Carlos G.

asked • 05/21/14

why 1/C ? V/R(e^-t/RC) dt is equal to V(1-e^-t/RC) Note: V,R and C are constants. I integrated but I got -Ve^-t/RC

perhaps I'm ommiting some detail in my integration and I'm eager to know why it doesn't match with the textbook result.

1 Expert Answer


Seyed Kaveh M. answered • 05/21/14

5.0 (642)

Math Tutor: Pre-Algebra to Calculus

Carlos G.

Hi Sir, well, that should be the missing fact that i was overlooking (the constant of integration). Actually, the expresion was: 1/C∫ i dt where i=V/R(e^−t/RC) what do you think? Thanks for answering.
1Ci dt


Seyed Kaveh M.

Hello again,
Sorry for the late reply. I know this is EE stuff as I had majored in electrical engineering, but had to review some stuff to be able to answer your question.
Yes you have done the integration correctly. But why is that answer different than yours? It is because when you want to find the step response of an RC circuit, the complete response is equal to the transient response plus steady-state response. What you have calculated is the transient response. For the complete response, you will need to add that with the steady state response (which is equal to Vs note: s is a subscript which I believe represents the source). So transient response= -Vs * (e^(-t/RC))  , steady state response= Vs. add the two up and you will have the right answer.


Carlos G.

Thank you so much for your reply Sir (I´ve just checked the page :) Well, I really appreciate your attention. Now I´m not in blank about this formula and its origin and procedure.
Have a nice day.


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