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Design >> Analog Design >> Energy stored in capacitor
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Message started by Rakesh on Feb 4^th, 2011, 1:28pm

Title: Energy stored in capacitor
Post by Rakesh on Feb 4^th, 2011, 1:28pm

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.

Title: Re: Energy stored in capacitor
Post by RobG on Feb 4^th, 2011, 2:57pm

If you were to calculate the energy loss due to a finite switch resistance you would find that it is equal to (CV²)/4. Note that it is independent of the resistance value -- so the energy gets used up by the resistance even if it is zero (in the case of zero resistance you get an infinite charging current: ∞²*0 = (CV²)/4 in this case when you take the limit).

Or it gets converted to dark matter. I sometimes get those two mixed up ;).

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 4^th, 2011, 7:03pm

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.

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 5^th, 2011, 1:15am

I do not understand where you get sin(x) and cos(x) from. The voltages and currents when re-charging the two capacitors will not be sin(x) or cos(x), exp(x) it will be if you have a resistors in-between.

I agree with the explaination from RobG. The energy loss in an assumed series resistor is independet of the value of it. It you take the limit r=>0 you will get the result. The problem can not be solved without a resistor. You can not mathematicly describe the current flowing from cap1 to cap2 without any resistor, the equation does not exist.

Title: Re: Energy stored in capacitor
Post by raja.cedt on Feb 5^th, 2011, 1:39am

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.

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 5^th, 2011, 5:28am

Yes, the current shape will look like a Dirac-pulse, it will be nil for all times except t=0 but the integral from -∞ to +∞ is finit and not zero. (Strange thing but that is the Dirac pulse)

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 8:10am

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

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 8:15am

at Rfidea.
Agreed current will be a dirac delta. I dont know tha magnitude of current.. it is undefined.
first of all its not possible to differentiate a discontinuos function so wat to talk of current.
So integral of that from -inf to inf is undefined and not zero.

my question is i dont have any lossy element in the system.. Then wer does the remaining power goes off. I agree half will get lost even if we have very very small value of resitor. the case here is no resistor..

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 5^th, 2011, 8:39am

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.

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 9:00am

Hi rf idea,
Agreed integral od dirac delta is one..its given this is heuristic idea not proven though...
agreed its impractical to have a circuit with zero resistance. this ques is for understanding only...

jus consider this case charging cap from ideal voltage source..same prob exists..so change the supply voltage as a ramp and make the slope tending to infinity..
we will get energy supplied from battery is 0.5CV^2 and not CV^2 as we wil get if we aasume a resistor and making r tending tonfinity...

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 5^th, 2011, 9:17am

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?

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 10:53am

Hi rfidea..
i have derived it... i il post it after drafting it sooner...its really an interesting problem.. lets figure it out :D

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 12:45pm

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

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 5^th, 2011, 1:42pm

Yes, that is without any resistor, I agree. But the math get a lot more complicated when you add the resistor.

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 1:45pm

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. :-*

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 5^th, 2011, 2:19pm

Your derivation is correct. It also match the energy "law" of a capacitor, E=0.5CV^2.

But you are saying that you get the same answer with the resistor. In earlier post you say you get CV^2, if the voltage is a ramp.

I think you have mixed up the energy provided from the supply when it applies a step voltage into the RC network. In that case the energi provided will be CV^2 and independed of R. You can not use this relationship with a ramp with zero transition time. Then you can not solve the i=C*dv/dt relationship of the capacitor and we are back where we started with the two capacitors.

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 2:36pm

I think there is some misunderstanding.. Jus see this..
Voltage across battery will be iR +1/c integral(i dt). and that we need to integrate over time . So i think u il get loss term in resistor if we include it.
Answer should not change in watever way we do.
When u give a ramp the volage across the cap will not be derivative of ramp due to resistor. we need to solve some first order differential equation..

is this ok with you

Title: Re: Energy stored in capacitor
Post by loose-electron on Feb 5^th, 2011, 5:09pm

as for the original post - the energy remains the same - remember that 2 variables are changings - voltage and capacitance.

As for the Ideal voltage source and ideal capacitor - sure why not? Just need infinite current for a brief period of time.

BTW - if you can give me a true ideal voltage source, we can solve the worlds energy problems.

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 5^th, 2011, 5:17pm

Hi tats true...think of condition of charging cap with ideal voltage source..
Some how we migrated to this question.
Energy supplied by battery and energy stored in capacitor are the same.. We wont get any loss
Is it corect to say this...

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 6^th, 2011, 1:45am

Rakesh wrote on Feb 5^th, 2011, 2:36pm:

The equations you are starting with is ok for me since you include the resistor. If you solving the differential system, please not forget that the voltage ramp is flatting out which influence the current.

You claiming that the answer will not change. I do not understand which answer will not change? And you can not say anything about the answer unless you have sokved the differential system and have an expression of i(t) and the integrate v(t)*i(t).

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 6^th, 2011, 10:47am

i = Vdd*C/(T)*(1-exp(-t/RC)).

This equation we need to integrate from 0 to infinity . keep in mind we need to consider the flattening of voltage at battery after t =T.

Is it correct way of doing...

Title: Re: Energy stored in capacitor
Post by rfidea on Feb 6^th, 2011, 12:24pm

Yes, I got the same equation when trying to solve the ramp problem with a resistor. It is valid up to t=T. I did not proceed, I'm to lazy I guess :)

But you are on the right track. After t=T the voltage is constant and there is a new differential equation to solve, with the initial condition set by the first equation.

Title: Re: Energy stored in capacitor
Post by Rakesh on Feb 6^th, 2011, 12:47pm

May i know your name and the place wer u belong to... jus for fun :D