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

  • 1
    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 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 at 14:27
  • 1
    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 at 14:31
  • Okey, so Rth will be equal to 8/3R right? – James Smith Sep 12 at 14:56
  • 1
    Yep, Rth = 8/3R – G36 Sep 12 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

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

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.