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I understand that the time constant in AC RC circuits is defined the value of resistance (in ohms) x the value of capacitance (in farads). Indeed, ohms(R) x farads(C) yields time (seconds).

$$\tau = RC $$

I also understand that when discussing continuous change (growth/decay) it is no surpise that e figures in. What I do not understand is why \$C * R\$ just happens to equal the time constant, \$tau\$, the point at which the voltage in a capacitor \$\approx63\%\$ of its final charged state C. In other words, why it it true that after one time constant, \$\tau\$, voltage in the capacitor (in an RC circuit) equals 63.2% of its starting voltage. Or, why does this hold:

$$0.63 \approx (1 - \frac{1}{e^1})$$ and $$V_\tau = V_0(1 - \frac{1}{e^1})$$

This is related to another question that asks why the RC time constant = 63.2% and not some other value: Why is the time constant 63.2% and not 50% or 70%?

This post included insightful answers describing that 63.2 is related to e (specifically \$\frac{1}{e}\$),and how e relates to continuous change, but not why capacitance x resistance yield this value.

It is not as if Georg Ohm or Michael Farady had e, or the time constant, in mind when developing these ideas (or units), correct? Am I missing something obvious?

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    \$\begingroup\$ Because 'e' is a magical number. It crops up everywhere. en.wikipedia.org/wiki/E_(mathematical_constant) \$\endgroup\$ – mkeith Feb 5 at 5:46
  • \$\begingroup\$ I don't understand your title question.. What do you think equals 1/e? \$\endgroup\$ – Curd Feb 5 at 10:36
  • \$\begingroup\$ I think you are going in circles. When we solve the ODE we get an exponential with power the time divided by a constant. So we define the time constant to be said constant in the exponential, which happens to be RC in this example. Then at one time constant the power of the exponential is one be design. \$\endgroup\$ – user110971 Feb 5 at 11:25
  • \$\begingroup\$ The time constant is not equal to 1/e. \$\endgroup\$ – Bart Jul 31 at 5:25
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Let's look at a simple circuit with an RC time constant:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, if

$$v(t) = \begin{cases} 0 & t < 0 \\ 1\ {\rm V} & t \ge 0 \end{cases}$$

then you can write a differential equation for the voltage across the capacitor for \$t > 0\$:

$$\frac{dv_c}{dt} = \frac{1-v_c}{RC}$$

where \$R\$ appears in the denominator because the resistor value limits the current supplied to the capacitor, and \$C\$ appears because a higher-valued capacitor needs more charge to reach a given voltage.

Which we solve essentially by knowing the answer (but of course you can go back and check if the solution satisfies the differential equation), with

$$ v_c(t) = 1 - e^{-\frac{t}{RC}}$$

So basically \$e\$ appears because it's the base of the exponential function that solves the differential equation \$\frac{df(t)}{dt}=f(t)\$, and the \$RC\$ term appears because of the way current relates to capacitor voltage in the circuit.

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  • \$\begingroup\$ This is not a really satisfactory answer, in the introductory part. You should start with a pair of Kirchoff's Laws, that I(t) along the loop of components (V, R and C) is the same, I = C*dVc/dt = Vr/R, and that the V(t) = Vr + Vc, which gives the differential equation you wrote. The equation is always valid regardless the initial condition V(t). The rest is good, the solution to the diff.eq to the step function gives the characteristic rise time, \$\endgroup\$ – Ale..chenski Feb 5 at 6:10
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The fact that the R*C = the time it takes for the voltage to decay to 63% is a fundamental feature of the universe. How you measure the resistance and capacitance doesn't change reality, so any valid way to define those units will come out with the same percentage. It may help you to remember that a percentage is unitless. The numerical values of R and C may change depending on what your units are, but the percentage will be the same after you work it all out.

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  • \$\begingroup\$ I think you meant to say "value" instead of "units"? Units for R and C don't change. \$\endgroup\$ – KingDuken Feb 5 at 5:32
  • \$\begingroup\$ That's sort of my point. the definition of Ohms and Farads could be different, but the resistance and capacitance will be the same. Just like you can measure in kilometers or miles and get a different numerical value, but the distance is unchanged. \$\endgroup\$ – Matt24 Feb 6 at 21:27
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The R and C, together with Vfinal, produce the initial Slewrate; that initial slewing would intercept the final voltage at TAU seconds, if the slewrate remained constant.

Of course, the slewrate is continually changing.

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enter image description here

The definition of this time constant tau=RC can be verified very easily using the graph of the step response g(t)=1-exp(-t/tau).

Taking the derivative of this function at t=0 we have:

g´(t)=(-exp(t/tau)(-1/tau) which for t=0 is g`(t=0)=1/tau

  • This is the first graphical interpretation: The time constant tau is the inverse of the slope of g(t) at t=0.

If we construct the tangent for g(t) at t=0 we have the function of a straight line with slope 1/tau. Hence the function is y=(1/tau)*t and for tau=t we have y=1 .

  • This the second graphical interpretation: The time constant tau is identical to crossing point between the tangent at t=0 and the final value of g(t) (in our case: "1").

Inserting this time constant into the step response g(t) we arrive at the mentioned value of 63% (as shown in the original question)

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After giving this some thought, I now realize my thinking about this was flawed at the outset. I was essentially asking why does 1 = 1. Should others engage in a similar fallacy, and find themselves perplexed by the incredible coincidence that R * C happens to equal this property of nature called the time constant \$t\$, consider this:

By definition, one time constant is, as a percentage of starting value, the point at which the thing being measured equals \$1/e*StartValue\$ if decreasing , or \$(1 - 1/e)*StartValue\$ if increasing. This is not specific or unique to RC circuits in any way. If we are to use this convention to describe the particular system in question here, the RC circuit, we simlply calculate a value such that R * C equals \$t\$. That is, since R * C = time, we can simply multiply R*C and take that product to be \$tau\$.

My mistake was, essentially, confusing cause and effect. The value of 1/e is what defines (in a sense 'causes') the specific values that need to be used as the time constant in a given system, not a magical coincidence resulting from some mysterious property of \$\Omega \$ and \$F\$. Quite literally, to ask why e/RC = 1 when RC is by definition the time constant (the point at which this holds true), is to ask why does 1 = 1 .

Perhaps had I seen it written as \$t\equiv R*C\$, I may have avoided this particular rabbit hole

(Thanks for the excellent clarifications above The Photon, LvW, et.al. I Tried to upvote, but I lack sufficient rep.)

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