This page deals with RC, RL, LC, and RLC circuits in the time domain.

## RC Circuits Edit

Generally we have a DC source Vs, a resistor R, and a capacitor C in a loop. We let the voltage across the capacitor be the principle variable V. From Kirchoff's voltage law we have

$ -V_s + iR + V = 0 $

We know that for capacitors $ i = C\frac{dV}{dt} $, so with this substitution and a little rearranging this becomes

$ \frac{1}{V_s - V}dV = \frac{1}{RC}dt $

Integrating, we have

$ \int \frac{1}{V_s - V}dV = \int \frac{1}{RC}dt $

$ -\ln (V_s - V) = \frac{t}{RC} + C $

$ V_s - V = e^{-t/RC - C} = e^{-t/RC}e^{-C} = Ae^{-t/RC} $

If we have initial conditions of V = 0 and Vs = some voltage, the capacitor is being charged through the resistor. In this case we find that A = V_s and so

$ V = V_s \left( 1 - e^{-t/RC} \right) $

If we have the initial conditions of Vs = 0 and V = some voltage Vi, the capacitor is discharging through the resistor to ground. In this case we find that A = Vi and so

$ V = V_i e^{-t/RC} $

### Time Taken to Charge/Discharge Edit

It takes an infinite amount of time for the capacitor to completely charge or discharge. However we can approximate the time taken for this to happen in realistic circuits by considering the time taken for the capacitor to charge to a certain percentage.

It's easiest to express this terms of the percentage of charge *remaining*. You can do it the other way, but the expressions are more complicated. Let p be the percentage of the remaining charge capacity to be charged/discharged, as a fraction. In both cases we have

$ e^{-t/RC} = p $

$ -\frac{t}{RC} = \ln (p) = - \ln \left( \frac{1}{p} \right) $

$ t = RC \ln \left( \frac{1}{p} \right) $

So for example, if you wanted the time for a 10pF capacitor connected to a 5K resistor to become 90% charged (that is, 10% or 0.1 charging left), you would do

$ t = 5K \cdot 10p \cdot \ln \left( \frac{1}{0.1} \right) = 50n \cdot \ln (10) = 50n \cdot 2.3 = 115 ns $

In most cases a useful approximation is $ t = 2RC $, the taken to get to 86% charge. If you're good at remembering magic numbers, you could try $ t = 2.3RC $, which gives you the time for 90% charge.