Scherz & Monk, Chapter 2

Circuit Theory

One idea runs through everything: a voltage — electrical "pressure" — pushes charge through materials. From that single root grow Ohm's law, Kirchhoff's conservation laws, dividers, Thévenin reduction, and — once we let the pressure oscillate — reactance, impedance, and resonance. This is the analytical toolkit the whole book runs on.

Prerequisites: basic algebra & ratios + metric prefixes (mA, µF, kΩ). Calculus optional — only three derivative facts matter.
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Chapters
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Simulations
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Assumed Knowledge

Chapter 0: 9 Volts In, 5 Volts Out

You have a fresh 9 V battery and a microcontroller whose input pin is rated for exactly 5 V. Wire the battery straight to the pin and you push 9 V across a part built for 5 — the magic smoke escapes and the chip is dead. You do not have a 5 V battery. What you do have is a drawer full of resistors.

Can two ordinary resistors turn 9 V into 5 V? They can. The arrangement is called a voltage divider, and it is the single most-used circuit pattern in all of electronics. But the divider also hides a trap: the moment your microcontroller actually draws current, the carefully-tuned 5 V quietly sags. Understanding exactly why — and how much — requires the entire chapter ahead.

Here is the idea in one line. Stack two resistors R1 and R2 across the 9 V supply. The current that flows is the same through both (they are in series). Each resistor "drops" a share of the 9 V in proportion to its resistance. Tap the junction between them, and you read out whatever fraction R2 claims:

Vout = Vin · R2 / (R1 + R2)

To get 5 V from 9 V we need R2/(R1+R2) = 5/9 = 0.556. Pick a convenient R2 = 10 kΩ and solve for the top resistor:

R1 = R2 · (Vin − Vout) / Vout = 10000 · (9 − 5) / 5 = 8000 Ω

Check: Vout = 9 · 10000 / (8000 + 10000) = 9 · 10000/18000 = 9 · 0.5556 = 5.0 V. It works. Drag the sliders below and hunt for the 5 V output yourself.

Why this is the perfect first problem. Solving it forces you to meet, in order: current (the same charge flows through both resistors), voltage (it splits proportionally), resistance (the splitting ratio), and the looming idea of a load (the chip drawing current ruins the ratio). Every tab in this chapter is really just one of those four ideas, examined closely.

The Voltage Divider — Find the 5 V Tap

Two resistors across a 9 V supply. Adjust R1 (top) and R2 (bottom). The tap voltage Vout is shown live. Goal: land on 5.00 V. Hint: R1 ≈ 8 kΩ with R2 = 10 kΩ.

R1 (top, kΩ) 8.0
R2 (bottom, kΩ) 10.0
Using a 9 V supply with R1 = 1 kΩ on top and R2 = 2 kΩ on the bottom, what is the tap voltage Vout?

Chapter 1: Current & Charge

Before voltage can push anything, there has to be something to push. That something is electric charge, carried in metals by electrons. Charge is measured in coulombs (C). A single electron carries a tiny negative charge of −1.602×10−19 C, so one coulomb is a staggering 6.24×1018 electrons — about six billion billion of them.

Current is simply the rate at which charge flows past a point. If a charge ΔQ passes in a time Δt, the current is:

I = ΔQ / Δt     (1 ampere = 1 coulomb per second)

So 1 A = 6.24×1018 electrons streaming past every second. Current is to charge what flow rate (litres per second) is to water (litres). The unit, the ampere, is named for André-Marie Ampère.

Worked example — counting electrons

An LED draws 20 mA = 0.020 A. How much charge moves through it in one minute, and how many electrons is that?

ΔQ = I · Δt = 0.020 · 60 = 1.2 C

Electrons: 1.2 C ÷ 1.602×10−19 C/electron = 7.5×1018 electrons in one minute — through a part the size of a lentil.

Conventional vs. electron flow

Benjamin Franklin guessed (wrong, as it turned out) that the moving charges were positive, and defined conventional current as flowing from + to −. Electrons actually drift the other way, from − to +. We have kept Franklin's convention for 250 years because the math comes out identical — just remember the two run opposite. Every arrow in this book points the conventional way.

The garden-hose-already-full trick. Electrons drift astonishingly slowly — a drift velocity of about 0.002 mm/s in a typical wire, slower than a snail. Yet the lamp lights the instant you flip the switch. How? The wire is already packed end-to-end with electrons. Push one in at the switch and one pops out at the lamp almost immediately — the field propagates near the speed of light. It is a hose already full of water: open the tap and water exits the far end at once, even though any single droplet barely crawls.
Charge Flow Counter

Set a current and a time interval. The widget computes the charge transferred and the electron count, and animates carriers drifting past a cross-section. Notice how doubling the current doubles the flow density.

Current (mA) 20
Time (s) 60

Current benchmarks worth memorizing

Device / situationTypical current
Indicator LED~20 mA
Smartphone (active)~200 mA
100 W incandescent bulb (120 V)~0.8–1 A
Car starter motor~200 A
Lightning strike (peak)~30,000 A
Dangerous through the heart100 mA – 1 A
If 0.5 C of charge flows past a point in 2 seconds, what is the current?

Chapter 2: Voltage — the Push

Charge will sit perfectly still forever unless something urges it to move. That urge is voltage — more precisely, a potential difference. Voltage measures how much energy each coulomb of charge gains or loses moving between two points:

1 volt = 1 joule per coulomb     V = Energy / Charge

A 9 V battery means: every coulomb that travels from the + terminal to the − terminal through your circuit delivers 9 joules of energy. Voltage is named for Alessandro Volta, inventor of the first chemical battery.

The water-pump analogy (load-bearing)

Picture a closed loop of pipe with a pump. The pump is the battery: it raises water to high pressure. Pressure difference is voltage. Flow rate is current. A narrow constriction in the pipe is resistance. Crucially, voltage is always a difference between two points — just as pressure only means something relative to somewhere else. There is no such thing as "the voltage at one point" until you pick a reference.

That reference is called ground (or 0 V). By convention we plant a flag, declare it 0 V, and measure every other node relative to it. Ground is not a magical sink — it is just the agreed-upon zero of the pressure scale, like sea level for altitude.

VAB = VA − VB     (voltage is always between two points)

Worked example — energy from a battery

A 9 V battery pushes 1.2 C through a circuit (our LED-for-a-minute from Chapter 1). How much energy did it deliver?

E = V · Q = 9 · 1.2 = 10.8 joules

That energy ends up as light and heat in the LED. The battery's chemistry supplied it.

Voltage does not flow. A common beginner error is to say "voltage flows through the wire." It does not — current flows; voltage is the difference in potential that drives the flow. You measure current by breaking the circuit and inserting the meter in series; you measure voltage by touching two points without breaking anything (in parallel). The water analogy: flow rate is measured along the pipe, pressure is measured across two taps.
Voltage as Height — the Reference Matters

Three nodes at fixed potentials. Move the ground reference between them and watch every node voltage relabel. The differences never change — only the labels do. That is what "voltage is relative" means.

Node A potential (V) 9.0
Ground reference C
A battery moves 3 C of charge and delivers 36 J of energy. What is its voltage?

Chapter 3: Resistance & Ohm's Law

Push charge through a material and the material pushes back. That opposition is resistance, measured in ohms (Ω). Georg Ohm discovered the beautifully simple relationship binding voltage, current, and resistance — the most important equation in this book:

V = I · R    →    I = V / R    →    R = V / I

One ohm is the resistance that lets exactly 1 A flow when 1 V is applied. Bigger R means more push needed for the same flow — a narrower pipe. Ohm's law is the lever you pull on constantly: know any two of V, I, R and the third falls out.

Where resistance comes from: shape and material

A resistor's value depends on what it is made of and its dimensions. The intrinsic "stickiness" of a material is its resistivity ρ (rho, in Ω·m). A wire of length L and cross-section A has:

R = ρ · L / A    →    ρ = R · A / L

Longer = more resistance (longer pipe). Fatter = less resistance (wider pipe). Copper's resistivity is about 1.72×10−8 Ω·m — conductors sit near 10−8, insulators near 1014. That is a span of twenty-two orders of magnitude, the widest range of any common physical property.

Resistance changes with temperature

Heat a metal and its atoms jiggle harder, scattering electrons more — resistance rises. The linear approximation:

ρ = ρ0 [1 + α(T − T0)]

where α is the temperature coefficient. This is exactly how a resistance-temperature sensor works, and why a cold incandescent filament draws a big inrush current the instant you switch it on.

Worked example — the Ohm's-law triangle

An LED needs about 2 V across it and 20 mA through it, fed from 9 V. The series resistor must drop the leftover 9 − 2 = 7 V while carrying 0.020 A:

R = V / I = 7 / 0.020 = 350 Ω  (use the standard 360 Ω)
The triangle mnemonic. Write V on top, I and R side by side below: a triangle. Cover the quantity you want and the remaining two show the formula. Cover V → I×R. Cover I → V/R. Cover R → V/I. The widget below is this triangle — and it also glows: as power P = V·I climbs, the resistor heats up, because resistance always converts electrical energy into heat.
Ohm's-Law Triangle — with Power Glow

Set the voltage across a resistor and its resistance. Current I = V/R and power P = V·I are computed live and shown as bars. The resistor body glows hotter (orange → red) as dissipated power rises — this is real Joule heating.

Voltage V (V) 9.0
Resistance R (Ω) 350
A 12 V source drives 3 A through a resistor. What is its resistance?

Chapter 4: Power & Heat

Voltage is energy per charge; current is charge per second. Multiply them and the charges cancel, leaving energy per second — that is power, measured in watts (W):

P = V · I     (1 watt = 1 joule per second)

Substituting Ohm's law gives two more forms, all equal, each handy when you happen to know a different pair of quantities:

P = V · I = I2 · R = V2 / R

The form P = I2R is special: it says heat dissipation grows with the square of current. Double the current and you quadruple the heat. This is Joule heating — the unavoidable conversion of electrical energy into thermal energy whenever current passes through resistance. It is the principle behind every toaster, kettle, and incandescent bulb, and the reason fat wires carry big currents without melting.

Worked example — sizing the LED resistor's power rating

Back to our 350 Ω LED resistor carrying 20 mA. How much heat must it shed?

P = I2R = (0.020)2 · 350 = 0.0004 · 350 = 0.14 W

A standard 1/4 W (0.25 W) resistor handles this with margin. Had we wanted 200 mA, the heat would be 100× larger (0.04×350 = 14 W) and would need a chunky power resistor on a heatsink. The square law bites hard.

Worked example — the quiz benchmark

A 12 V device draws 8.3 A. Its power consumption:

P = V · I = 12 · 8.3 = 99.6 W ≈ 100 W
Why high-voltage transmission lines exist. To deliver a fixed power P = V·I, you can use high voltage and low current, or low voltage and high current. But the heat wasted in the wires is I2Rwire — it depends only on current. So the grid ships power at hundreds of thousands of volts and tiny current, slashing I2R losses, then transformers step it down for your house. The entire architecture of the power grid is one application of P = I2R.
Power Dissipation Explorer

Set voltage and current; watch all three power formulas agree and the implied resistance. The bar fills toward the resistor's power rating — cross it and the component overheats (red zone).

Voltage V (V) 12
Current I (A) 2.0
A 12 V device draws 8.3 A. What is its power consumption?

Chapter 5: Resistors in Series & Parallel

Real circuits have many resistors. Two simple rules collapse any ladder of them into a single equivalent — and those rules follow directly from the conservation laws we will name properly in Chapter 7.

Series: same current, voltages add

Resistors are in series when they form a single chain with no branches — the same current must flow through every one (charge has nowhere else to go). Their resistances simply add:

Rseries = R1 + R2 + R3 + …

The supply voltage divides among them in proportion to their values — which is exactly the voltage divider from Chapter 0. Three 1 kΩ resistors in series make 3 kΩ.

Parallel: same voltage, currents add

Resistors are in parallel when both ends connect to the same two nodes — each sees the full voltage, and the total current splits among them. Here the reciprocals add:

1/Rparallel = 1/R1 + 1/R2 + …     (two only:) R = R1R2/(R1+R2)

The crucial intuition: parallel resistance is always smaller than the smallest resistor. Adding another path can only make it easier for current to flow. Two equal resistors in parallel give exactly half their value.

Worked example — 1k ∥ 2k ∥ 4k

1/R = 1/1000 + 1/2000 + 1/4000 = 0.001 + 0.0005 + 0.00025 = 0.00175
R = 1 / 0.00175 = 571 Ω

As promised, 571 Ω is less than the smallest member (1 kΩ). And two equal 1 kΩ in parallel: R = (1000·1000)/(2000) = 500 Ω, exactly half.

The shortcut everyone forgets. N identical resistors of value R in parallel give R/N — no reciprocal arithmetic needed. And a quick sanity check for any parallel combination: the answer must be smaller than your smallest resistor but larger than (smallest)/N. If your calculator says otherwise, you inverted something.
Series / Parallel Equivalent Resistance

Three resistors. Toggle between series and parallel wiring and watch the equivalent resistance recompute. The schematic redraws to match. Confirm: series sums, parallel falls below the smallest.

R1 (Ω) 1000
R2 (Ω) 2000
R3 (Ω) 4000
Three resistors 1 kΩ, 2 kΩ, and 4 kΩ are wired in parallel. The equivalent resistance is approximately:

Chapter 6: Dividers & the Loading Trap

We can finally close the loop on Chapter 0. With R1 = 8 kΩ and R2 = 10 kΩ the divider reads a clean 5.0 V — as long as nothing is connected to the tap. The instant your microcontroller pin actually draws current, reality intrudes.

Why the load sags the output

Your load (call it RL) appears in parallel with R2. From Chapter 5, parallel resistance is always smaller than either resistor alone. So R2 effectively shrinks, it now claims a smaller share of the supply, and Vout drops below 5 V. The lighter the load (bigger RL), the less the sag.

R2′ = R2 · RL / (R2 + RL)     Vout = Vin · R2′ / (R1 + R2′)

Worked example — a 20 kΩ load on the 5 V tap

Load RL = 20 kΩ parallels R2 = 10 kΩ:

R2′ = (10k · 20k)/(30k) = 6.667 kΩ
Vout = 9 · 6667 / (8000 + 6667) = 9 · 0.4546 = 4.09 V

Our perfect 5 V collapsed to 4.09 V — an 18% error — just from a modest load. A heavier load (smaller RL) would be even worse.

The 10% rule of thumb

Designers tame loading with a simple rule: make the divider's current at least 10× the load current, equivalently keep R1, R2 roughly ≤ RL/10. Stiffer (lower-value) dividers waste more power but hold their voltage. A divider is a reference, not a power source — for real current, you follow it with a transistor or a voltage regulator (Chapter 11).

This is the chapter's quiet lesson. A formula that is exactly right on paper can be badly wrong in a circuit because the act of measuring/using the output changes it. The voltage divider is the gateway to a deeper idea — every real source has internal resistance, and connecting a load forms a divider with it. That insight is Thévenin's theorem, two tabs from now.
Voltage Divider with Load Toggle (10% Rule)

The 9→5 V divider. Toggle the load on, then drag RL. Watch Vout sag away from 5 V as the load gets heavier. The meter turns red when the error exceeds 10%.

Load RL (kΩ) 20
You connect a load to a divider's output and Vout drops below its no-load value. Why?

Chapter 7: Kirchhoff's Two Laws

Ohm's law handles one resistor at a time. To analyze a whole network we need two conservation principles, stated by Gustav Kirchhoff. They are nothing more than "charge is conserved" and "energy is conserved," dressed for circuits — and every series/parallel rule you have met is a special case of them.

KCL — the current (node) law

At any junction (node), charge cannot pile up or vanish, so:

∑ Iin = ∑ Iout     (current into a node = current out)

Water-pipe version: whatever flows into a T-junction must flow out. If 3 A enters a node and one branch carries 1 A away, the other branch must carry exactly 2 A. This is why parallel branch currents add up to the total.

KVL — the voltage (loop) law

Walk around any closed loop and the voltage rises and drops must cancel — you return to where you started at the same potential:

∑ V = 0     (around any closed loop)

The battery raises potential; each resistor drops it. KVL says the sum of the resistor drops equals the battery rise. This is why series voltages add — and it is the engine of the voltage divider.

Worked example — a single loop

A 9 V battery in series with R1 = 8 kΩ and R2 = 10 kΩ (our divider). KCL: the same current I flows through both (one loop, one current). KVL around the loop:

9 = I·R1 + I·R2 = I(8000+10000) → I = 9/18000 = 0.5 mA

Then the drop across R2 is VR2 = I·R2 = 0.0005·10000 = 5.0 V — the divider formula, derived from Kirchhoff rather than memorized. Check: VR1 = 0.0005·8000 = 4.0 V, and 4.0 + 5.0 = 9.0 V. KVL closes.

Everything reduces to these two laws. Series and parallel formulas, dividers, Thévenin, Norton, superposition — all are bookkeeping shortcuts for solving the system of equations that KCL and KVL produce. For any network: write one KCL equation per node and one KVL equation per loop, plug in Ohm's law, and solve the simultaneous equations. The "tricks" just save you from doing that by hand every time.
KCL Node Balancer

A node with two currents flowing in and one branch out. KCL fixes the second outgoing branch automatically. Drag the inputs; the dependent branch always balances so ∑in = ∑out.

Iin,1 (mA) 30
Iin,2 (mA) 20
Iout,1 (mA) 18
A node has 30 mA and 20 mA flowing in, and one branch carrying 18 mA out. What must the second outgoing branch carry?

Chapter 8: Thévenin, Norton & Superposition

Suppose you face a tangled box of batteries and resistors with just two wires sticking out. You want to know how it behaves when you hook up a load. Must you re-solve the whole mess for every load? No. Two theorems let you replace any linear two-terminal network with a trivially simple equivalent.

Thévenin's theorem

Any linear network of sources and resistors, seen from two terminals, behaves exactly like a single voltage source VTh in series with a single resistor RTh.

Norton's theorem

The dual statement: the same network equals a single current source IN in parallel with a resistor RN. The two are interchangeable:

RN = RTh     IN = VTh / RTh     VTh = IN · RN

Worked example — our divider as a Thévenin source

The 9 V / 8 kΩ / 10 kΩ divider seen from its output terminals:

VTh = open-circuit Vout = 9 · 10k/18k = 5.0 V
RTh = R1 ∥ R2 = (8k·10k)/18k = 4.44 kΩ

So the divider is a 5 V source with 4.44 kΩ of internal resistance. Now reconnect that 20 kΩ load — it forms a divider with RTh:

Vout = 5 · 20k/(4.44k+20k) = 5 · 0.818 = 4.09 V

Exactly the loaded value we computed the hard way in Chapter 6 — now it falls out in one line. That is the power of Thévenin: the loading sag is just the divider formed by RTh and RL.

Superposition

When several sources act at once in a linear circuit, the response is the sum of the responses to each source acting alone (others zeroed). Analyze one source at a time, then add. It turns a multi-source headache into several easy single-source problems.

Internal resistance is everywhere. A real 9 V battery is a Thévenin source: an ideal 9 V behind a small RTh (an alkaline 9 V is roughly 1–2 Ω fresh, tens of ohms when worn). That is why a tired battery still reads 9 V on a meter (no load → no drop) yet collapses under a motor (big current → big I·RTh drop). "Measures fine, works terribly" is always a Thévenin-resistance story.
Thévenin Equivalent & Loading

A Thévenin source VTh behind RTh feeding a load RL. Watch terminal voltage vs. delivered current as the load varies — the classic drooping source characteristic. Maximum power transfer occurs at RL = RTh.

VTh (V) 5.0
RTh (kΩ) 4.4
RL (kΩ) 20
A real battery reads exactly 9 V on a meter but droops badly when a motor is attached. The best explanation is:

Chapter 9: AC & RMS

Everything so far assumed a steady, one-direction DC (direct current). But the wall socket, radio signals, and audio all use AC (alternating current) — voltage that swings positive and negative, over and over, tracing a sine wave. Generalizing our toolkit to AC is the gateway to the entire second half of electronics.

Describing a sinusoid

An AC voltage is fully described by three numbers: its peak amplitude VP, its frequency f (cycles per second, hertz), and its phase φ (where in the cycle it starts):

V(t) = VP · sin(2πf·t + φ)

The period T is the time for one full cycle, and f = 1/T. The angular frequency ω = 2πf packages the 2π for convenience. A 60 Hz mains supply repeats every T = 1/60 = 16.7 ms.

The RMS problem — what is the "voltage" of a wave?

An AC voltage spends as much time negative as positive, so its average is zero — useless. What we want is the value that delivers the same heating power as an equivalent DC voltage. That is the root-mean-square (RMS) value. For a sine wave it works out to a clean factor:

VRMS = VP / √2 = 0.707 · VP     VP = 1.414 · VRMS

Worked example — the wall outlet's real peak

A "120 VAC" outlet quotes the RMS value (because RMS is what determines heating and what your meter reads). Its actual peak swing is much larger:

VP = 1.414 · 120 = 170 V

So that outlet really swings between +170 V and −170 V, 60 times a second. The 120 V figure is the DC-equivalent for power. This is why mains insulation must be rated well above 120 V.

RMS is the honest average for power. Why root-mean-square? Because heating goes as voltage squared (P = V²/R). Square the wave (now everything is positive), take the mean, then square-root back to a voltage. For a sine this ritual yields exactly peak/√2. The 0.707 factor is not arbitrary — it is the geometry of a sine wave, and it is why a 170 V peak heats a toaster like a steady 120 V DC would.
AC Waveform Builder — with RMS Line

Build a sinusoid: set amplitude, frequency, and phase. The dashed lines mark ±VRMS = 0.707·VP. Toggle a second wave to see phase difference. Read off the period as 1/f.

Amplitude VP (V) 100
Frequency (Hz) 2.0
Phase φ (deg) 0
A wall outlet is rated 120 VAC (RMS). What is its approximate peak voltage?

Chapter 10: Capacitors, Inductors & Reactance

Resistors burn energy. The two remaining passive components — capacitors and inductors — instead store energy and give it back, and they do something a resistor never does: they care about how fast the voltage or current is changing. This is where time and frequency enter.

The capacitor — it fights voltage change

A capacitor stores charge on two plates. Charge stored is proportional to voltage, the constant being capacitance C (farads):

Q = C · V     and so     I = C · dV/dt

The second form is the whole story: current flows only when voltage is changing. A capacitor resists sudden voltage changes — it takes time to charge. Put a resistor in series and the voltage approaches the supply along an exponential curve governed by the time constant τ:

VC(t) = VS(1 − e−t/RC)     τ = R · C

Worked example — the RC time constant

R = 100 kΩ, C = 10 µF:

τ = RC = 100000 · 0.00001 = 1.0 second

After one τ the capacitor reaches 63.2% of the supply; after 2τ, 86.5%; after 3τ, 95%; after 5τ it is essentially full (>99%). Those percentages are universal — they come from e−1, e−2, … and never change.

The inductor — it fights current change

An inductor (a coil) stores energy in a magnetic field. Its defining law is the mirror image of the capacitor's:

VL = L · dI/dt     E = ½L·I2     τ = L/R

An inductor resists sudden current changes. Inductance L scales with the square of the number of turns (L ∝ N²).

Reactance — resistance's AC cousin

Under AC, "how hard is it to push current" becomes reactance X (ohms), and it depends on frequency:

XC = 1 / (2πfC)       XL = 2πfL

A capacitor's reactance falls with frequency (it passes high frequencies, blocks DC). An inductor's reactance rises with frequency (it passes DC, blocks high frequencies). They are exact opposites.

XC example: 220 pF at 10 MHz = 1/(2π·10×106·220×10−12) ≈ 72 Ω
XL example: 100 µH at 120 Hz = 2π·120·100×10−60.075 Ω; at 15 MHz ≈ 9425 Ω
One mantra to keep them straight. "Capacitors pass change, inductors pass steadiness." A cap blocks DC and passes high-frequency wiggle (that is why it is a coupling/bypass part). An inductor passes DC and chokes off high-frequency wiggle (that is why it is a "choke"). Memorize the slopes: XC slides down with frequency, XL climbs up. The next tab is what happens where those two lines cross.
RC Charge / Discharge Curve with τ Markers

Set R and C; τ = RC updates. Press Charge or Discharge to animate VC(t). Dotted markers sit at 1τ (63%), 2τ (86.5%), 3τ (95%), 5τ (99%). The vertical line sweeps with real time scaled to τ.

R (kΩ) 100
C (µF) 10
τ = 1.00 s
After exactly one time constant τ, a charging capacitor reaches what fraction of the supply voltage?

Chapter 11: Impedance, Resonance & the Whole Picture

We have a frequency-independent resistance R and two frequency-dependent reactances XC and XL that move in opposite directions. Impedance Z unifies them — it is the total opposition to AC, combining resistance and net reactance. Because resistive and reactive effects are 90° out of phase, they combine like the sides of a right triangle:

|Z| = √(R2 + X2)     where X = XL − XC

Resonance — where the two reactances cancel

Since XC falls with frequency and XL rises, there is exactly one frequency where they are equal. There the net reactance XL − XC is zero, they cancel completely, and the circuit is purely resistive. This special frequency is resonance f0:

XL = XC ⇒ 2πf0L = 1/(2πf0C) ⇒ f0 = 1 / (2π√(LC))

Worked example — an LC tank

L = 5 µH, C = 35 pF:

f0 = 1/(2π√(5×10−6 · 35×10−12)) = 1/(2π√(1.75×10−16)) ≈ 12 MHz

This is precisely how a radio tunes: change C (the tuning knob) and you move f0 to select a station. The sharpness of the peak is the quality factor Q = X/R — high Q means a narrow, selective resonance.

The punchline of the whole chapter. Capacitors fight voltage change and pass high frequencies; inductors fight current change and pass low frequencies; put them together and they fight each other. At resonance the fight is a perfect draw — energy sloshes back and forth between the cap's electric field and the inductor's magnetic field, and the circuit rings like a bell. Every oscillator, filter, and radio in this book lives or dies by f0 = 1/(2π√LC).
SHOWCASE — Reactance Sweep & Resonance Finder

The payoff sim. XC (teal, falling) and XL (orange, rising) plotted against frequency on a log axis. Their crossing is resonance f0. Drag L and C to move it; drag R to see |Z| (purple) and its dip at f0. Toggle the |Z| trace and the resonance marker.

Inductance L (µH) 5
Capacitance C (pF) 35
Resistance R (Ω) 50
f₀ ≈ 12 MHz

The complete cheat sheet

Every key equation from this chapter, in one place:

ConceptEquationIn one sentence
CurrentI = ΔQ/ΔtCharge per second; 1 A = 1 C/s.
VoltageV = E/QEnergy per coulomb; 1 V = 1 J/C.
Ohm's lawV = I·RThe master relation between push, flow, opposition.
ResistivityR = ρL/AResistance from material and shape.
PowerP = VI = I²R = V²/REnergy per second; heat goes as I².
Series RR = R1+R2+…Same current, resistances add.
Parallel R1/R = ∑1/RiSame voltage; always below the smallest.
DividerVout = VinR2/(R1+R2)Voltage splits by resistance ratio.
KCL / KVL∑Iin=∑Iout; ∑Vloop=0Charge and energy conserved.
ThéveninVTh, RTh = R seriesAny 2-terminal box = source + resistor.
NortonIN = VTh/RThThe current-source dual of Thévenin.
ACV(t)=VPsin(2πft+φ)f = 1/T; ω = 2πf.
RMSVRMS = VP/√2The DC-equivalent heating voltage (0.707·peak).
CapacitorQ=CV; I=C·dV/dtFights voltage change; passes high f.
RC constantτ = RC63% in 1τ, >99% in 5τ.
InductorV=L·dI/dt; τ=L/RFights current change; passes low f.
ReactanceXC=1/(2πfC); XL=2πfLAC opposition; opposite slopes.
Impedance|Z|=√(R²+X²)Total AC opposition.
Resonancef0=1/(2π√LC)Where XL=XC and they cancel.

Where this goes next

Components (Ch 3) puts physical bodies on every symbol here — resistor color codes, capacitor types, real inductor parasitics.
Semiconductors (Ch 4) break linearity: diodes and transistors obey nonlinear I–V curves, but you still analyze them with Thévenin and Ohm around the device.
Filters (Ch 9) are dividers where one leg is a reactance — the frequency-dependent Vout from this chapter, made useful.
Oscillators (Ch 10) are resonance turned into a sustained ring: f0 = 1/(2π√LC) sets the tone.

Connections to sibling chapters

"From one push on a charge, all of circuit theory unfolds."

You can now turn 9 V into 5 V, predict when it sags, collapse any network to a source-plus-resistor, and follow a signal from DC steadiness all the way to resonance. Chapter 3 hands you the real components to build with.

As frequency increases in an LC circuit, how do the two reactances change — and what happens at resonance?
← Chapter 1: Introduction Chapter 3: Components →