This is the circuit I got from an example in a book "Practical Electronics for Inventors", on page 48:

enter image description here

How can one understand that \$0=R\frac{dI}{dt}+\frac{1}{C}I\$ implies \$I=I_0e^{\frac{-t}{RC}}\$?

  • \$\begingroup\$ Fast forward to the Laplace transform chapters, it'll make sense that way. \$\endgroup\$
    – Matt Young
    Mar 30, 2013 at 6:13
  • \$\begingroup\$ I can't see where the 1st equation came from - there is no mention of Vo. The 2nd equation presumably incorporates Vo into Io, Io being the current at t = 0. There is an error - why not scan the page and show it that way. \$\endgroup\$
    – Andy aka
    Mar 30, 2013 at 10:28

2 Answers 2


This should help (it helped me!!): -

enter image description here

The final formula is written as V/R instead of Io but it means the current at time = 0.

Copied from http://www.intmath.com/differential-equations/6-rc-circuits.php in case you need the full article. This took me back a few years!


First by algebraic manipulation, then by integration.

For the first step, divide all terms by R.
For the second step, subtract one of the RHS terms from both sides.
Then the equation should be in a form with a standard integral, so just integrate it. You should now have the result.

  • \$\begingroup\$ I think his first equation is incorrect - there is nothing that accounts for the supply voltage Vo \$\endgroup\$
    – Andy aka
    Mar 30, 2013 at 10:54
  • \$\begingroup\$ Good point. I think the equation was correct in itself. Whether it had anything to do with the circuit is another matter :-) \$\endgroup\$
    – user16324
    Mar 30, 2013 at 11:04
  • \$\begingroup\$ OK I see how he got there - he differentiated wrt i to get rid of Vo!! (I can be dumb a lot) \$\endgroup\$
    – Andy aka
    Mar 30, 2013 at 11:09

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