Electrical Circuits and Sensors · Module 4 of 10
Capacitors, Inductors, and Transient Response
Resistors respond instantly; capacitors and inductors remember. They store energy and resist sudden change, and the result is a circuit whose voltages and currents evolve in time with a single time constant.
Readiness check
This module adds time to the circuit. Tick only what you can do closed-notes.
- Apply Ohm's law and KVL to a series loop.
- Recall that e0 = 1 and e-1 ≈ 0.368.
- Evaluate a decaying exponential at a given time.
- Recall that energy is stored in fields, not just dissipated.
- Convert µF, ms, and kΩ to base units.
The core idea
A capacitor stores energy in an electric field and resists sudden voltage change; an inductor stores energy in a magnetic field and resists sudden current change. Together with a resistor they make a first-order circuit that settles over a time constant.
iC = C dv/dt, vL = L di/dtRC circuit: τ = RC; RL circuit: τ = L/Rx(t) = xfinal + (x0 − xfinal) e−t/τA capacitor's current is proportional to how fast its voltage changes, i = C dv/dt, so its voltage cannot jump. An inductor's voltage is proportional to how fast its current changes, v = L di/dt, so its current cannot jump. When either is connected to a resistor and a source, the circuit obeys a first-order differential equation whose solution is a single exponential. The time constant τ, equal to RC for a capacitor circuit or L/R for an inductor circuit, sets the pace: after one τ the response has moved about 63 percent of the way to its final value, and after about five τ it has effectively arrived. Every measured signal that rises or decays smoothly is governed by this same law.
The skills, taught in order
Five skills introduce the two storage elements, the energy they hold, and the transient they produce.
4.1 Capacitance
A capacitor stores charge on two plates: q = Cv, so its current is i = C dv/dt. Capacitance C is measured in farads, almost always microfarads or smaller in practice. Because the voltage is the integral of current, it changes smoothly and cannot step instantaneously.
4.2 Inductance
An inductor opposes changes in current through it: v = L di/dt, with L in henries. The stored energy lives in the magnetic field of its coil. Its current is the integral of voltage, so the current changes smoothly and cannot step instantaneously.
| Element | v-i relation | Stored energy | Cannot jump |
|---|---|---|---|
| Capacitor | i = C dv/dt | ½ C v2 | voltage |
| Inductor | v = L di/dt | ½ L i2 | current |
The duality of the two elements: swap voltage for current, C for L, and one row becomes the other.
4.3 Energy storage
A capacitor holds ½Cv2 of energy; an inductor holds ½Li2. Unlike a resistor, which only dissipates, these elements return their stored energy to the circuit. That stored energy is what powers the transient when a source is switched.
4.4 The time constant
For a resistor and capacitor the time constant is τ = RC; for a resistor and inductor it is τ = L/R. A larger τ means a slower response. The number of time constants, not the absolute time, tells you how far the transient has progressed.
4.5 First-order transient response
Any first-order response is x(t) = xfinal + (x0 − xfinal)e−t/τ: start at the initial value, head exponentially toward the final value, governed by τ. Identify those three quantities and the whole waveform is written.
Engineering connection: a sensor's RC filter, in Module 9, is exactly this transient, used deliberately to smooth a noisy signal.
Worked example 1: charging an RC circuit
A 10 V source charges a 2 µF capacitor through a 5 kΩ resistor, starting from zero. Find the time constant, the capacitor voltage at t = τ, and the current at that instant.
- ProblemFind τ, vC(τ), and i(τ) for the charging circuit in Figure 1.
- Given / findVs = 10 V, R = 5 kΩ, C = 2 µF, vC(0) = 0. Find τ, vC(τ), i(τ).
- AssumptionsIdeal source and elements; capacitor initially uncharged.
- Modelτ = RC; vC(t) = Vs(1 − e−t/τ); i(t) = (Vs/R)e−t/τ.
- Equationsτ = RCvC(t) = Vs(1 − e−t/τ)i(t) = (Vs/R) e−t/τ
- Solveτ = 5000 × 2×10−6 = 10 ms. At t = τ: vC = 10(1 − e−1) = 10 × 0.632 = 6.32 V. i = (10/5000)e−1 = 0.002 × 0.368 = 0.736 mA.
- CheckAt t = τ the voltage is the standard 63.2 percent of 10 V, and the current is 36.8 percent of its initial 2 mA, as the exponential demands.
- ConclusionOne time constant moves the capacitor 63 percent of the way to the supply; the current dies away as the voltage builds.
Worked example 2: decay of an RL circuit
An inductor of 2 H carrying an initial current of 4 A is allowed to decay through an 8 Ω resistor. Find the time constant and the current after 250 ms.
- ProblemFind τ and the current at 250 ms for the decaying inductor in Figure 2.
- Given / findL = 2 H, R = 8 Ω, I0 = 4 A. Find τ and i(0.250 s).
- AssumptionsSource removed at t = 0; the inductor current is continuous, starting at 4 A.
- Modelτ = L/R; the source-free current is i(t) = I0e−t/τ.
- Equationsτ = L/Ri(t) = I0 e−t/τ
- Solveτ = 2/8 = 0.25 s = 250 ms. At t = 250 ms = τ: i = 4 e−1 = 4 × 0.368 = 1.47 A.
- CheckSince t equals exactly one τ, the current must be 36.8 percent of 4 A, which is 1.47 A. The result is independent of units once τ is in seconds.
- ConclusionThe RL decay mirrors the RC charge: the same exponential, now with τ = L/R, draining the stored magnetic energy into the resistor.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Letting a state jump | Capacitor voltage steps at t = 0 | "Which quantity must be continuous?" | Capacitor voltage and inductor current cannot jump. |
| Wrong time constant | τ built as L·R or R/C | "Is this RC or RL?" | Use τ = RC for capacitors, τ = L/R for inductors. |
| Confusing charge and decay | A rising curve used for a decay | "What are the initial and final values?" | Write xfinal + (x0 − xfinal)e−t/τ and let the signs sort it out. |
| Prefix slip in τ | τ off by orders of magnitude | "Are R, C, L in base units?" | Convert kΩ, µF, and mH before forming τ. |
Practice ladder
Find the time constant of a 10 kΩ resistor with a 4.7 µF capacitor.
Show answer
τ = RC = 10 000 × 4.7×10−6 = 0.047 s = 47 ms.
A capacitor charges from 0 toward 5 V with τ = 20 ms. What is its voltage at t = 40 ms?
Show answer
t/τ = 2, so v = 5(1 − e−2) = 5(1 − 0.135) = 5 × 0.865 = 4.32 V.
An inductor of 0.5 H decays through 100 Ω from 200 mA. Find τ and the current after 10 ms.
Show answer
τ = L/R = 0.5/100 = 5 ms. t/τ = 10/5 = 2, so i = 0.2 e−2 = 0.2 × 0.135 = 27.1 mA.
You want a sensor's RC filter to settle to within 1 percent in 50 ms. Roughly what time constant do you need, and why?
What good work looks like
Settling to 1 percent takes about five time constants (e−5 ≈ 0.0067), so τ ≈ 50/5 = 10 ms. Choosing R and C to give τ ≈ 10 ms trades noise smoothing against response speed.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Build or simulate an RC circuit, record its charging curve, and compare the measured time to reach 63 percent against the computed τ = RC.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the v-i relations for C and L.
i = C dv/dt for a capacitor; v = L di/dt for an inductor.
2. Which quantity cannot jump for each?
Capacitor voltage and inductor current are both continuous.
3. Give the two time constants.
τ = RC for an RC circuit; τ = L/R for an RL circuit.
4. How far does a transient move in one τ?
About 63 percent of the way from its initial to its final value.
5. Write the general first-order solution.
x(t) = xfinal + (x0 − xfinal)e−t/τ.
Textbook mapping
This module follows Alexander and Sadiku, Fundamentals of Electric Circuits, 4th edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Capacitors and inductors | Alexander & Sadiku, Chapter 6 |
| First-order RC and RL circuits | Alexander & Sadiku, Chapter 7 |
| The time constant and natural response | Alexander & Sadiku, Chapter 7 |
Chapter numbers refer to the 4th edition. The transient laws are standard, so any recent edition will align closely.