Appearance
1.5 — Capacitors, Inductors and Transients
Switch on a torch and the light appears instantly. Switch on a camera flash and it takes a second to charge, then dumps everything in a millisecond. Switch off a motor and a spark jumps across the switch contacts, even though the supply was only 12 V.
Resistors cannot do any of that. A resistor's current is set by the voltage right now and it has no memory. The three behaviours above — slow filling, sudden dumping, and a voltage far higher than the supply — all come from components that store energy and therefore remember what happened a moment ago.
The equations here are differential equations, and Volume II Chapter 6.1 is where the solving method comes from. The physics behind the two components is Volume IV Chapter 4.3 for the capacitor's stored electric field and Chapter 4.6 for the inductor's magnetic one; this chapter takes both as given and works out what they do in a circuit.
1. The capacitor: storing charge on two plates
Take two metal plates, put them close together with an insulator between, and connect them to a battery. Electrons pile onto one plate; an equal number are pulled off the other. No charge ever crosses the gap — the insulator prevents that. But charge has accumulated, one plate is negative and the other positive, and there is now an electric field in the gap holding energy.
The amount of charge you can pile on for a given voltage is the capacitance:
C = \frac{Q}{V} \qquad\text{so}\qquad Q = CV
The unit is the farad (F), after Michael Faraday. One farad is one coulomb per volt, which is a colossal amount — most capacitors you will meet are picofarads to microfarads, twelve to six orders of magnitude smaller. A one-farad capacitor was exotic laboratory equipment until supercapacitors arrived.
For a parallel-plate capacitor:
C = \frac{\varepsilon A}{d}
where A is the plate area, d the separation, and \varepsilon (Greek epsilon) the permittivity of the material between them. Bigger plates, closer together, and a better dielectric all raise capacitance. That single formula explains why real capacitors are built the way they are.
A film capacitor is metres of metallised plastic rolled up into a cylinder. Rolling is how you fit a large plate area into a small volume, since the formula rewards area directly. An electrolytic capacitor attacks the other term instead: it grows an oxide layer only a few tens of nanometres thick on aluminium foil, making d absurdly small. Both are the same equation, optimised from opposite ends.
Permittivity is written as two factors multiplied together, \varepsilon = \varepsilon_r\varepsilon_0. The second, \varepsilon_0 = 8.854\times10^{-12} F/m, is the permittivity of empty space and never changes. The first, \varepsilon_r, is the relative permittivity: a plain multiplier saying how much better than vacuum a particular material is at this job. Older books call it the dielectric constant.
The numbers span a wide range. Air is about 1, meaning it is barely better than nothing. Plastics are 3 to 4. The ceramics used in modern surface-mount capacitors reach into the thousands, which is how a component the size of a grain of rice holds a microfarad.
The defining equation, and why it has a derivative in it
Differentiate Q = CV with respect to time. C is constant, so:
\frac{dq}{dt} = C\frac{dv}{dt}
and since current is dq/dt (Chapter 1.1):
\boxed{i = C\,\frac{dv}{dt}}
Read aloud: "i equals C dee v by dee t." The current into a capacitor is proportional to how fast its voltage is changing, not to the voltage itself.
Two enormous consequences fall straight out of this equation, and everything else about capacitors is a special case of them.
First: the voltage across a capacitor cannot change instantly. If v jumped from 0 to 5 V in zero time, dv/dt would be infinite, so the current would be infinite. Nothing supplies infinite current. Therefore capacitor voltage is always continuous — it can rise fast, but never in a step. This one fact is what makes a capacitor a memory element, and it is the starting condition for every transient calculation.
Second: to a steady DC voltage, a capacitor is an open circuit. If v is constant, dv/dt = 0, so i = 0. Once it has finished charging, no current flows at all. This is why a capacitor blocks DC and why it can sit across a power rail forever without drawing current.
Energy stored
Work out how much energy goes in while charging. Power is p = vi = vC\frac{dv}{dt}, and energy is power integrated over time:
E = \int_0^T vC\frac{dv}{dt}\,dt = C\int_0^V v\,dv = C\cdot\frac{V^2}{2}
\boxed{E = \tfrac{1}{2}CV^2}
Energy goes as the square of voltage, which is why a camera flash capacitor is charged to 300 V rather than 3 V: a hundred times the voltage is ten thousand times the energy in the same component. It is also why a charged high-voltage capacitor is dangerous long after the equipment is unplugged, and why service manuals tell you to discharge them through a resistor before touching anything.
Series and parallel — the opposite way round from resistors
In parallel, capacitors share voltage and their charges add, so their capacitances add:
C_{\parallel} = C_1 + C_2 + \cdots
Which makes physical sense: putting two capacitors side by side is the same as building one with double the plate area.
In series, the same charge is pushed onto every one (there is nowhere else for it to go), and the voltages add:
\frac{1}{C_{\text{series}}} = \frac{1}{C_1}+\frac{1}{C_2}+\cdots
Also physical: two in series is like one with double the plate separation.
Note the reversal. Series adds for resistors and reciprocals for capacitors; parallel is reciprocals for resistors and adds for capacitors. If you remember the physical pictures — area and separation — you will never have to remember which is which.
2. The inductor: storing energy in a magnetic field
Wind a wire into a coil and pass current through it, and a magnetic field appears through the coil's middle. Change the current, and the changing field induces a voltage across the coil that opposes the change. That is Faraday's law of induction plus Lenz's law, and the opposition is the entire personality of the component.
The inductance L is defined by how much voltage a given rate of change of current produces:
\boxed{v = L\,\frac{di}{dt}}
Unit: the henry (H), after Joseph Henry, who discovered induction independently of Faraday but published later. Practical inductors run from nanohenries (a straight piece of wire) to henries (a mains transformer winding).
Look at that equation next to the capacitor's. They are the same equation with v and i swapped, and C replaced by L. That is duality, and it means every capacitor result has a mirror inductor result:
First: the current through an inductor cannot change instantly. A step change in current would need infinite voltage. Inductor current is always continuous.
Second: to a steady DC current, an inductor is a short circuit. Constant current means di/dt = 0, so v = 0. An ideal inductor with DC through it has no voltage across it at all.
Energy stored
By the same integration as before, but with the roles swapped:
\boxed{E = \tfrac{1}{2}Li^2}
And here is the spark at the switch, explained. A motor winding might hold \frac12\times0.1\ \text{H}\times(5\ \text{A})^2 = 1.25 J. Open the switch and you are demanding that the current fall from 5 A to 0 in the microsecond it takes the contacts to part. The equation v = L\,di/dt then produces a voltage of
v = 0.1 \times \frac{5}{10^{-6}} = 500{,}000\ \text{V}
or as much as the air gap will stand before it breaks down and arcs. The inductor will find a way to keep its current flowing, and if you do not give it a path it will make one through the air.
The engineering fix is the flyback diode (also called a freewheeling diode): a diode connected across the coil, backwards, so it does nothing during normal operation but provides a loop for the current to circulate through when the switch opens. Every relay, every motor driver and every switching power supply has one, and Chapter 2.2 explains the diode that makes it work. Leave it out and you destroy the transistor doing the switching, usually within a few operations.
Series and parallel
Inductors follow the resistor pattern, not the capacitor one:
L_{\text{series}} = L_1+L_2+\cdots, \qquad \frac{1}{L_{\parallel}} = \frac{1}{L_1}+\frac{1}{L_2}+\cdots
provided their magnetic fields do not couple. When they do couple, they become a transformer, which is Chapter 9.1.
3. The RC transient — charging and discharging
Now put a resistor and capacitor in series with a switch and a battery, and ask what actually happens over time. This is the circuit behind every timing, filtering and debouncing job in electronics.
Deriving the curve
Close the switch at t = 0. KVL around the loop:
V = iR + v_C
The current is the capacitor's own current, i = C\,\dfrac{dv_C}{dt}, so:
V = RC\frac{dv_C}{dt} + v_C
Rearranged:
RC\frac{dv_C}{dt} = V - v_C
Read that in words before solving it: the rate at which the voltage rises is proportional to how far it still has to go. Far from the target, it climbs fast; close to the target, it crawls. That description alone tells you the shape is an exponential approach, and it is the same equation as a cooling cup of tea (Volume II, Chapter 6.1).
Separate the variables and integrate:
\int\frac{dv_C}{V-v_C} = \int\frac{dt}{RC}
-\ln(V-v_C) = \frac{t}{RC} + k
Exponentiate and apply the initial condition v_C(0) = 0, which we are entitled to use because capacitor voltage cannot jump:
\boxed{v_C(t) = V\left(1 - e^{-t/RC}\right)}
The current follows from i = (V - v_C)/R:
i(t) = \frac{V}{R}e^{-t/RC}
The current starts at its maximum V/R — at the first instant the uncharged capacitor looks like a short circuit — and decays to zero.
The time constant
The product RC has units of seconds (ohms times farads really does come out in seconds; check it if you like) and is given its own symbol:
\tau = RC
\tau is the Greek letter tau, and it is called the time constant. Substituting t=\tau:
v_C(\tau) = V(1 - e^{-1}) = V(1-0.368) = 0.632V
One time constant gets you 63.2% of the way. The table everyone in electronics knows by heart:
| Time | Fraction of final value | Remaining |
|---|---|---|
| 1τ | 63.2% | 36.8% |
| 2τ | 86.5% | 13.5% |
| 3τ | 95.0% | 5.0% |
| 4τ | 98.2% | 1.8% |
| 5τ | 99.3% | 0.7% |
The pattern is that each time constant removes another 63% of whatever gap remains. Strictly the capacitor never finishes charging, since the exponential only reaches its target at infinity, but after five time constants the error is under one percent and no real measurement would notice.
Discharging
Remove the source and connect the charged capacitor across the resistor. Now KVL gives v_C + iR = 0 with i = C\,dv_C/dt, so:
RC\frac{dv_C}{dt} = -v_C \quad\Rightarrow\quad \boxed{v_C(t) = V_0e^{-t/RC}}
A pure decay with the same time constant. The same 63% rule applies, counting downward: after one τ, 36.8% of the original voltage remains.
Worked example
R = 10\ \text{k}\Omega, C = 100\ \mu\text{F}, supply 5 V.
\tau = 10{,}000 \times 100\times10^{-6} = 1\ \text{second}
After 1 s: 5(1-e^{-1}) = 3.16 V. After 3 s: 5(1-e^{-3}) = 4.75 V. Time to reach 4 V:
4 = 5(1-e^{-t}) \Rightarrow e^{-t} = 0.2 \Rightarrow t = \ln 5 = 1.61\ \text{s}
Peak current at t=0: 5/10{,}000 = 0.5 mA.
This is exactly the circuit inside a 555 timer, Chapter 2.6, where a comparator watches the capacitor voltage and flips a switch when it crosses a threshold. The whole family of blinking-LED, tone-generating and pulse-width circuits is this equation plus a comparator.
4. The RL transient
Duality gives it to you for free. A resistor and inductor in series across a source, switch closed at t=0:
V = iR + L\frac{di}{dt}
which is the same shape of equation with i in place of v_C. The solution is:
i(t) = \frac{V}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}
The time constant is now L/R, not LR. This is the one place duality does not simply swap symbols, and it is worth checking why: bigger inductance means more opposition to change, so slower — L on top. Bigger resistance means the current settles at a smaller value which it reaches sooner — R on the bottom.
The voltage across the inductor is the mirror image:
v_L(t) = Ve^{-t/\tau}
Full supply voltage across the inductor at the first instant, decaying to zero. At t=0 an inductor with no initial current looks like an open circuit; after a long time it looks like a wire. Precisely opposite to the capacitor.
5. Two rules that solve almost every transient without calculus
For any single-energy-storage circuit driven by DC sources, you can write the answer down without solving a differential equation, using this:
\boxed{x(t) = x_\infty + \left(x_0 - x_\infty\right)e^{-t/\tau}}
where x is any voltage or current, x_0 is its value just after the switching instant, and x_\infty is the value it settles at.
To use it you need three numbers, and each has a trick:
- x_0 — the initial value. Find it using continuity: capacitor voltage and inductor current cannot jump, so their values just after the switch equal their values just before. Everything else in the circuit follows from those.
- x_\infty — the final value. Redraw the circuit with capacitors as open circuits and inductors as short circuits (the DC steady-state rules from sections 1 and 2) and solve it as a plain resistive problem.
- \tau — the time constant. It is R_{Th}C or L/R_{Th}, where R_{Th} is the Thevenin resistance seen by the storage element with sources killed — exactly the calculation of Chapter 1.4.
Worked example using the rule
A 12 V source, R_1 = 4\ \text{k}\Omega in series, feeding a node with R_2 = 6\ \text{k}\Omega to ground and $C = 1\ \mu$F to ground. The capacitor starts uncharged; the switch closes at t=0. Find v_C(t).
Initial value. Capacitor uncharged, so v_C(0) = 0 V.
Final value. Capacitor is an open circuit, so it is just a divider:
v_C(\infty) = 12\cdot\frac{6}{4+6} = 7.2\ \text{V}
Time constant. Kill the source (short it). The capacitor then sees 4\text{k}\parallel6\text{k} = 2.4\ \text{k}\Omega:
\tau = 2400 \times 10^{-6} = 2.4\ \text{ms}
Assemble:
v_C(t) = 7.2 + (0-7.2)e^{-t/2.4\text{ms}} = 7.2\left(1-e^{-t/2.4\text{ms}}\right)\ \text{V}
Done in four lines, no integration. This is the method to use in every exam and every real design.
6. RLC — when both storage elements are present
Put a resistor, inductor and capacitor in a series loop and the equation gains a second derivative, because you now have two independent energy stores that can hand energy back and forth:
L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{1}{C}i = 0
This is precisely the damped mass-on-a-spring equation from Volume II, Chapter 6.2, with L playing the mass, R the friction, and 1/C the spring stiffness. The mathematics is identical, so the behaviour is identical, and the three regimes have the same names.
Define two quantities:
\omega_0 = \frac{1}{\sqrt{LC}} \quad\text{(natural frequency, rad/s)}, \qquad \zeta = \frac{R}{2}\sqrt{\frac{C}{L}} \quad\text{(damping ratio)}
\zeta is the Greek letter zeta and is a pure number with no units. It alone decides the character of the response:
| Damping ratio | Name | What you see |
|---|---|---|
| \zeta \lt 1 | underdamped | rings, overshoots, decaying oscillation |
| \zeta = 1 | critically damped | fastest arrival with no overshoot |
| \zeta \gt 1 | overdamped | slow, no overshoot |
Underdamped is the interesting one, because it means the circuit oscillates on its own at frequency
\omega_d = \omega_0\sqrt{1-\zeta^2}
with the amplitude shrinking as e^{-\zeta\omega_0 t}. Energy sloshes between the capacitor's electric field and the inductor's magnetic field, losing a bit to the resistor each cycle, exactly as a swinging pendulum trades height for speed and loses a bit to air.
Where you see each one: a digital signal arriving down a long trace often rings, which is an underdamped RLC formed by the trace's inductance and the receiver's input capacitance, and the fix is a small series resistor to raise \zeta. A power supply output is designed close to critically damped so it recovers from a load step fast without overshooting into the load's maximum voltage rating.
If the resistance is zero the damping is zero, and the circuit oscillates forever at \omega_0. That is the LC tank, the heart of every oscillator and every radio tuner, and it is where Chapter 1.6 goes next.
Everything in this chapter has been about the transient — the once-off settling after a switch flips. If instead you drive a circuit with a sine wave that never stops, the differential equations become unbearable and there is a much better way: replace them with algebra using complex numbers. That is Chapter 1.6, and it is the trick that makes all of AC engineering possible.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Capacitors
C = \frac{Q}{V}, \qquad Q = CV, \qquad C = \frac{\varepsilon_r\varepsilon_0A}{d}
i = \frac{dq}{dt} = C\frac{dv}{dt}
Derivation: differentiate Q = CV with respect to time, holding C constant.
E = \int vi\,dt = \int vC\frac{dv}{dt}dt = C\int_0^V v\,dv = \tfrac12CV^2
C_\parallel = \sum C_k, \qquad \frac{1}{C_{series}} = \sum\frac{1}{C_k}
Why parallel adds: side-by-side plates behave as one capacitor with the areas summed, and C \propto A. Why series takes reciprocals: stacked capacitors behave as one with the separations summed, and C \propto 1/d.
Inductors
v = L\frac{di}{dt}, \qquad E = \tfrac12Li^2
L_{series} = \sum L_k, \qquad \frac{1}{L_\parallel} = \sum\frac{1}{L_k}
Energy derivation: E = \int vi\,dt = \int Li\frac{di}{dt}dt = L\int_0^I i\,di = \tfrac12LI^2, the identical integration to the capacitor with v and i swapped.
Transients
RC charging
V = RC\frac{dv_C}{dt} + v_C
Separate and integrate:
\int\frac{dv_C}{V-v_C} = \int\frac{dt}{RC} \;\Rightarrow\; -\ln(V-v_C) = \frac{t}{RC}+k
With v_C(0)=0:
v_C = V\left(1-e^{-t/RC}\right), \qquad i = \frac{V}{R}e^{-t/RC}
RC discharging
v_C = V_0e^{-t/RC}
RL
i = \frac{V}{R}\left(1-e^{-t/\tau}\right), \qquad v_L = Ve^{-t/\tau}, \qquad \tau = \frac{L}{R}
The universal shortcut
x(t) = x_\infty + (x_0-x_\infty)e^{-t/\tau}
with x_0 from continuity (capacitor voltage and inductor current never jump), x_\infty from the DC steady state (capacitor open, inductor short), and \tau from the Thevenin resistance seen by the storage element.
Time constant table
\frac{v}{V} = 1-e^{-n} \ \text{after } n\tau: \quad 0.632,\ 0.865,\ 0.950,\ 0.982,\ 0.993
Second order — RLC
L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac{i}{C}=0
Guess i = e^{st}; each derivative multiplies by s:
Ls^2+Rs+\frac{1}{C}=0 \quad\Rightarrow\quad s = \frac{-R\pm\sqrt{R^2-4L/C}}{2L}
\omega_0 = \frac{1}{\sqrt{LC}}, \qquad \zeta = \frac{R}{2}\sqrt{\frac{C}{L}}, \qquad \omega_d = \omega_0\sqrt{1-\zeta^2}
\zeta \lt 1 underdamped and ringing; \zeta = 1 critically damped and fastest without overshoot; \zeta \gt 1 overdamped and slow.
What the next chapter fixes
Solving a differential equation for every circuit is exhausting, and almost nothing in the real world is a single switching event — it is a sine wave, repeating forever. Chapter 1.6 shows that for a steady sine, the calculus collapses into ordinary algebra with complex numbers, which is the single most useful simplification in electrical engineering.