Newton's second law for a falling object and Kirchhoff's voltage law for a simple circuit both produce the same kind of linear first-order equation, $y' + ay = b$. Learning to read that equation once pays off in both places.
Key idea
$y' + ay = b$ with $a > 0$ has the equilibrium $y = \frac{b}{a}$, and every solution approaches it like $e^{-at}$: $$y = \frac{b}{a} + \left(y_0 - \frac{b}{a}\right)e^{-at}.$$ The time constant $\frac{1}{a}$ is the time to close about 63% of the remaining gap.
| System | Equation | Long-run value | Time constant |
|---|---|---|---|
| Falling with drag $kv$ | $mv' = mg - kv$ | $v_T = \frac{mg}{k}$ | $\frac{m}{k}$ |
| RC circuit | $Rq' + \frac{q}{C} = E$ | $q = CE$ | $RC$ |
| RL circuit | $Li' + Ri = E$ | $i = \frac{E}{R}$ | $\frac{L}{R}$ |
Why it works
For the falling object, take down as positive. Gravity pulls down with force $mg$; air resistance $kv$ pushes against the motion. So $mv' = mg - kv$. As $v$ grows, the drag grows until it balances gravity at $v_T = \frac{mg}{k}$, the terminal velocity. In a circuit, the battery voltage $E$ equals the sum of the voltage drops: $Ri$ across a resistor, $\frac{q}{C}$ across a capacitor, $Li'$ across an inductor, with $i = q'$.
Worked example: Terminal velocity
A 2 kg object falls from rest with drag $4v$ newtons ($g = 9.8$ m/s²). Find $v(t)$ and the time to reach half the terminal velocity.
- $2v' = 19.6 - 4v$, so $v' + 2v = 9.8$.
- Equilibrium $v_T = 4.9$ m/s; from rest, $v = 4.9(1 - e^{-2t})$.
- Half of $v_T$: $e^{-2t} = \frac{1}{2}$, so $t = \frac{\ln 2}{2}$.
Answer: $v = 4.9(1 - e^{-2t})$ m/s; half speed after $\frac{\ln 2}{2} \approx 0.35$ s
Worked example: Charging a capacitor
An RC circuit has $R = 100$ Ω, $C = 0.001$ F and a 9 V battery; the capacitor starts uncharged. Find $q(t)$ and the current.
- $100q' + 1000q = 9$, so $q' + 10q = 0.09$.
- Equilibrium $q = CE = 0.009$ C; time constant $RC = 0.1$ s.
- $q = 0.009(1 - e^{-10t})$, and $i = q' = 0.09e^{-10t}$ A.
Answer: $q = 0.009(1 - e^{-10t})$ C; the current starts at $\frac{E}{R} = 0.09$ A and dies away
Worked example: Switching on an inductor
An RL circuit has $L = 0.5$ H, $R = 10$ Ω and $E = 12$ V, with $i(0) = 0$. Find $i(t)$.
- $0.5i' + 10i = 12$, so $i' + 20i = 24$.
- Steady current $\frac{E}{R} = 1.2$ A, time constant $\frac{L}{R} = 0.05$ s.
Answer: $i = 1.2(1 - e^{-20t})$ A
Common mistake
Sign errors in the force balance. Drag always opposes motion; with down positive it is $-kv$. If your terminal velocity comes out negative, or if the speed grows without limit, a sign is wrong. A second slip is forgetting to divide by $m$ (or $R$, or $L$) before reading off the rate $a$.