# Electric Circuit Question - RC Circuit

## Homework Statement

http://img163.imageshack.us/img163/6339/rccircuit.png [Broken]

## The Attempt at a Solution

I'm trying to find the node voltage. I was going to try and solve the problem knowing that the net charge is zero.

Q = CV

I'm having problems the voltage across each capacitor though because some branches are in series with resistors and power sources.

Thanks for any help!

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gneill
Mentor
Are you looking for the node voltage as a function of time, or the steady state value (t → ∞)?

I'm just looking for the steady state value of the voltage at that node.

At steady state, are the resistors of any use there?

They do absolutely nothing. So when I was trying to solve it using that the net charge must be zero.

(2*10^(-9))(Vx-12)+(6*10^(-9))(Vx-6)+(4*10^(-9))(Vx-0)+(8*10^(-9))(Vx-2)=0

which solves out to be 3.8 however when I build in circuit labatory I get just about 5 volts even.

I think I'm doing something wrong but am not sure what.

gneill
Mentor
Here's a suggestion. You can see that at steady state no current will flow, and so no voltage drops will exist across the resistors. So redraw the circuit without the resistors (replace with shorts). Next, use superposition (one supply at a time).

which solves out to be 3.8 however when I build in circuit laboratory I get just about 5 volts even.

I think I'm doing something wrong but am not sure what.
Had you grounded the circuit like it is shown in the circuit diagram?

Well that makes since and I don't know why instead of using superposition I can't just use the fact that the net charge must be zero at the node. I checked that the voltage before capacitor one is 12 volts, the voltage before the second capacitor is 6 volts, the voltage before capacitor 3 is 0, the voltage before capacitor 4 is 2 volts.

This is how I got my net charge equation and then just solve for the voltage at the node. The voltage across each capacitor should just be Vx - the numbers are mentioned above and the charge at the capacitor should be the voltage across it times it's capacitance.

gneill
Mentor
Well, your result Vx = 3.8V looks fine (more precisely, Vx = (19/5)V). If you use the simulator's Time Domain analysis, set the parameter "Skip Initial" to "yes" before running the analysis. This prevents the simulator from futzing about trying to set or use some initial circuit conditions on the components before the simulation begins; you want the supplies to "turn on" and the capacitors to be uncharged when the simulation begins at time 0.