Entropy change in an RC circuit

  • Thread starter yamata1
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Hello,I would like some help for a problem

Homework Statement


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

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

Homework Equations


[itex]\Delta S_{thermostat}=\frac{Q_{exchanged}}{T_0}[/itex]

The Attempt at a Solution


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

Answers and Replies

  • #2
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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?
 
  • #3
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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.
 
  • #4
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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.
 
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rude man
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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 = [itex]\lambda dq[/itex]/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.
 
  • #6
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I think I have an explanation :

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

ile_TEX.cgi?q_{f}.e=c.gif


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

ile_TEX.cgi?E_{Joule}=\frac{1}{2}q_{f}.e=\frac{1}{2}c.gif


This energy is received through heat by the heat reservoir at fixed temperature ,which increases the entropy by :
ile_TEX.cgi?\frac{q_{f}^{2}}{2c.gif


Accouting for the resistance, we have :
[tex] \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}}[/tex]

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 :

ile_TEX.cgi?P_{Joule}=r.i^{2}=r.gif


ile_TEX.cgi?\frac{dS_{univ}}{dt}=\frac{P_{Joule}}{T_{0}}=\frac{r}{T_{0}}.gif
 

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