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If we plot and compare v-t or I-t curves with actual exponential curve we can see that they are same, but why? Is there any proof?

From the equation we can see that e is creating bridge between electrical property and time. Why e?

I have a good text book that covers the curve’s characteristics and how it can be used but doesn’t show any mathematical proof.

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    \$\begingroup\$ It's exponential because that's the only curve that satisfies the differential equation that describes the instantaneous behavior of the R-C circuit. \$\endgroup\$
    – Dave Tweed
    Aug 31, 2014 at 18:09
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    \$\begingroup\$ In plain terms - because as the capacitor charges (or discharges) the voltage difference becomes less over time which reduces current which means it takes longer and longer to charge (or discharge) the cap further \$\endgroup\$
    – squarewav
    Aug 31, 2014 at 19:06
  • \$\begingroup\$ These equations relate to a specific circuit. However simple the circuit is, you should link it in your question, because it is the key to the equations. \$\endgroup\$
    – jippie
    Aug 31, 2014 at 19:49

2 Answers 2


Capacitance is defined as:

\$C = \frac{Q}{V}\$

and equally:

\$C = \frac{\delta Q}{\delta V}\$

Since current is defined as the rate of change of charge:

\$I(t) = \frac{\delta Q(t)}{\delta t}\$


\$I(t) = C \frac{\delta V(t)}{\delta t}\$

If we set up a circuit with a voltage source, Resistor and a capacitor

\$V = V_r + V_c\$ and this creates a differential equation

\$v = i(t)R + \frac{1}{C}\int i(t) \delta t \$

Solving such differential equations produces an equation:

\$I(t) = \frac{V}{R}e^\frac{-t}{RC}\$

Which equally can be re-arranged to be

\$V(t) = V(1-e^\frac{-t}{RC}) \$ taking into account initial conditions

  • \$\begingroup\$ Great. I will go through the mathematical process. \$\endgroup\$
    – Amit Hasan
    Aug 31, 2014 at 18:40
  • \$\begingroup\$ Though I am still learning this mathematical phenomena, this answer is a great start point for me. \$\endgroup\$
    – Amit Hasan
    Sep 17, 2014 at 17:46

Intuitively, the relationship between capacitor voltage and current is simply i = (E - Vc)/R, by Ohm's law, so as the capacitor voltage rises the current must drop as kind of a mirror.

Getting one of the two functions wrt time requires solving the differential equation,


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