Electronics · Math shelf · Optional — for the curious · build 2026.09.23-1644

Kirchhoff's Laws

The idea

Two conservation laws run every circuit ever built: energy (the voltage law) and charge (the current law). You've already measured both. Gustav Kirchhoff wrote them down in 1845 — as a 21-year-old student.

KVL — Kirchhoff's voltage law

The version you proved in lesson 4: around any loop, the drops sum to the supply voltage. The formal dress is one notch sharper — give every voltage a sign (a rise through the supply is +, a drop across a part is −) and walk any closed loop:

around any closed loop: the signed voltages sum to 0

5 V 220 Ω +5 V rise −3.0 V drop −2.0 V drop walk the loop +5 − 3.0 − 2.0 = 0 ✓
The bookkeeping: pick a direction, walk the whole loop, sign every voltage. If the sum isn't zero, you mis-measured — or discovered perpetual motion.

Why must it be zero? Voltage is energy per coulomb (the What's a Coulomb? sheet): a coulomb arriving back where it started must have spent exactly what the supply gave it. A loop that summed high would mint free energy; one that summed low would leak it into nowhere. Conservation of energy, wearing circuit clothes.

And the law says any closed loop — including a loop through two parallel branches that never touches the supply. Walk it and the sum forces a famous fact: parts in parallel share one identical voltage. Hold that thought for practice problem 4.

KCL — Kirchhoff's current law

The version you've used since the junction dot: current into a node equals current out. Formal dress:

at any node: current in = current out

10 mA in 6 mA out 4 mA out 10 = 6 + 4 ✓
Charge doesn't pool and doesn't vanish. Whatever arrives at a node this second leaves it this second — through some combination of the other wires.

This one is conservation of charge: coulombs aren't created or destroyed at a junction, and they can't pile up on a wire — so the per-second flow in must equal the per-second flow out. It's why branch currents add (the whole parallel resistance law is KCL in disguise), and why a 0 V resistor reading condemns an entire single loop: no current here means no current anywhere on the one path.

The complete toolkit

Here's the formalization worth savoring: KVL + KCL + Ohm's law is the whole game. Every circuit-analysis trick this shelf has taught is those three laws pre-solved for a common case:

The shortcutWhat it secretly is
drops sum to the supplyKVL, one loop
series resistances addKVL + Ohm, same current
parallel: reciprocals addKCL + Ohm, same voltage
the divider formulaKVL + KCL + Ohm, solved once for two resistors

Worked, to prove the divider claim: the 1 kΩ / 1 kΩ divider. KCL at the tap (nothing loading it): the current in R1 continues into R2 — one current, I. KVL around the loop: 5 − I·1000 − I·1000 = 0, so I = 2.5 mA. The tap sits above ground by R2's drop: 2.5 mA × 1 kΩ = 2.5 V. Same answer the divider formula hands you — because the formula is this derivation, done once and bottled.

The person

Gustav Kirchhoff (Germany, 1824–1887) published these laws in 1845 while still a student. He wasn't done: with Robert Bunsen he founded spectroscopy and explained the dark lines in sunlight — telling humanity, for the first time, what the sun is made of. The same mind behind "your drops must sum to five" also read the chemistry of the stars.

The math, for the curious — always optional

Write KCL for every node and KVL for every loop of a big circuit and you get a system of simultaneous linear equations — one unknown per current and node voltage. That's not just theory: circuit-simulation software (engineers use a family called SPICE) analyzes a design by building exactly that equation system and solving it, thousands of times over. Every chip in your laptop was rehearsed as Kirchhoff bookkeeping before it was ever manufactured.

Practice

Answers at the bottom. Problems 3 and 4 are buildable — predict, then measure.

  1. Walk the blue-LED loop: 5 V supply, 220 Ω, blue LED (3.2 V). What must the resistor drop, by KVL alone?
  2. A node has 12 mA flowing in and two wires out; one carries 7 mA. The other?
  3. The dark red+blue series circuit from lesson 4: the resistor reads 0 V and the red LED reads 1.7 V. Without touching the meter again — what is the blue LED reading?
  4. New build: red and blue LEDs in parallel (same two nodes), one 220 Ω feeding the pair. The meter reads ≈ 2.0 V across both LEDs. Use KVL's parallel fact to explain what the blue LED is doing — before you build it and look.

Answers

1. 5 − 3.2 = 1.8 V — no Ohm's law needed. 2. 12 − 7 = 5 mA. 3. 5 − 0 − 1.7 = 3.3 V — KVL works on dark circuits too; the whole supply still lands somewhere. 4. Parallel parts share one voltage, and the red LED conducts at ≈ 2.0 V, clamping the shared node there. Blue needs ≈ 3.2 V to open, never gets it, stays dark: red hogs the branch. (Series red+blue: both dark. Parallel: red only. Both are Kirchhoff bookkeeping.)