# Determine the voltage drops across the capacitor

I am having troubles with one type of tasks when preparing for an exam. I think I get the way to solve it when t = 0, but I don't really know what my approach should be when t>0.

That is how I solved it:

//Edit Updated version for t>0. I think it might be correct now. Is it?

• Why your Rth is 3R in the second case? Also as you can see because of 2/3E > 2/5E the capacitor will be charged from 2/5E to 2/3E. Not discharged. – G36 Sep 12 '18 at 14:14
• @G36 As far as I remember we don't have to take these resistors that are in branch where capacitor is placed, so we are left with 2 resistors (R and 2R that are serially connected), but I can be wrong. – James Smith Sep 12 '18 at 14:27
• Why? to find the time constant you need to find the resistance seen from the capacitor terminals. And from the capacitor point of view, the R and 2R are not connected in series. electronics.stackexchange.com/questions/377467/… – G36 Sep 12 '18 at 14:31
• Okey, so Rth will be equal to 8/3R right? – James Smith Sep 12 '18 at 14:56
• Yep, Rth = 8/3R – G36 Sep 12 '18 at 15:02

$$V_0+ V_fe^{-f(r,c)t}$$