Time Constant of a simple RC circuit (RC are connected in series) is R (in Ohm) x C (in Farad). The wiki 's explanation of "RC time constant" is very simple.

"
The RC time constant, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads), i.e. $\tau$ = R × C.
It is the time required to charge or discharge the capacitor, through the resistor, by ≈ 63.2 percent of the difference between the initial value and final value. This value is derived from the mathematical constant e, specifically $1-e^{-1}$ .
"
However, wiki explains it further in more broadly in "Time constant". Time constant is universal not limited to electric/electronic circuits. Generally "Time constant" is defined as

"
Physically, the constant represents the time it takes the system's step response to reach $1-1/e \approx 63.2\,\%$ of its final (asymptotic = infinitely close to a certain) value. In radioactive decay the time constant is called the decay constant (λ), and it represents both the mean lifetime of a decaying system (such as an atom) before it decays, or the time it takes for all but 36.8% of the atoms to decay. For this reason, the time constant is longer than the half-life, which is the time for only 50% of the atoms to decay. (wiki: Time Constant)
"
Here " step response" is

"step response" is the time behavior of the outputs of a general system when its inputs change from zero to one in a very short time. The concept can be extended to the abstract mathematical notion of a dynamical system using an evolution parameter. (wiki: Step Response)
"

Now we think about "RC time constant" in somehow different ways from the most text books.

1. The unit of Time Constant (usually denoted by $\tau$ ) is second (or time) as the name shows where as decay constant (usually denoted by λ) is 1/sec (or 1/ time). So " t / λ" is just number, no unit. So $e$ ^-t/RC makes sense, no unit, where RC = $\tau$ .

2. The unit of RC

The unit of RC is second (or time). See below (wiki: Ohm and Farad).

$\Omega = \dfrac{\mbox{V}}{\mbox{A}} = \dfrac{\mbox{m}^2 \cdot \mbox{kg}}{\mbox{s} \cdot \mbox{C}^2} = \dfrac{\mbox{J}}{\mbox{s} \cdot \mbox{A}^2}=\dfrac{\mbox{kg}\cdot\mbox{m}^2}{\mbox{s}^3 \cdot \mbox{A}^2}= \dfrac{\mbox{J} \cdot \mbox{s}}{\mbox{C}^2} = \dfrac{\mbox{1}}{\mbox{S}}$

in which the following units appear: volt (V), ampere (A), siemens (S), watt (W), second (s), joule (J), coulomb (C), kilogram (kg), and meter (m).

Please do not confuse siemens (S) with second (s)

$\mbox{F} = \,\mathrm{\frac{A \cdot s}{V} = \dfrac{\mbox{J}}{\mbox{V}^2} = \dfrac{\mbox{W} \cdot \mbox{s}}{\mbox{V}^2} = \dfrac{\mbox{C}}{\mbox{V}} = \dfrac{\mbox{C}^2}{\mbox{J}} = \dfrac{\mbox{C}^2}{\mbox{N} \cdot \mbox{m}} = \dfrac{\mbox{s}^2 \cdot \mbox{C}^2}{\mbox{m}^{2} \cdot \mbox{kg}} = \dfrac{\mbox{s}^4 \cdot \mbox{A}^2}{\mbox{m}^{2} \cdot \mbox{kg}} = \dfrac{\mbox{s}}{\Omega}}$

The unit of ＜RC time constant＞, as the name suggests, is s = second (time)

R : unit Ω (Ohm)
C : unit F (Farad)

so unit wise, by using the above unit formulas

RC: ΩF: Ω x s/Ω = s, or (s/F) x F = s

This is a bit strange and anti-tuition

3. Why $e$ $?$

Natural response (wiki: RC Circuit)

The simplest RC circuit is a capacitor and a resistor in series. When a circuit consists of only a charged capacitor and a resistor, the capacitor will discharge its stored energy through the resistor.

The voltage across the capacitor, which is time dependent, can be found by using Kirchhoff's current law, where the current through the capacitor must equal the current through the resistor.

(the same current runs through each element (current is a through variable): i_r (in) = i_{c (out)} although the current value changes with time or dynamically)

This results in the linear differential equation

$C\frac{dV}{dt} + \frac{V}{R}=0$ .

Solving this equation for V yields the formula for exponential decay: (*)

$V(t)=V_0 e^{-\frac{t}{RC}} \ ,$

where V₀ is the capacitor voltage at time t = 0.

The time required for the voltage to fall to $\frac{V_0}{e}$ is called the RC time constant and is given by

$\tau = RC \ .$

(*) You need some knowledge of solving this differential equation. See below.

Exponential decay (wiki)

A quantity undergoing exponential decay. Larger decay constants make the quantity vanish much more rapidly. This plot shows decay for decay constants of 25, 5, 1, 1/5, and 1/25 for x from 0 to 5.

A quantity is subject to exponential decay if it decreases at a rate proportional to its value. Symbolically, this process can be expressed by the following differential equation, where N is the quantity and λ (lambda) is a positive number called the decay constant:

$\frac{dN}{dt} = -\lambda N.$

(My note: The underlined parts are key. Exponential, either positive (growth) or negative (decay) has this nature. (Ref: ttp://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/)
Also note that N should be N (t), which value changes with time, dynamical changes)

The solution to this equation (see Solution of the differential equation below) is: Exponential rate of change

$N(t) = N_0 e^{-\lambda t}. \,$

Here N(t) is the quantity at time t, and N₀ = N(0) is the initial quantity, i.e. the quantity at time t = 0.

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Solution of the differential equation The equation that describes exponential decay is

or, by rearranging with respect time (t),

Integrating, we have

(**)

( N should be N (t) ) where C is the constant of integration, and hence

where the final substitution, N₀ = e^C, is obtained by evaluating the equation at t = 0, as N₀ is defined as being the quantity at t = 0.

This is the form of the equation that is most commonly used to describe exponential decay. Any one of decay constant, mean lifetime, or half-life is sufficient to characterise the decay.

(**) Again you need some knowledge of integration. See below.

Right hand side:

dN(t) 1
---- = ----- d N(t)
N (t) N (t)

The integrating the right hand side it becomes

ln N(t) + C

as

Then take exponential the both sides becomes

as

ln N(t)
e = N(t)

You must understand this. This seems tricky but this comes from the following definition or the inverse of this.
ln e^x= x

-----

Now back to the above result - RC time constant:

where V0 is the capacitor voltage at time t = 0.

The time required for the voltage to fall to is called the RC time constant and is given by

Relations between Capacitor Voltage, Capacitor Charge and Resistor current (or more generally the circuit current). Please note that the relation between Capacitor Voltage and Resistor Voltage is Vr+Vc=0 or Vr=-Vc.

Chart form Hyperphysics

Examples

When R = 1 Ohm and C = 1F, or more realistically R = 1K Ohm and C = 1000uF (1F/1,000) or R = 1M Ohm and C = 1uF (1F/1,000,000), RC = 1 (sec).

So t/RC --> 1 (sec)/ 1(sec) = 1 (no unit), then

V(1) = Vo / e ---> Vo / 0.3686...

Of course it is not necessary to be 1 and usually not 1. When R = 1K Ohm and C = 1000uF then RC = 1 /1000 (sec), very short period. Then

V(0.001) = Vo / e ---> Vo / 0.3686...

The original question in here is ＜Why e ?＞.

The answer is that this is a natural phenomenon. The nature has been made this way or God made it this way. Shall we leave science?

What is the cause of this rate, exp (-1) or 1 / e is made? It mat relate more with a function of capacitor, not resistor (which does not include differentiation itself). It may relate with how (in terns of time) a capacitor is charged and discharged. Discharge may be a reverse or inverse case of being charged. It is unrealistic to start with an already charged capacitor and it may be less difficult to think how a capacitor is charged and then reverse or inverse it to a case of being discharged. To do this we must add a battery to the above simple circuit.

From wiki: RC Circuit
These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is charging; for discharging, the equations are vice-versa. These equations can be rewritten in terms of charge and current using the relationships C=Q/V and V=IR (see Ohm's law).

Thus, the voltage across the capacitor tends towards V as time passes, while the voltage across the resistor tends towards 0, as shown in the figures. This is in keeping with the intuitive point that the capacitor will be charging from the supply voltage as time passes, and will eventually be fully charged.
These equations show that a series RC circuit has a time constant, usually denoted being the time it takes the voltage across the component to either rise (across C) or fall (across R) to within of its final value. That is, is the time it takes to reach and to reach .

The rate of change is a fractional per . Thus, in going from to

, the voltage will have moved about 63.2% of the way from its level at toward its final value. So C will be charged to about 63.2% after , and essentially fully charged (99.3%) after about . When the voltage source is replaced with a short-circuit, with C fully charged, the voltage across C drops exponentially with t from towards 0. C will be discharged to about 36.8% after , and essentially fully discharged (0.7%) after about . Note that the current, , in the circuit behaves as the voltage across R does, via Ohm's Law.

These results may also be derived by solving the differential equations describing the circuit:

and

. The first equation is solved by using an integrating factor and the second follows easily

Capacitor voltage step-response.

Resistor voltage step-response.

Chart form Hyperphysics

Charging a Capacitor When a battery is connected to a series resistor and capacitor, the initial current is high as the battery transports charge from one plate of the capacitor to the other. The charging current asymptotically approaches zero as the capacitor becomes charged up to the battery voltage. Charging the capacitor stores energy in the electric field between the capacitor plates. The rate of charging is typically described in terms of a time constant RC.

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Now we think about ＜the voltage will have moved about 63.2% of the way from its level at toward its final value. So C will be charged to about 63.2% after ＞.

When there is no charge on the capacitor plates charges will be built up quickly after the the battery voltage applied and the current starts to flow. As time passes more and more charges exist and therefore voltage become being built up on the plates but at the rate of

As time passes once charges (voltage) built up they tend to resist against further building up of new charges (voltage) so the the rate of building up slows down again at the rate of

because less and less space for charges (geometrically in 2D and 3D) or some other force (physically, chemically, etc) or say voltage against further building up of charges become stronger and stronger (higher and higher for voltage). The voltage relates as CV= Q (charge) or

where C is constant (-> Capacitance, sptt)

Now how about in case of being discharged. Discharging quickly at fast and then slowly at the rate of

This may be because that in the beginning many charges leave away from the plate quickly (due to higher voltage) once the exit wide open (short) and the time passes less and less charges (lower voltage) leave away from the plates. Or the magnitude of some force (say voltage like and relate with the number of the charges on the plates) which pushes the charges on the plates away from the plates becomes less and less strong as time passes.
Again the voltage relates as CV= Q (charge) or

Re-written in 2023.

sptt

Asnetcomp Tutorials

Wednesday, January 23, 2013

RC time constant - conceptual undertanding

Exponential decay (wiki)