Finding q(t) on a capacitor in an RC circuit

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Homework Help Overview

The discussion revolves around analyzing an RC circuit involving capacitors C1 and C2, where the switch is initially closed until C2 reaches maximum charge, after which it is opened. The goal is to find the charge q1(t) on capacitor C1 over time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss applying Kirchhoff's rules to derive relationships between the charges and currents in the circuit. There are attempts to set up differential equations to express q1 in terms of time, while some participants question the constancy of total charge in the circuit.

Discussion Status

Some participants have offered guidance on using Kirchhoff's laws and suggested differentiating equations to derive new relationships. There is an ongoing exploration of how to relate the charges on the capacitors and the currents through them, with no explicit consensus on the best approach yet.

Contextual Notes

Participants note the challenge of eliminating certain variables and the implications of differentiating equations to find new insights. There is also mention of formatting issues when attempting to present equations clearly.

Robert S
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The following circuit is given:
[PLAIN]http://img822.imageshack.us/img822/6369/image2lz.jpg

Switch S is closed until the charge q2 on C2 reaches its maximum, then at t=0 the switch is opened. Find q1(t).

Using Kirchkoff's rules I found:
EMF= (q1)/(C1+I2R
I2R= q2/C2
I1= I2+I3

In order to solve for q1 I need to substitute [itex]I_1=\frac{dq_1}{dt}[/itex] and solve the differential equation. But I need one more equation to eliminate I3 and q2.
I was thinking of relating the charge q2 to the total charge in the circuit, but after some calculation I found that the total charge isn't constant.

At t=0 the total charge q2 on C2 is EMF*C2 and the charge on C1 is zero.
At t=[itex]\inf[/itex] the total charge on both capacitors is [itex]EMFC_1C_2 / (C_1 + C_2)[/itex] (since the current is zero you can combine the two capacitors using [itex]C_3^{-1}=C_1^{-1} + C_2^{-1}[/itex])

any help is appreciated :-)

edit: I tried to put the kirchkoff equations in Latex, but for some reason the rest of the post becomes unreadable.
 
Last edited by a moderator:
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Robert S said:
The following circuit is given:
[PLAIN]http://img822.imageshack.us/img822/6369/image2lz.jpg

Switch S is closed until the charge q2 on C2 reaches its maximum, then at t=0 the switch is opened. Find q1(t).

Using Kirchkoff's rules I found:
EMF= (q1)/(C1+I2R
I2R= q2/C2
I1= I2+I3

In order to solve for q1 I need to substitute [itex]I_1=\frac{dq_1}{dt}[/itex] and solve the differential equation. But I need one more equation to eliminate I3 and q2.
I was thinking of relating the charge q2 to the total charge in the circuit, but after some calculation I found that the total charge isn't constant.

At t=0 the total charge q2 on C2 is EMF*C2 and the charge on C1 is zero.
At t=[itex]\inf[/itex] the total charge on both capacitors is [itex]EMFC_1C_2 / (C_1 + C_2)[/itex] (since the current is zero you can combine the two capacitors using [itex]C_3^{-1}=C_1^{-1} + C_2^{-1}[/itex])

any help is appreciated :-)

edit: I tried to put the kirchkoff equations in Latex, but for some reason the rest of the post becomes unreadable.

I would label the output of the voltage source as V1, and the voltage between the caps as V2. Write the KCL equation at that V2 node, and solve for V2(t). Then use Q=CV to calculate the charge across C1 as a function of time.
 
Last edited by a moderator:
Robert S said:
Using Kirchkoff's rules I found:
EMF= (q1)/(C1+I2R

The correct form is: EMF= q1/C1+I2R

Robert S said:
I2R= q2/C2
I1= I2+I3

In order to solve for q1 I need to substitute [itex]I_1=\frac{dq_1}{dt}[/itex] and solve the differential equation. But I need one more equation to eliminate I3 and q2.
You can write the first equation also as EMF=q1/C1+q2/C2, and differentiating it, you get

I1/C1+I3/C2=0

I2 is obtained from the second equation: I2=q2/(C2R)

I3=dq2/dt.

Now you can apply the Nodal Law:

I1=I2+I3

I suggest to solve for q2(t), it is easy to get q1(t) from it.

ehild
 
I didn't realize I could differentiate one equation to get a new one. Thanks! :-)
 

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