I am learning about RC transient responses and if a charged capacitor is connected to a single resistor, its voltage will decrease exponentially. I understand this and how its equation is derived. I also understand how the energy stored in a capacitor is obtained.

So, given an infinite amount of time, all of the energy will be dissipated in the resistor. I would like to prove that the integral of p(t)dt between 0 and infinity is equal to 0.5CV₀². I've tried substituting in i²(t)R for p(t) and then equating i to -Cdv(t)/dt but I end up with something that isn't even an integral anymore, namely of the form RC²∫dv(t)/dt between 0 and infinity.

What sort of substitution should I make to allow this to be integrated? Thank you for reading.

  • 1
    \$\begingroup\$ What is the integral of dV/dt in this circuit over infinite time? \$\endgroup\$ Jan 3, 2017 at 18:33
  • \$\begingroup\$ The integral of a derivative is the function, so v(t)? The voltage in the across the capacitor at infinite time would tend to 0. \$\endgroup\$
    – LCHL
    Jan 3, 2017 at 18:57

2 Answers 2


The voltage across a capacitor discharging (exponentially) through a resistor is: -

\$V = V_0\times e^{\dfrac{-t}{CR}}\$ where Vo is the voltage at t = 0

The current is the above voltage divided by R

So, power is \$\dfrac{V_0^2}{R}\times e^{\dfrac{-2t}{CR}}\$

Note that the exponential term now has a 2 in it because it became squared.

Now if you integrate power you accumulate the energy: -

\$\dfrac{V_0^2}{R}\times \dfrac{CR}{-2}\left[e^{\dfrac{-2t}{CR}}\right]_0^\infty\$

If you resolved the integral between 0 and infinity it comes to -1 hence energy taken by resistor is simply: -


  • \$\begingroup\$ Very informative. Thank you. As I'm sure you know, integrating between 0 and infinity (from the perspective of the resistor) will yield a positive number instead. \$\endgroup\$
    – LCHL
    Jan 3, 2017 at 20:02
  • \$\begingroup\$ @LCHL I got the terms back to front when resolving the integral. I'll fix later - been a long day! \$\endgroup\$
    – Andy aka
    Jan 3, 2017 at 20:50
  • \$\begingroup\$ @LCHL fixed!!!! \$\endgroup\$
    – Andy aka
    Jan 3, 2017 at 21:19

The equation \$i = C\frac{dv}{dt}\$ describes the relationship between voltage and current for the capacitor. As you have found, you can't simply substitute this expression for the resistor current. You have to solve the circuit for the current (or voltage).

To solve for the voltage, sum the resistor and capacitor currents: \$0 = \frac v R + C\frac{dv}{dt}\$

This is a differential equation having the solution \$ v=V_0e^{-t/RC}\$ where \$V_0\$ is the voltage at \$t=0\$

Alternately, you could solve for the resistor current:
\$v = Ri\$, so \$\frac{dv}{dt} = R\frac{di}{dt}\$ and therefore \$-i = CR\frac{di}{dt}\$

The negative sign is needed here because the capacitor current is the negative of the resistor current.

This has the same form as the voltage equation; its solution is \$i=I_0e^{-t/RC}\$ where \$I_0 = V_0/R\$

Now you can square either of these and multiply or divide by the resistance to get an equation for instantaneous power, and integrate that to get the energy.


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