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.

Task: enter image description here

That is how I solved it: my solution

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

after t>0

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    \$\begingroup\$ 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. \$\endgroup\$ – G36 Sep 12 '18 at 14:14
  • \$\begingroup\$ @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. \$\endgroup\$ – James Smith Sep 12 '18 at 14:27
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    \$\begingroup\$ 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/… \$\endgroup\$ – G36 Sep 12 '18 at 14:31
  • \$\begingroup\$ Okey, so Rth will be equal to 8/3R right? \$\endgroup\$ – James Smith Sep 12 '18 at 14:56
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    \$\begingroup\$ Yep, Rth = 8/3R \$\endgroup\$ – G36 Sep 12 '18 at 15:02

I think you have the first part (switch closed) right. Then, the top R, the two R's in parallel, and the final R all form a voltaage divider with the C at that voltage.

When you open the switch, however, the C will charge to the full E; the current flowing through the two resistors now in 'parallel' with the C don't matter. However, the rate at which it charges will be determined by all the resistors in series with C.

And, as usual, the solution will look like

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


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