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

The Voltage Divider Formula

The setup

One path from 5 V to ground through R1 then R2; the tap sits between them. "Voltage at the tap" = measured tap-to-ground.

Building the formula

Three short moves, each one a fact you've already measured. (The V = IR sheet on this shelf is the tool used in each step: read it first if you haven't.)

One path, one current. Everything flowing through R1 must continue through R2 — there's nowhere else to go. The supply pushes 5 V across the pair, so I = 5 ÷ (R1 + R2).
The tap reads R2's share. The tap-to-ground measurement is exactly the drop across R2: V_tap = I × R2.
Combine. Substitute step 1's current into step 2:

The divider formula

V_tap = 5 V × R2 ÷ (R1 + R2)

The tap voltage is the supply times R2's fraction of the total resistance. (For a supply other than 5 V, swap in its voltage.)

Sanity-check it at the extremes

An engineer's habit: before trusting a formula, push it to its limits and see if it behaves.

R2 enormous → the fraction R2/(R1+R2) creeps toward 1 → tap ≈ 5 V. ✓ matches "bigger bottom resistor, higher tap."
R2 tiny → fraction ≈ 0 → tap ≈ 0 V. ✓ the tap is practically wired to ground.
R1 = R2 → fraction = ½ → tap = 2.5 V, no matter what the equal value is. ✓ exactly what your 1 k/1 k and 4.7 k/4.7 k builds both read.

Worked: lesson 5's table, predicted

R1 (top)	R2 (bottom)	The fraction	Tap
1 kΩ	1 kΩ	1 ÷ 2 = 0.50	2.50 V
220 Ω	1 kΩ	1000 ÷ 1220 ≈ 0.82	≈ 4.1 V
1 kΩ	220 Ω	220 ÷ 1220 ≈ 0.18	≈ 0.9 V
4.7 kΩ	4.7 kΩ	0.50	2.50 V
4.7 kΩ	10 kΩ	10 ÷ 14.7 ≈ 0.68	≈ 3.4 V — the "build the 3.3" winner

Notice the second and third rows: swap the resistors and the two shares trade places, so the two taps sum to 5 V — the same split, read from opposite ends.

The pot is this formula with a knob

A potentiometer's wiper splits one resistance into R1-above and R2-below. Turning the knob trades resistance between them while the total stays fixed — so the fraction, and the tap, sweep smoothly from 0 to the full supply. Every volume knob ever made is this equation with a shaft through it.

What dividers are for

Half of electronics is making the right voltage appear at a node, and the divider is the cheapest machine that does it. The same fraction trick wears four different hats:

Set a reference. A chip that needs a steady comparison voltage — "flag me when the input crosses 2.5 V" — gets that 2.5 V from a divider. Your "build the 3.3" challenge was exactly this job.
Read a sensor. Swap one resistor for a part whose resistance changes with the world — a photoresistor (light), a thermistor (temperature) — and the tap becomes a voltage that reports the world: dark room, one reading; bright room, another. This is the heart of the sensor build coming later in the course.
Scale a signal down. A 5 V signal meeting a chip that only tolerates 3.3 V gets stepped through a divider first — the same fraction, applied to a voltage that's changing.
Put it on a knob. The pot: volume controls, dimmers, joystick axes — a divider whose fraction you turn.

What dividers are not for: power

A divider makes a voltage to be measured or compared — by a meter, a chip input, anything that draws almost nothing. Try to power something from the tap and the load becomes a second path across R2: the fraction shrinks and the tap sags (the loading note below computes it). Delivering real current at a lower voltage is a regulator's job — which is what's actually inside your supply making its rock-steady 3.3 V. Your challenge recreated the number; the regulator holds it under load.

Practice

Answers at the bottom. Every one of these is buildable — predict, then measure.

R1 = 10 kΩ, R2 = 10 kΩ. Tap?
R1 = 1 kΩ, R2 = 4.7 kΩ. Tap?
You need ≈ 1.0 V at the tap and R1 is 4.7 kΩ. Roughly what R2 — bigger or smaller than R1, and about how much? (Estimate first, then compute.)
Design your own "3.3 V from the kit" pair different from 4.7 k/10 k. How close can your kit's values get?

Why the LED sagged the tap — a preview

Lesson 5's stretch: hang an LED + 220 Ω across the tap and the voltage drops. The load offers the current a second path alongside R2, which behaves like a smaller R2 — a smaller fraction, a lower tap. Two resistances side by side acting like one smaller one is called parallel resistance — the Series & Parallel sheet works out the law, including this exact sag.

Answers

1. 2.50 V — equal halves, always. 2. 5 × 4.7 ÷ 5.7 ≈ 4.1 V. 3. You want a fraction of 1/5, so R2 must be much smaller: R2 ÷ (R2 + 4.7 k) = 0.2 → R2 ≈ 1.2 kΩ (a 1 k gives ≈ 0.88 V — close). 4. Any pair with R2 ÷ (R1+R2) ≈ 0.66 works: 1 k over 2.2 k gives 5 × 2.2 ÷ 3.2 ≈ 3.44 V; 4.7 k over 10 k ≈ 3.40 V. Real resistors are ±5 %, so "close" is the honest target.