Where did the energy from the capacitor go?

Where did the energe go?

hi,guys
i have a question as following:

Charging or discharging a capacitor may cause energy loss even if no dissipative elements are apparent.
Figure 1 (a) shows a capacitor C1 charged to voltage Vi and no voltage on capacitor C2 before switch closure. C1 is equal to C2 and the energy in the system is:

Energy = (C1*V1^2)/2

After switch closure (b), the charge and voltage is divided equally between the two capacitors (conservation of charge) and the total energy in the system is:

Energy = (C1*(V1/2)^2)/2 + (C2*(V2/2)^2)/2 = (C1*V1^2)/4

Half the energy has disappeared. Where did it go?

hope for ur replies!

Where did the energe go?

There is a glitch in your equation:

Energy = (C1*(V1/2)^2) + (C2*(V2/2)^2)

We have C1=C2 that since I can write:

Energy = (C1*(V1/2)^2) + (C1*(V2/2)^2) =>

Energy = (2*C1*(V1/2)^2) =>

Energy = (2*C1*(V1^2))/4 =>

Energy = (C1*(V1^2))/2 and we have the same result, therefore there is no energy loss.

See ya

Re: Where did the energe go?

claudiocamera said:

There is a glitch in your equation:

Energy = (C1*(V1/2)^2)/2 + (C2*(V2/2)^2)/2

We have C1=C2 that since I can write:

Energy = (C1*(V1/2)^2)/2 + (C1*(V2/2)^2)/2 =>

Energy = (2*C1*(V1/2)^2)/2 =>

Energy = (2*C1*(V1^2))/4 =>
------------------------------------

Here, energy should be equal to (2*C1*(V1^2))/8 ??

Added after 43 seconds:

claudiocamera said:

There is a glitch in your equation:

Energy = (C1*(V1/2)^2)/2 + (C2*(V2/2)^2)/2

We have C1=C2 that since I can write:

Energy = (C1*(V1/2)^2)/2 + (C1*(V2/2)^2)/2 =>

Energy = (2*C1*(V1/2)^2)/2 =>

Energy = (2*C1*(V1^2))/4 =>
------------------------------------

Here, energy should be equal to (2*C1*(V1^2))/8 ??

Click to expand...

Click to expand...

Where did the energe go?

The message was edited to the correct way. There was a divided by 2 error.

the correct is:
Energy = (C1*(V1/2)^2) + (C2*(V2/2)^2)

We have C1=C2 that since I can write:

Energy = (C1*(V1/2)^2) + (C1*(V2/2)^2) =>

Energy = (2*C1*(V1/2)^2) =>

Energy = (2*C1*(V1^2))/4 =>

Energy = (C1*(V1^2))/2 and we have the same result, therefore there is no energy loss.

Ok now ?

Re: Where did the energe go?

Half the energy is lost in the process of physically moving the charges from one capacitor to the other.

Let's look at some math:

The energy it requires to move a charge q (in other words, the work done by the electric field) is

\[W = q\Delta V\].

However, the value \[\Delta V\] is the voltage difference between the two capacitors and is not constant over time. Therefore, we must set up an integral over the charge transfer.

Let's call dq the incremental charge element that will be moved over from C1 to C2. And let's define \[\Delta V\] to be \[V_1 - V_2\]. If we call q the amount of charge moved so far, we can find \[\Delta V\] as a function of q.

\[V_1(q) = \frac{Q_o - q}{C_1}\], where \[Q_o = C_1 V_1\] is the initial charge on C1.

\[V_2(q) = \frac{q}{C_1}\]

\[\Delta V(q) = V_1 - V_2 = \frac{Q_o - 2q}{C_1}\]

Now we are ready to set up the integral:

\[W = \int\limits_0^{\frac{1}{2}V_1 C_1} \Delta V(q) dq\]
\[W = \int\limits_0^{\frac{1}{2}V_1 C_1} \frac{Q_o - 2q}{C_1} dq\]
\[W = \frac{C_1V_1^2}{4}\]

The limits of this integral are from 0 to \[\frac{1}{2}V_1 C_1\], because as you stated in the original post, the amount of charge moved over is \[\frac{1}{2}V_1 C_1\].

This is work done by the electric field and is equal to the energy lost.

Added after 2 minutes:

claudiocamera said:

The message was edited to the correct way. There was a divided by 2 error.

the correct is:
Energy = (C1*(V1/2)^2) + (C2*(V2/2)^2)

We have C1=C2 that since I can write:

Energy = (C1*(V1/2)^2) + (C1*(V2/2)^2) =>

Energy = (2*C1*(V1/2)^2) =>

Energy = (2*C1*(V1^2))/4 =>

Energy = (C1*(V1^2))/2 and we have the same result, therefore there is no energy loss.

Ok now ?

Click to expand...

claudio, your initial equation is incorrect. You dropped a factor of 1/2 in your energy equations.

Where did the energe go?

hi,jayc
i am very appreciated for ur answer, and ur reply really help me. another question, can we say capacitor is not a lossless element, and only inductor is lossless among pasitive elements? charging a capacitor cause energe loss anyway? if we use a voltage source to charge a capacitor, the input energe is twice of the energe storing in the capacitor?

Re: Where did the energe go?

wanily1983 said:

hi,jayc
i am very appreciated for ur answer, and ur reply really help me. another question, can we say capacitor is not a lossless element, and only inductor is lossless among pasitive elements? charging a capacitor cause energe loss anyway? if we use a voltage source to charge a capacitor, the input energe is twice of the energe storing in the capacitor?

Click to expand...

I believe an element can be considered lossless if it conserves energy over time. For example, if we apply a DC voltage accross a capacitor, once it is charged, the current supplied goes to 0 and the power dissipation is 0. Similarly, if we apply an DC current through an inductor, over time, the voltage drop on the inductor will go to 0, and again, power dissipation is 0.

A resistor, on the other hand always has a power dissipation of IR^2, and is only 0 if the input is 0.

This also holds true for non-zero frequency inputs. This can be seen through a phasor diagram, if you are familiar. The impedance of a capacitor is \[Z_C = \frac{1}{j\omega C}\] and an inductor is \[Z_L = j\omega L\]. Both of these elements have purely imaginary impedances, so on the phasor diagram, their output currents (given an input voltage) either lag or are ahead by 90 degrees. If we integrate the product of current and voltage (to get power) over time, they will always be 0. This is the same as integrating a sin and cos product.

However, note that we are talking about ideal devices. In general, all capacitors will have some series resistance which introduces loss at non-zero frequencies. Similarly, inductors have series resistance and parasitic capacitance as well.

In your case, you are actually reconfiguring the circuit by closing the switch, so you can expect some abnormal behavior. That particular situation you described is a well-studied topic called charge sharing. It really has not much to do with the determining if an element is lossy in terms of circuits.


jayc

Where did the energe go?

hi jayc:
we know that the energey lost by moving the Charges according ur post

but i want to know what the energy lost convert into???
It convert into magnetic wave ,hot,.........?????

thanks

Re: Where did the energe go?

As for the question of where the energy goes?

Believe it or not the energy is lost to heat in the resistive element of the capacitor or any series resistance. If the resistance is very low, the current is very high, but the energy lost is the same. If the resistance is low enough, the current will be high enough to spark over the switch and energy will be lost to radiation and air ionization and heating.

If you want to look at ideal elements, the current will approach infinity as the resistance approaches zero, however the equation for power will converge on the exact amount of energy needed. It is a weird effect and not at all intuitive, but has been well studied as mentioned before.

Where did the energe go?

Thanks newelltech
Thanks for ur clear explaination!!!!

Re: Where did the energe go?

Ditto to newelltech's explanation. That is the correct explanation.
If you want to better understand that, add a resistor in series with the switch. Then calculate the power lost in that resistor. You will find that the power dissipated in the resistor from t0 to infinity is equal to the power lost by the initially charged cap.
More interestingly, you will find that the power dissipated in the resistor is the same, regardless of the value of the resistor.

Re: Where did the energe go?

That's really a interesting phenomenon !!!!!!!!

Radiation,well,would some one please explain that:

when the circuit have a resistence,we can calculate that all the energy lost is Equivalent to the energy consumed by the resistence,so why the energy lost are all consumed by the resistence,and without no radiation such as magnetic wave????
i my view i think the energy lost may be Equivalent to the energy consumed by the resistence plus the energy convert into Radiation wave........
but when i calculated,i found the energy lost is Equivalent to the energy only consumed by the resistence,so i got confused~~~~~~~~~~~because we know that when there is a current flow through a Traverse it must cause a magnetic field and we have magnetic-wave,~~~~~~~

thanks~~~~~~~~