Assume a System with two capacitors. one is charged to voltage V and other has voltage 0 on it. So the energy in the system is (0.5CV^2).

Now lets says i connect the two capacitors in parallel. So the voltage across them will be V/2. Energy in the system is  0.5*(0.5CV^2).

Question is where does the remaining energy go as i dont have any resistor in charging capacitor.
Secondly is it possible to charge a cap with ideal voltage source.

I agree tat half of the energy is lost in resistor however small it is.

But if we dont have resistor how can we say that we have a loss. We get a infinete current for 0 time interval. Is it that capacitor will not get charged.
We can take the limit provided the limit exists. we have a discontinuity at the instant we turn on the switch. The node voltage is not known as the first capacitor will force it to be V and second zero at the same time.

I think like this..  
Impulse  in time domain transforms to constant in freuency domain. And the energy will remain constant. The current will be quadrature in phase to the current . So the integral of voltage times current gives zero. integral(sin(x)cos(x)=0). So we dont lose any real power or we dont get real current to charge the capacitor.

hi rfid,
what rakesh intention (i guess) is when you charge cap without resistor current looks like impulse. So impulse Fourier transform is white.

Thanks.

ok Assume the condition like this.. u r charging a cap with ideal voltage source.
I can sat that voltage in the source is a unit step function. We will get dirac delta function in current which is difficult to handle.

So assume that voltage in the voltage source as a ramp with slope mt. it become step when m tends to infinity.

In this case if we find the energy supplied by the battery is 0.5CV^2 in the limit m tending to infinity.and not CV^2 which we will get wen we have resistor.


This tells that if we dont have any resistor we wont have any loss in charging a cap. However even if we have a small resistor we will lose half of the energy if we charge at one go.

Similarly is it possible to explain that two capacitor problem

Hi Rakesh!

About the Dirac delta pulse. Yes, the value at t=0 is inifinity or not defined. But the integral is! That is the Dirac delta pulse, see also

http://en.wikipedia.org/wiki/Dirac_delta_function

This is more or less exactly what happens when you using r=>0. If you try to solve the electronic problem with two caps you will not be able to get expressions how the current and voltage looks like during the recharging, whithout putting in a resistor. (If you are a skilled math pro, I'm not, you maybe can use the Dirac delta function) This is like saying that the problem without a resistor is non-physical and the basic relationship of a capacitor, i=C*dv/dt, can not be solved for this problem.

Hi Rakesh!

Yes, the charging with a slope seems interesting. I do not have any explaination. You claim that the energy from the battery when a resistor is used is CV^2. I was trying to solve that problem with some differential equations. It seems to be a hard labor to get it the whole way... Have you done it or have you a simpler way to prove it?

Hi all,
      just see the derivation attached  for the problem ideal battery charging the capacitor.

True agreed. Math gets complicated with resistor, but u il get the same answer. This is because once the capacitor is charged u il get current through the cap as zero.

Wat do you think of the derivation that without resistor energy supplied by battery is 0.5CV^2 or there is any flaw.