why does RC time constant happen to equal 1/e?
Let's look at a simple circuit with an RC time constant:







simulate this circuit &ndash; Schematic created using CircuitLab

Now, if 

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

then you can write a differential equation for the voltage across the capacitor for \$t &gt; 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.