# RC circuit

1. Apr 4, 2006

### Cyrus

Here is the circuit, everything is in PARALLEL,

First its a 1mA source, then its a switch, then its a resistor of 10k-ohm then its a capacitor 1uF. Sorta like below:

----------------------------------
| ...........| ............. |............. |
O.........Switch ........ R ............ C
|.............| .............|.............|
----------------------------------

initial conditions, t=0, switch is CLOSED.

I said that the closed switch shorts out the rest of the circuit, so the inital current/voltage across the resistor is zero.

Then when its opened, I said that its like a current source, in parallel with the resistor and the capacitor. But the current source and the resistor is the norton circuit with a load of a capactior. So I changed it to a thevenin eq. circuit, by using Vt=InRt = 10V

Now my circuit is a 10V source, in series with a 10k-ohm resistor, and a load of the capacitor, and this is charging the circuit

Then the voltage across the resistor as a function of time, (For the inital circuit!) is the same as that of the capacitor,

Vr(t)= 10-10e^(-t/tau)

Yes, no? Maybe so?

Last edited: Apr 4, 2006
2. Apr 4, 2006

### Staff: Mentor

I believe, with the switch open the current source is in series with the R and C, which are in parallel. Part of the current goes through the resistor, and part of the current goes to charge the capacitor, so the effective voltage on the capacitor is less than 10 V, until the capacitor is fully charged.

As the capacitor charges, the current through the resistor increases since the voltage across the capacitor increases with increasing charge, and the voltage across the resistor increases.

VR(t)= 10 - 10 e-t/tau appears to be correct.

Last edited: Apr 4, 2006