Electrical Circuits
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 7.3: “Applications of Second-Order Differential Equations” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/7-3-applications-of-second-order-differential-equations |
| Textbook used in class | Stewart, Calculus, Section 17.3: “Applications of Second-Order Differential Equations” (RLC circuit example) |
Quick Reference
For a series RLC circuit, Kirchhoff’s voltage law gives: $$L\,Q'' + R\,Q' + \frac{1}{C}\,Q = E(t),$$ where $Q(t)$ is the charge on the capacitor, $L$ = inductance (H), $R$ = resistance ($\Omega$), $C$ = capacitance (F), $E(t)$ = electromotive force (V). Current $I = Q'$.
Spring-RLC analogy:
| Spring-mass | RLC circuit |
|---|---|
| mass $m$ | inductance $L$ |
| damping $c$ | resistance $R$ |
| spring constant $k$ | $1/C$ |
| displacement $x$ | charge $Q$ |
| velocity $x'$ | current $I = Q'$ |
| external force $F(t)$ | EMF $E(t)$ |
Motivation
An RLC circuit obeys the same second-order linear DE as a spring-mass system. Every qualitative feature -- oscillation, decay, resonance -- has an exact electrical counterpart. Resistance $R$ plays the role of damping: a purely $LC$ circuit oscillates forever (no resistance = no damping), while adding resistance damps the oscillation. Engineers exploit this to tune circuits to specific frequencies (radio tuners) or to suppress oscillation (filters).
Because the equations are identical in form, everything you know about springs applies directly to circuits. If you can solve a spring problem, you can solve the corresponding circuit problem by substituting the analogous quantities.
Key Concept: Transient vs Steady State
When $E(t)$ is a sinusoid, the particular solution (steady-state response) is also a sinusoid at the same frequency. The complementary solution (transient) involves $e^{\alpha t}$ with $\alpha < 0$ and decays to zero.
After a long time, only the steady-state survives. Engineers care most about the steady-state amplitude and phase shift relative to the driving EMF, particularly near the resonant frequency $\omega_0 = 1/\sqrt{LC}$, where the amplitude peaks.
Worked Example
An RLC circuit has $L = 1$ H, $R = 6\ \Omega$, $C = 1/9$ F, $E(t) = 0$. If $Q(0) = 1$ C and $I(0) = Q'(0) = 0$, find $Q(t)$.
The DE is: $$Q'' + 6Q' + 9Q = 0.$$
Characteristic equation: $r^2 + 6r + 9 = (r+3)^2 = 0 \Rightarrow r = -3$ (repeated).
General solution: $Q(t) = (c_1 + c_2 t)e^{-3t}$.
Apply initial conditions: $Q(0) = c_1 = 1$. $Q'(t) = c_2 e^{-3t} + (c_1+c_2 t)(-3)e^{-3t} = (c_2 - 3c_1 - 3c_2 t)e^{-3t}$. $Q'(0) = c_2 - 3c_1 = c_2 - 3 = 0 \Rightarrow c_2 = 3$.
$$Q(t) = (1+3t)e^{-3t}.$$
This is critically damped: the charge returns to zero without oscillating.
the voltage across the resistor equals $E(t)$. In a series RLC circuit, the EMF equals the sum of the voltages across all three elements: $E = V_L + V_R + V_C = LQ'' + RQ' + Q/C$. The resistor alone does not carry the full driving voltage. This is why Kirchhoff’s voltage law produces the full second-order DE, not a simpler relation.
Leveled Practice
Problem 1. An LC circuit (no resistance) has $L = 1$ H and $C = 1/4$ F, with $E = 0$. Classify the behavior and write the natural frequency.
Show answer
DE: $Q'' + 4Q = 0$. Characteristic equation: $r^2 + 4 = 0$, $r = \pm 2i$.
Undamped: $Q(t) = c_1\cos 2t + c_2\sin 2t$.
Natural frequency: $\omega_0 = 2$ rad/s, period $\pi$ s.
Problem 2. Using the analogy table, what value of $R$ makes the RLC circuit in Problem 1 critically damped?
Show answer
Critical damping: $R^2 = 4L/C = 4(1)(4) = 16$, so $R = 4\ \Omega$.