RC ideal discharge energy - critical parameters

[t_maggot]

RC discharge energy

In the ideal RC discharge circuit (with finite C and R values)as in

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/capdis.html#c2

I want to know this :

The total energy (joule) consumed in the resistor (in this transient discharge phase)depends on its own value?

I think not because all the capacitor energy is "lost" as heat to end in a zero energy condition with V=0, but I appreciate a second opinion and if one tells me how to derive this from the equations.

 

[brmadhukar]

Re: RC discharge energy

If you integrate the total "energy" dissipated by resistor youu will get the answer. The power is not constant.
brmadhukar

 

[FvM]

Re: RC discharge energy

Calculating E = ∫ P = ∫I•U dt you get V0²•C/2, according to the total energy.

 

[Kral]

Re: RC discharge energy

t_maggot,
FvM is correct. However, you don't need to do any integration to get the answer. The energy stored in a capacitor is equal to CV^2/2. The energy dissipated in the resistor must equal the energy initially stored in the capacitor, which is CV^2/2.
Regards,
Kral

 

[FvM]

Re: RC discharge energy

However, you don't need to do any integration to get the answer.

I knew it before.

 

[t_maggot]

Re: RC discharge energy

Ok, thanks everyone, i think that is clear:

In the RC discharge circuit (with finite R≠0 and C values), the Energy dissipated in the R does not depend on its value.

So, the dissipated discharge energy (integrated power) for a given C and V is constant.

Now my real problem is not for this RC circuit but for a C||-R-||C circuit.
Discharge of one capacitor "inside" the other, trough an R. I talk for paralleling a charged with a discharged capacitor through one resistor.
**broken link removed**
In this circuit topology, what is the role of the R?
The same principle applies?
The dissipated energy (as heat) in the resistor is constant and does not depends on its value?

Or more practically, (if we assume that I replace the R with a mosfet) the power dissipated on the mosfet does not depend on its own Rds(on)?(!)

How I can calculate the amount of energy dissipated in the resistor and the amount transfered to the (discharged) capacitor ?
What are the equations describing this circuit topology? (I,V,C,E ?)

Any help or reference appreciated

 

[iandme25]

Re: RC discharge energy

As said earlier the energy stored by any capacitor is given by CV^2/2. now when u connect the charged capacitor c1, then according to the time constatnt of rc1 and rc2 the capacitor c2 will start charging through c1, untill its fully charged or the voltage of the both the capacitors becomes same. then the process will stop.(depends on the parameters of the both the capacitors).


now the time taken to charge c2 to the voltage when both the capacitors will have same voltage can be calculated by the time constants of rc1 and rc2.

now upto this time the c2 will keep charging and hence the losses in r will be only for that time, after that there will be no current flow.


hence the total energy that is dissipated in resistor and transfer to the c2 can be calculated by the integration of I^2R(for the time interval equal to the charging time of c2 untill the voltage of both the capactor becomes same) + C2V^2/2.

i hope i'm right and if not correct me.

but all these depends on the parameters of the both the capacitors, otherwise there could be a second solution, so for better answer post the parameters of the c1 and c2.

 

[t_maggot]

Re: RC discharge energy

The circuit I am talking about is theoretically and the parts C1,R,C2 are perfect. There is no inductance in, so I think the oscillation you describe, does not take place, simply the C1 discharge through C2 until there voltages become equal. And of course I am pretty sure that not all but only a part of the Initial condition energy is dissipated as heat in the resistor. Anyway my questions for the "C||R||C" circuit are still open.

 

[Kral]

Re: RC discharge energy

t_maggot,
The following development, based on conservation of charge, shows that the energy dissipated in R is independent of the value of R.
Let:
Edissipated = energy dissipated in R
Vi = the intital voltage on C1
Vf = final voltage on both C1 and C2
Ei = the initial energy stored in C1
Q1i = initial charge in C1
Q1f = final charge on C1
Q2f = final charge on C2
.
l) Ei = C1Vi^2/2
2) Q1i = C1Vi
3) Q1f = C1Vf
4) Q2f = C2Vf
. Based on conservation of charge:
5) Q1f + Q2f = Q1i = ViC1
6) Q2f = ViC1-Q1f
. Substitute 6) into 4)
7) ViC1 – Q1f = VfC2
8) Q1f = ViC1 – VfC2
. Substitute 8) into 3)
9) ViC1 – VfC2 = C1Vf
10) Vf(C1 + C2) = ViC1
11) Vf = Vi{C1/(C1 + C2)]
. Solve for the total final energy in both caps:
12) Ef = E1 + E2 = C1Vf^2/2 + C2Vf^2/2
. Substitue 11) into 12)
13) Ef = C1[C1/(C1 + C2)]^2 Vi^2 + C2[C1/(C1 + C2)]^2 Vi^2
14) Edissipated = Ei – Ef
15) Edissipated = C1Vi^2 – {C1/(C1 + C2)]^2 Vi^2 (C1 + C2)

Note that R does not enter into the equation for Edissipated.

I'm interested if anyone can spot a flaw in my logic or algebra.
Regards,
Kral

Added after 2 minutes:

I'm not sure how the "smileys" got in there, but they should both be 8 followed by a parenthesis.

 

[flanello]

Re: RC discharge energy

Karl,
I have looked through your algebra, and I think it is correct till line 12).
In 13) you have forgotten the /2 from 12).
Thus the correct form ist:

13) Ef = 1/2 {C1[C1/(C1 + C2)]^2 Vi^2 + C2[C1/(C1 + C2)]^2 Vi^2}
14) Edissipated = Ei – Ef
15) Edissipated = 1/2 {C1Vi^2 – {C1/(C1 + C2)]^2 Vi^2 (C1 + C2) }

R determines how fast the charging process takes place

 

[malizevzek]

Re: RC discharge energy

I agree with flanello,

I would, however, write the solution in prettier form:

Edissipation=½C1C2/(C1+C2)Vi^2

From here you can see that C1C2/(C1+C2) is equivalent capacitance seen from the resistor, so the expression can be written as:

Edissipation=½CeqVi^2, which is analog for the case of one capacitor with resistor.

The resistor value influences only the time constant RCeq, not the dissipation energy or final voltage Vi.

Djordje

 

[Kral]

Re: RC discharge energy

Good eye, flanello! Also, I agree that malizevzek's form is more elegant and compact. Thanks to you both.
Regards,
Kral

 

[t_maggot]

Re: RC discharge energy

Thanks to all, but I think that the above conclusion conflicts with the conservation law of energy
Please correct me if I am wrong:

If we asume we have this three parts

C1 = 1F/1V
R = 1Ω
C2 = 2F/0V

Before we connect the parts to a circuit we have:

QC1 = 1Coulomb,
EC1 = 1/2Joule

So the initial total energy we have is only 1/2Joule and it is from the Charged C1.
(C2 and R energy = 0Joule)

If now connect the parts in this cicuit
**broken link removed**
and according to the equations we have, the discipated energy in the R is

EDIS = ½[C1C2/(C1+C2)]Vi² = .... => EDIS = 1/3Joule

But from the law of charge conservation we have Q1F + Q2F = Q1i

So we have

VF = Vi[C1/(C1 + C2)] = .... => VF = 1/3Volt
Q1F = C1VF = .... => Q1F = 1/3Coulomb
Q2F = C2VF = .... => Q2F = 2/3Coulomb

Ok with the charges, now we can calculate the final enegies of C1 and C2

E1F = ½Q1FVi = ... => E1F = 1/6Joule
E2F = ½Q2FVi = ... => E2F = 2/6Joule

now, the total energy we have stored in the capacitors after the discharge phase is complete:

E1F + E2F = 1/2Joule

But we have already discipate 1/3Joule in the resitor so the total energy we have (EF + EDIS) is 5/6Joule > 1/2Joule witch is the initial energy!

 

[Kral]

Re: RC discharge energy

t_maggot,
The final energies in C1, C2 are:
E1 = C1Vf^2/2 = [1(1/3)^2]/2 = 1/18 J
E2 = C2Vf^2/2 = [2(1/3)^2]/2 = 2/18 J
.
Final energy stored in the capacitors is 3/18 = 1/6 J
.
So the total energy transferred is 1/3 + 1/6 = 1/2, which is equal to the inital energy stored in C1
.
Regards,
Kral

 

[flanello]

Re: RC discharge energy

I agree with Karl.

t_maggot,
the error of your calculation is in

Ok with the charges, now we can calculate the final enegies of C1 and C2

E1F = ½Q1FVi = ... => E1F = 1/6Joule
E2F = ½Q2FVi = ... => E2F = 2/6Joule

there you must calculate with the final voltage Vf and not with the initial voltage Vi:
E1F = ½Q1FVf = ... => E1F = 1/18 Joule
E2F = ½Q2FVf = ... => E2F = 2/18 Joule

 

[t_maggot]

Re: RC discharge energy

Oh! yes! This is a "small" mistake.

(and I have almost break the energy conservation law...)

Anyway thanks again Kral, flanello and malizevzek, I think that I have clear the things in my mind now.