# Entropy change in an RC circuit

Hello,I would like some help for a problem

## Homework Statement

Initially:At t=0 [/B]the cylindrical capacitor of capacitance $c=\frac{\epsilon s}{d}$ (d the distance between the 2 electrodes and s their surface; $\epsilon = \epsilon(T)$ is the dielectric permittivity) is discharged and we close the circuit and charge isothermally the cylindrical capacitor until it has a $c*v$ charge.
The current is $i=\frac{dq}{dt}$ and the voltage is v(t), the electric charge of the capacitor is $q(t)=c*v(t)$.
We have an RC circuit receiving work $W$ from a generator of voltage $v$ and energy $Q_{exchanged}$ from a thermal reservoir (thermostat) of $T_0$ Kelvin,the circuit has its entropy change$\Delta S_{in}$ during the charging.
We have $C_q$ the heat capacity for a constant charge q and $\lambda= \frac{qT \epsilon'}{\epsilon c}$ another coefficient in the equation :
$TdS_{in}= C_q dT + \lambda dq$ (1)

What is the entropy of the universe $\Delta S_{univ}$ equal to ?
Does $C_q$ depend on $q$ ?

## Homework Equations

$\Delta S_{thermostat}=\frac{Q_{exchanged}}{T_0}$

## The Attempt at a Solution

If we integrate $C_q$ over T ,I think q does appear in the formula for $C_q$ but it's not a variable I think.To find $\Delta S_{in}$ we divide the equation (1) by T then integrate $\frac{\lambda}{T}$ over q from 0 to q=cv.
$\Delta S_{univ}=\Delta S_{thermostat}+\Delta S_{in}$
I would like to know how to express $\Delta S_{thermostat}$ and it's sign. Thank you.

## Answers and Replies

What physical physical picture you are trying to work with? That is, what phenomenon are you interested in? It seems you have the capacitor bathed in a temperature reservoir. Why?

What physical physical picture you are trying to work with? That is, what phenomenon are you interested in? It seems you have the capacitor bathed in a temperature reservoir. Why?

I want to know how to express the entropy change , I think the entropy change of the universe is 0 but I'm not sure what the equation is.The RC circuit is getting heat from the reservoir and work from the generator,I'm guessing it's to make it a reversible process.

I am not sure, but I suspect you are confused. Ordinarily one does not consider heat when analyzing an RC circuit, and the capacitance of a capacitor does not vary much with temperature. I am not so sure that the net change in entropy of the universe is zero; For irreversible processes, entropy increases.

Generally the flow of current through a circuit with resistance is not a reversible process. The resistors give off heat but never absorb heat.

yamata1
rude man
Homework Helper
Gold Member
This is pretty advanced thermodynamics. I can only give you some ideas:
Since the capacitor is being charged isothermally, dT = 0 and you are left with dS = $\lambda dq$/T = (ε'/εc)q dq. You didn't say what ε' and c are. But your eq. (1) is of course a form of the Maxwell "2nd T dS equation" so there must exist a coefficient representing the change of polarization with T ( related by the state equation). So maybe ε' = dε/dT and ΔSdielectric = (ε'/εc) q dq from q=0 to q = Cv.
Then the total change in entropy of the universe is just ΔSdielectric - ΔSthermostat.

Sorry, not very definite; hope others will do better.

I think I have an explanation :

The generator gives the system {resitance,capacitor} the electric energy :

The capacitor charges itself by en absorbing half ( the work ##W=\int vdq##). The difference represents the lost energy through the Joule effect :

This energy is received through heat by the heat reservoir at fixed temperature ,which increases the entropy by :

Accouting for the resistance, we have :
$$\Delta S_{univ}=\frac{\varepsilon'}{2\varepsilon.c}\cdot q_{f}^{2}-\frac{\varepsilon'}{2\varepsilon.c}\cdot q_{f}^{2}+\frac{q_{f}^{2}}{2c.T_{0}}=\frac{q_{f}^{2}}{2c.T_{0}}$$

there is a creation of entropy caused by the Joulet effect which is an irreversible phenomenon as Gene Naden wrote
I am not sure, but I suspect you are confused. Ordinarily one does not consider heat when analyzing an RC circuit, and the capacitance of a capacitor does not vary much with temperature. I am not so sure that the net change in entropy of the universe is zero; For irreversible processes, entropy increases.

Generally the flow of current through a circuit with resistance is not a reversible process. The resistors give off heat but never absorb heat.
.
The power received by the heat reservoir is the power produced by the Joule effect :

#### Attachments

• ile_TEX.cgi?q_{f}.e=c.gif
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• ile_TEX.cgi?E_{Joule}=\frac{1}{2}q_{f}.e=\frac{1}{2}c.gif
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• ile_TEX.cgi?\frac{q_{f}^{2}}{2c.gif
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• ile_TEX.cgi?P_{Joule}=r.i^{2}=r.gif
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• ile_TEX.cgi?\frac{dS_{univ}}{dt}=\frac{P_{Joule}}{T_{0}}=\frac{r}{T_{0}}.gif
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