# Capacitor charging

1. Aug 24, 2010

### seahs

Consider the two scenarios below:
1) charging a capacitor from 0v to 1v
2) charging a capacitor from 1v to 2v

energy stored on a capacitor is 0.5*C*(voltage)^2. So the energy increase for capacitor is 0.5*C in 1) and 1.5*C = 0.5*C*(2^2-1) in 2)

Assume the ohmic loss are both 0.5*C since the voltage differences in both cases are the same.

The energy consumption should be C in 1) and 2*C in 2)

However, I'ave done the simulation and the energy consumption seem to be the same

Is there something wrong with the analysis?

2. Aug 24, 2010

### n.karthick

Yes this is wrong.
The energy increase in case 2 is 0.5*C*(2-1)^2 and not 0.5*C*(2^2-1).

As per your simulation, energy consumption is same.

3. Aug 24, 2010

### seahs

The initial energy is 0.5*C
The final energy is 2*C = 0.5*C*2^2

Why the energy increase is 0.5*C*(2-1)^2???

4. Aug 24, 2010

### n.karthick

Sorry, I was wrong

Energy increase=final energy - initial energy
= 2C - 0.5C
= 1.5 C
and not 0.5 C as I mentioned earlier.

5. Aug 24, 2010

### willem2

You're talking about ohmic loss, but a capacitor will have no ohmic loss.

I suppose you charged your capacitor through a resistor in your simulation. The ohmic loss in the resistor R is equal to QR, where Q is the charge of the capacitor. Q is equal to CV, so the ohmic loss (electric engergy converted to heat) in the resistor is indeed proportional to the voltage, and the ohmic loss in your 2 scenarios will be equal.

6. Aug 24, 2010

### Naty1

It takes the same energy to increase the voltage, that is further charge a capacitor, in any one volt increment.....so the work is the same in going from 0 to 1 volt as from 100 to 101 volts.

W = 1/2cv2 where v is the potential difference across the plates....

A way to think about this is that the charge increase is q = (it).....= CV....

so for a fixed difference in voltage (V) you have to push the same amount of charge to the plates...do the same amount of work....

7. Aug 24, 2010

### seahs

ohmic loss = QR = i*t*R = V*t???

8. Aug 24, 2010

### seahs

I thought stored energy E = 1/2cv2 instead of work W = 1/2cv2

9. Aug 24, 2010

### willem2

If you use a resistor R in series with a capacitor C, and a voltage source V, it will take twice as long to charge to a certain voltage if you double C, because

the voltage across the capacitor is $$V e ^ {\frac { -t } {R C}$$

And the energy loss in the resistor is twice as big.

10. Aug 25, 2010

### Naty1

it is not "instead of"....they are the same thing...both are correct........you should be aware that when a capacitor is charged the WORK done in moving charge onto the plates IS the change in potential also referred to as a change in potential energy and the symbol U [for potential energy] (as distinguished from kinetic energy) may also be used.

Suggest you carefully reread my prior post; moving electrons around, in this case onto the plates of a capacitor requires WORK....those electrons do not collect on the plates with their electrical repulsion voluntarily...they must be pushed....

It takes a while to get a feel for this stuff as you learn about it....

Last edited: Aug 25, 2010
11. Aug 25, 2010

### Naty1

Post #9 reflects a confusing perspective.....at least to me....much too complicated....

q = CV, so it you double the capacitance (C) then of course it holds double the charge (q) at some voltage V....and will actually take twice as long to charge whether or not there is an external resistor....because you have to push double the number of electrons onto the negative plate.

A capacitor does have ohmic loss, but it is so small we assume an "ideal" capacitor, that is a circuit element with ONLY capacitance.....To say a capacitor has no ohmic loss is like saying a capactor has no inductive loss..it does, but we can neglect that tiny loss because it is negligible in comparison to the impedance of the capacitance....