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

1. Jul 15, 2010

### Robert S

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

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 $I_1=\frac{dq_1}{dt}$ 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=$\inf$ the total charge on both capacitors is $EMFC_1C_2 / (C_1 + C_2)$ (since the current is zero you can combine the two capacitors using $C_3^{-1}=C_1^{-1} + C_2^{-1}$)

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: May 4, 2017
2. Jul 15, 2010

### Staff: Mentor

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: May 4, 2017
3. Jul 17, 2010

### ehild

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

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

4. Jul 18, 2010

### Robert S

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