Or we can use this equivalent circuit



When I replace the capacitor with a voltage source at t = 1ms

 

I always found it easier to solve these problems, and check my answer, if I first examined the two extremes of t=0 and t=infinity. At t=0, the capacitor is a short. At infinity it's not passing current and is charged to the voltage across R2. Both of these are easy to calculate.

In between, you know that the transition from one state to the other HAS to follow the RC curve for the capacitor. The only wrinkle in this problem is calculating what R is relevant to charging the capacitor.

 

Or we can use this equivalent circuit

View attachment 170032

When I replace the capacitor with a voltage source at t = 1ms

Yes thanks that's what I got with my calculations also
So now I know that the Voltage across R2 at t=0.001 is 0.87 V, and that at t=0, the voltage is 3.43 across R3 or Ux.
But what is the voltage at R3 at t=0.001?

I need that also since Ux is R2 + R3 at t = 0.001 :/

 

My thinking so far

To get the supply voltage for the capacitor, Vs , I look at the circuit at t = infinity

R(total) = 4200, I(total) = 1.19 mA
V(R2) = 1000 * 1.19 mA = 1.19 V
V(R1) = Vs

The voltage across the capacitor is then: (at t = 0.001)
Vc = Vs(1-e^(-t/RC))
Vc = 0.87 V

To get the Voltage Ux I want to calculate the voltage across R1 at t=0.001, then subtract that voltage from 5V.


I use superposition to remove the voltage source 0.87 V (the capacitor).

With current division I(R1) = 762 / (3200 + 762) * 0.00656 = 0.00126 A

I don’t know how to get the current across R1 if I remove the voltage source 5V.

However if I take 0.00126 * 1000 = 1.261
5V - 1.261V = 3.738

So maybe I’m thinking right here? My guess is if I also could do the superposition without the 5V (and subtract the currents across R1 at t = 0.001) I would be closer to 3.71?

 

Or we can use this equivalent circuit

View attachment 170032

When I replace the capacitor with a voltage source at t = 1ms

Hi,

Very good, but shouldn't you show how you got the time constant too?

 

Very good, but shouldn't you show how you got the time constant too?

I simply find the resistance seen from the capacitor point of view.

So maybe I’m thinking right here?

Try to use the superposition and try to find Vx for two cases. The first one without 5V source (short circuit) and the second one without Vc1 source (short circuit instead).

Vx = Vx1 + Vx2 = 0.272V + 3.4375V = 3.7095V

 

I simply find the resistance seen from the capacitor point of view.

Hi,

Yes but simply stating that does not show how to do it. See what i mean?

 

Hi,

Yes but simply stating that does not show how to do it. See what i mean?

I solved the problem anyways. Thanks for the good help all.
Hope someone else might find this thread educational.

 

I solved the problem anyways. Thanks for the good help all.
Hope someone else might find this thread educational.

Hi,

That's great. You can also see there are different ways to do this problem too.

 

To get the final answer I got the current across R2 at t=0.001

With only 0.87V source
IR2 = 0.87 / 1000 = 0.00087
ITotal = 0.87 / (R1 + R3) // R2
I1 = ITotal - IR2 (I1 Is across R1 and R3)

With only 5V source
I2 = ITotal = 5 / (R1 + R3) = 5 / 3200 (no current across R2 here since there's no voltage)

With 0.87V and 5V source
I = I1 - I2
V1 = R * I
Ux = 5 - V1

Ux = 3.71 V