Circuits with a single reactive element (inductor or capacitor).

1. Capacitor

Recall that “capacitor’s law” is

\[i_C(t) = C \frac{\mathrm{d} \, V_C(t)}{\mathrm{d} t} ,\]

or

\[v_C(t) = \frac{1}{C} \int\limits_0^t i_C(x)\,\mathrm{d}x \; + v_C(0^-) .\]
Remember that the x in the integral is a dummy variable and disappears during integration. Time shows up in the integral’s upper limit, so we can’t also legally use t as the variable of integration.

2. Inductor

Recall that “inductor’s law” is

\[v_L(t) = L \frac{\mathrm{d} \,i_L(t)}{\mathrm{d} t} ,\]

or

\[i_L(t) = \frac{1}{L} \int\limits_0^t v_L(x)\,\mathrm{d}x \; + i_L(0^-) .\]
Remember that the \(t=0\) in the lower limit and the initial condition is just a placeholder that sets the boundary between “then” and “for any time afterwards”. The initial condition doesn’t need to be at 0 seconds.

3. Single time-constant solution

a.k.a. a first-order circuit

Single time constant refers to the τ in the \(\exp(-t/\tau)\) term. First-order refers to the circuit solution being a first order differential equation. It is always true that a circuit with N total inductors and capacitors yields an Nth-order differential equation.[1]

A circuit with a reactive element (i.e. a capacitor) has a circuit analysis solution that involves a differential equation. Since there is only one reactance, there is only one time constant. The solution to this first-order differential equation is an exponential: \(A \exp(-t/\tau) + B\)

The R in the τ=R·C is an effective resistance and may NOT be the value of a single physical resistor! Think "Thevenin-equivalent resistance seen by the capacitor" instead.

The general solution from your differential equations course is correct, but you always then need to solve for the two constants using the initial conditions of the particular circuit. A more practical and easier form is:

The only first-order differential equation solution you need to know
\[v(t) = \underbrace{\biggl( V_{\text{final}} - V_{\text{initial}} \biggr)}_{\text{change}} \underbrace{\left[ 1 - \exp{\left(\dfrac{-t}{\tau}\right)} \right]}_{\text{diff eq magic}} + \underbrace{V_{\text{initial}}}_{\text{don't forget}}\]

  • \(\tau = R\,C \text{ or } L / R\)

  • \(t_{\text{initial}} = 0\)

  • replace every V with I for the current solution

If t is in seconds, then this forces τ to be in units of seconds because it makes no sense to raise a number to a power having units, for example “2.7 to the 3 meters power”.

Does it seem weird that resistance × capacitance equals time? See how the units work out once you expand them into their more basic units:

R C units
\[\begin{align*} R C \rightarrow & \, \Omega \cdot F\\ &\frac{V}{A} \cdot \frac{C}{V} = \frac{C}{A} = \frac{A\cdot s}{A}\\ R C \rightarrow & \, \text{seconds} \end{align*}\]
L/R units
\[\begin{align*} \frac{L}{R} \rightarrow \; & \frac{H}{\Omega} \\ &\frac{Wb}{A} \cdot \frac{A}{V} = \frac{V \cdot s}{A} \cdot \frac{A}{V} \\ \frac{L}{R} \rightarrow & \, \text{seconds} \end{align*}\]
The ampere is an SI base unit while the coulomb is a derived unit (A·s). But it is common to think of the reverse: an ampere is a coulomb per second.
1. …​and compels you to love Laplace transforms more than differential equations. Note that “more than” doesn’t necessarily mean you love either very much :)