# Time of charge of capacitor of complex circuit

I'm a newbie in this area and was reading about the time it takes to a capacitor to get charged known as τ = RC.

The question is, how do you calculate such time when a circuit is formed by more elements? Say I have a voltage attached to R1 which is attached to another R2 in parallel with a capacitor. What would the value of R be in this case?

• What you are describing is a textbook example of Thévenin's theorem. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem But calculating time constant of an arbitrary circuit can be much more complicated than that an cannot easily be explained in a question like this. It requires knowledge of Thévenin, Kirchhoff and complex numbers. – jippie Dec 22 '12 at 14:07
• If you add (a link to) a specific diagram, we might be able to explain a bit. I may sound a bit demotivating in my previous comment, but I didn't mean to. It's just that it is a complex matter and you have to gain the knowledge step by step. – jippie Dec 22 '12 at 14:16
• 1/Req = 1/R1 + 1/R2 for parallel R where inverse of R is called admittance and they add for parallel R and for series R's... Req= R1+R2+... – Tony Stewart Sunnyskyguy EE75 Dec 22 '12 at 15:05
• @Richman, inverse of R is called conductance. Admittance is the inverse of impedance. – The Photon Dec 22 '12 at 17:06
• A good way to empirically model this stuff is to use something like LTspice. You could whip up this example very easily and model it at the nanosecond scale. – Toby Lawrence Dec 22 '12 at 17:38

Ultimately how much any particular capacitor is charged depends on the total Coulombs it holds. The capacitance and the voltage level at which you consider it charged dictate the number of Coulombs you need to run thru it to "charge" it from the totally discharged state.

Charging a capacitor thru a resistor from a fixed voltage source is only one of many possible ways to dump the required Coulombs onto the capacitor. In that case, the voltage is a exponential approaching the voltage source value. In some cases the circuit may look more like a current source, in which case the capacitor voltage would rise linearly. In other cases the circuit could be non-linear, so all kinds of unusual voltage profiles are possible.

For the case of a resistor (R) charging an initially discharged capacitor (C) from a voltage source Vs, the voltage on the capacitor (Vc) vs. time is...

$V_C = V_S \cdot (1-e^{(\frac{t}{R \cdot C})})$

Note from the equations that the rate of charge slows as the capacitor approaches its final voltage and that the capacitor never really reaches that final value. The time constant T = R * C describes how long it takes to get to (1 - 1/e) (about 63%) of the final value. In general when one wants to calculate time until the capacitor is "charged" one needs to define how close to the final value they want to be.

To some percentage P of the final value the time is...

$T = - R \cdot C \cdot ln(1 - P)$, where P=0 is 0% and P=1.0 is 100%

For the case of a resistor (R1) charging a capacitor (C) and parallel resistor (R2) from supply voltage (Vs) the equation is the same as above except that R is replaced with the parallel combination of the two resistors (Rp), and the supply voltage is replaced with the value created by the voltage division of the two resistors (Vd).

$V_C = Vd \cdot (1 - e^{(\frac{-t}{Rp \cdot C})})$

$Vd = \frac {Vs \cdot R2}{R1 + R2}$
$Rp = \frac {R1 \cdot R2}{R1 + R2}$

For more complex circuits you typically need to write out a system of differential equations (mesh equations) that mathematically describes the circuit using Kirchhoff's voltage and current laws.

Next solve the set of differential equations to find the voltages and currents vs time.

This process is described in most linear circuit analysis textbooks.

Here comes a nice and helpful "trick" which - in most cases - allows us to quickly calculate (or estimate) the relevant time constant:

Imagine that the capacitor is charged and evaluate/analyse the circuitry which is connected to the capacitor for DISCHARGING (set the voltage sources to zero or remove the current source).

In your example, it is easy to see that discharging takes place - at the same time (which means: in parallel) - through both resistors.