2 can not get out of this (circular) circuit. So in the end the electrons will be divided equally over both capacitors, so the charge and voltage will be the same. Voltage and charge behave linear. So half the number of electrons, or half the charge, will result in half the voltage. So each capacitor will have 5 Volts in the end. WIth a high impedance Voltmeter you can verify this, and it's not a spectacular result. What is interesting though, it makes no difference what is the value of the resistor. Perhaps also not very spectacular to see, but you can take 10 Ohms, 1000 Ohms, or any value, it always end ups with 5 Volts on each capacitor. Even try it without resistor, just touch an empty capacitor with a full one, and will give a little spark, and again exactly 5V on each capacitor. It is sure, the spark cost some energy, which was taken from the full capacitor.
Lets make the calculations. When we begin, the energy of the charged capacitor equals: 0.5 * C * U^2 . So at 10V, 100uF, that is 0.005 Joule.
After discharge, down to 5V, the energy of each of the two capacitors is: 0.5 * C * U^2, giving at 5V, 100uF: 0,00125 Joule. For the two capacitors, we find 0.0025 Joule together.
So, half the energy is lost. We can expect, it must have went into the resistor, or into the spark which is also a resistor of some kind.
This is also the reason why capacitors are not easy to use as an efficient way to store energy. A full charge and discharge cycle would destroy a large part of the energy. Unlike batteries, because these do not change their voltage very much. A battery charge and discharge cycle would loose no energy at all, if the voltage would stay constant. Such batteries do not exist, but we can approach this situation by simply not discharge the battery completely, and by keeping them on the constant temperature. In the same way capacitors CAN be used as efficient energy storage medium, but you should only charge or discharge them very little.
I have asked this question, just for fun, to Karel Elbert Kuijk, who worked at Philips labs, he has such an excellent technical background. Immediately he wrote me the mathematical proof (in Dutch) which I translated here:
As we can see, indeed half of the energy went into the resistor. But it gives an additional result as well: The value of the resistor plays no role for this, because it gets eliminated from the equations. With this last item, the curious part starts, but first comes the calculation:
From Karel Elbert Kuijk:
The current behaves like:
(1)
Where V is our start voltage of 10 Volt.
The charge on the second capacitor, if charged, equals:
(2)
The power, dissipated in the resistor R equals:
(3)
As you can see, the resistance gets eliminated in equation (3)
The missing energy
was dissipated in resistor R, and like proven by equation (3): regardless of the value of R.
* * * *
All the above is logical and normal. Now comes the curious question.
What happens of we do the experiment with theoretical ideal components? So that would be with capacitors of zero inner resistance, a switch without contact resistance, and closes endlessly fast, and all of these things. Still of course half of the energy gets absorbed in the resistor, the proof is written above here.
What happens if R is an ideal resistance, of zero ohms?
Perheaps this would come down to "limit" calculation, which is something virtual. It is something we can sometimes calculate, but even then, not understand it really. Like if you devide zero by zero, is the outcome zero, or endless?
But... is this really happening? Consider the following:
The moment, the switch closes, so at t=0, we have all electrons in Capacitor C1. Will these move in endless short time to C2, causing endless high current? If endless short time, they must l have endless high velocity, and this much we know: No velocity can be above light speed. So, a short time yes, but not endlessly short. It will take a real, discrete amount of time. Moreover, no phenomena can pass a distance in space faster than the speed of light. So when the "ideal" switch closes, it will take a very small amount of time, caused by the speed of light or somewhere below. During that very short time we have 10 Volt over a resistor of.... zero ohms.