A series circuit with an inductor $L$, a resistor $R$ and a capacitor $C$ obeys the same equation as a damped spring. Everything from free vibrations and resonance carries over, with new names.
Key idea
With charge $q(t)$ and current $i = q'$: $$Lq'' + Rq' + \frac{1}{C}q = E(t).$$ Natural frequency $\omega_0 = \frac{1}{\sqrt{LC}}$; underdamped when $R^{2} < \frac{4L}{C}$, critically damped when $R^{2} = \frac{4L}{C}$, overdamped when $R^{2} > \frac{4L}{C}$.
| Spring | Circuit |
|---|---|
| mass $m$ | inductance $L$ |
| damping $c$ | resistance $R$ |
| stiffness $k$ | $\frac{1}{C}$ |
| position $x$ | charge $q$ |
| velocity $x'$ | current $i$ |
| force $F(t)$ | voltage $E(t)$ |
Why it works
Kirchhoff's voltage law: around the loop, the voltage drops add up to the source voltage. The drops are $L\frac{di}{dt}$ across the inductor, $Ri$ across the resistor and $\frac{q}{C}$ across the capacitor. With $i = q'$ this is $Lq'' + Rq' + \frac{q}{C} = E$. The inductor resists changes in current the way mass resists changes in velocity, and the capacitor pushes back in proportion to its charge the way a spring does.
Worked example: Charging with oscillation
$L = 1$ H, $R = 2$ Ω, $C = \frac{1}{5}$ F, a 10 V battery, $q(0) = 0$, $i(0) = 0$. Find $q(t)$ and $i(t)$.
- $q'' + 2q' + 5q = 10$. Particular solution: $q_p = CE = 2$.
- Roots of $r^{2} + 2r + 5$: $-1 \pm 2i$, so $q$ $= 2$ $+ e^{-t}(c_1\cos 2t + c_2\sin 2t)$.
- $q(0) = 2 + c_1 = 0$, so $c_1 = -2$; $q'(0) = -c_1 + 2c_2 = 0$, so $c_2 = -1$.
- Differentiate: $i = q' = 5e^{-t}\sin 2t$.
Answer: $q = 2 - e^{-t}(2\cos 2t + \sin 2t)$ C and $i = 5e^{-t}\sin 2t$ A
Worked example: Critical resistance
With $L = 0.5$ H and $C = 0.02$ F, which $R$ makes the circuit critically damped?
- Set $R^{2} = \frac{4L}{C} = \frac{2}{0.02} = 100$.
Answer: $R = 10$ Ω
Worked example: Tuning a circuit
A radio circuit has $L = 0.1$ H and $C = 0.001$ F. At what angular frequency does it resonate?
- $LC = 10^{-4}$, so $\omega_0 = \frac{1}{\sqrt{LC}} = 100$.
Answer: 100 rad/s: a source at that frequency builds the largest current
Common mistake
Writing the damping test as $R^{2} < 4LC$. In the dictionary, $k$ corresponds to $\frac{1}{C}$, so $c^{2} < 4mk$ becomes $R^{2} < \frac{4L}{C}$. A units check helps: $\frac{L}{C}$ has units of ohms squared.