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

[Robert S]

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.

 

[berkeman]

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.

 

[ehild]

Using Kirchkoff's rules I found:
EMF= (q₁)/(C₁+I₂R

The correct form is: 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.

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 q_2(t), it is easy to get q1(t) from it.

ehild

 

[Robert S]

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

 

[rwisz]

Based on the information given, it seems that you are trying to solve for the charge on C1 in an RC circuit with a switch that is initially closed and then opened at t=0. In order to solve for q1, you will need to use Kirchhoff's laws and the equations you have already derived. However, as you mentioned, you will need one more equation to eliminate I3 and q2.

One possible way to find this additional equation is to use the conservation of charge in the circuit. Since no charge is added or removed from the circuit, the total charge at any time should remain the same. This means that the sum of the charges on C1 and C2 should be equal to the total charge in the circuit at any given time.

In other words, we can write the equation q1(t) + q2(t) = q_total. Since we know the initial and final values of q2, we can substitute those in and solve for q1. This will give us the additional equation we need to solve the differential equation and find q1(t).

Another approach could be to use the fact that the total current in the circuit is equal to the sum of the currents through each component. This means that I1 = I2 + I3, which can be rewritten as dq1/dt = dq2/dt + dq3/dt. Again, using the initial and final values of q2, we can substitute those in and solve for q1.

I hope this helps and good luck with your calculations!