Capacitor Charging: Is There Something Wrong with the Analysis?

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    Capacitor Charging
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Discussion Overview

The discussion revolves around the analysis of energy consumption and storage in capacitors during charging processes. Participants explore the energy increase when charging a capacitor from different voltage levels and the implications of ohmic losses in the circuit.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the energy increase when charging a capacitor from 1V to 2V should be calculated as 0.5*C*(2-1)^2, while others initially proposed 0.5*C*(2^2-1).
  • There is a discussion about the energy consumption being equal in both scenarios despite differing voltage increases, with some participants suggesting that ohmic losses should be considered.
  • One participant mentions that the work done to charge a capacitor is the same for any one-volt increment, emphasizing that the energy required is consistent regardless of the starting voltage.
  • Another participant clarifies that the stored energy and the work done in charging the capacitor are conceptually the same, but there is confusion regarding the terminology used.
  • Concerns are raised about the complexity of the analysis and whether ohmic losses in capacitors can be neglected, with some arguing that while ideal capacitors have negligible losses, real capacitors do exhibit some level of ohmic loss.

Areas of Agreement / Disagreement

Participants express differing views on the calculation of energy increase and the role of ohmic losses, indicating that there is no consensus on these points. Some participants agree on certain aspects of energy storage and work done, but disagreements persist regarding the implications of ohmic losses and the correct formulas to use.

Contextual Notes

There are unresolved mathematical steps regarding the energy calculations and assumptions about ohmic losses in real versus ideal capacitors. The discussion reflects varying interpretations of the principles involved in capacitor charging.

seahs
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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 :confused:

Is there something wrong with the analysis?
 
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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.
 
n.karthick said:
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.


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

Sorry, I was wrong :frown:

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

Is there something wrong with the analysis?

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.
 
It takes the same energy to increase the voltage, that is further charge a capacitor, in anyone 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...
 
willem2 said:
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.

ohmic loss = QR = i*t*R = V*t?
 
Naty1 said:
It takes the same energy to increase the voltage, that is further charge a capacitor, in anyone 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...


I thought stored energy E = 1/2cv2 instead of work W = 1/2cv2
 
seahs said:
ohmic loss = QR = i*t*R = V*t?

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 [tex]V e ^ {\frac { -t } {R C}[/tex]

And the energy loss in the resistor is twice as big.
 
  • #10
I thought stored energy E = 1/2cv2 instead of work W = 1/2cv2

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:
  • #11
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...
 

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