Build and analyze a voltage divider: Worked examples

This is the "throw every possible input at it" companion to the main voltage-divider note. Before touching numbers we map out every kind of situation a divider question can be, so that after these examples you never meet a case you haven't already seen.

Reminder of the one formula everything comes from — derived in the parent note from Ohm's Law applied twice:

The scenario matrix

Each row is a class of situation. The last column names the example that nails it.

#	Case class	What's special about it	Covered by
A	Equal resistors	$R_{1} = R_{2}$ → split in half	Ex 1
B	Unequal resistors	ratio ≠ 1 → uneven split	Ex 2
C	Degenerate: $R_{2} = 0$	bottom resistor is a wire	Ex 3
D	Degenerate: $R_{1} = 0$	top resistor is a wire	Ex 3
E	Limiting: $R_{2} ≫ R_{1}$	output → full $V_{in}$	Ex 4
F	Limiting: $R_{2} ≪ R_{1}$	output → near $0$	Ex 4
G	Design (solve for a resistor)	given $V_{o u t}$ , find $R_{1}$	Ex 5
H	Loading effect	load $R_{L}$ across $R_{2}$	Ex 6
I	Real-world word problem	sensor / potentiometer	Ex 7
J	Exam twist (multi-stage / KVL)	three resistors, tap the middle	Ex 8

We'll walk them in order. Figures accompany the cases where geometry of the split helps you see it.

The split is literally a length divided in two — see the bar below (each resistor "owns" a length equal to its resistance, and $V_{o u t}$ is the fraction from the bottom).

Figure — Build and analyze a voltage divider

Compare the two split bars side by side (Ex 1 vs Ex 2) to see why the fraction moves:

Worked example Ex 3 — Cases C & D: degenerate (a resistor becomes a wire)

$V_{in} = 12 V$ . (C) $R_{1} = 2 k Ω$ , $R_{2} = 0$ . (D) $R_{1} = 0$ , $R_{2} = 2 k Ω$ .

Forecast: a $0 Ω$ resistor is just a plain wire with no voltage drop. If $R_{2}$ is a wire, the tap sits directly on ground → $0$ V. If $R_{1}$ is a wire, the tap sits directly on $V_{in}$ → full 12 V.

(C) $V_{o u t} = 12 \cdot \frac{0}{2000 + 0} = 12 \cdot 0 = 0 V$ . Why this step? Numerator is $R_{2} = 0$ ; nothing to drop across a wire to ground.
(D) $V_{o u t} = 12 \cdot \frac{2000}{0 + 2000} = 12 \cdot 1 = 12 V$ . Why this step? When $R_{1} = 0$ the fraction becomes $\frac{R _{2}}{R _{2}} = 1$ ; the whole input appears at the tap.

Verify: both extremes are what a wire physically does — no drop across $0 Ω$ . Sanity: (C)+(D) numerators sum to the full resistance idea, and each answer sits at an endpoint $0$ or $V_{in}$ . ✓

⚠️ Practical note: if $R_{1} = 0$ the divider is a dead short from $V_{in}$ to $V_{o u t}$ but current still flows freely through $R_{2}$ to ground — real supplies dislike this. Formula still works.

This "output vs. $R_{2}$ " curve shows the whole journey between the two degenerate endpoints:

Worked example Ex 5 — Case G: design (solve for a resistor)

Make $V_{o u t} = 1.8 V$ from $V_{in} = 5 V$ . You pick $R_{2} = 10 k Ω$ . Find $R_{1}$ .

Forecast: we need $R_{2}$ 's share to be $1.8/5 = 0.36$ . Since $R_{2}$ must be less than half the total (share below 0.5), $R_{1}$ should be bigger than $R_{2}$ . Guess $R_{1}$ somewhere around $17$ – $18$ k $Ω$ .

Required ratio: $\frac{R _{2}}{R _{1} + R _{2}} = \frac{1.8}{5} = 0.36$ . Why this step? Only the ratio sets $V_{o u t}$ ; absolute values just set current.
Substitute $R_{2} = 10 k$ : $0.36 = \frac{10}{R _{1} + 10}$ (in k $Ω$ ). Why this step? Plug the chosen $R_{2}$ so one unknown remains.
Cross-multiply: $0.36 (R_{1} + 10) = 10 \Rightarrow R_{1} + 10 = \frac{10}{0.36} = 27.78$ . Why this step? Isolate $R_{1}$ .
$R_{1} = 27.78 - 10 = 17.78 k Ω$ . Why this step? Subtract $R_{2}$ to peel off $R_{1}$ .

Verify: $V_{o u t} = 5 \cdot \frac{10}{17.78 + 10} = 5 \cdot \frac{10}{27.78} = 5 \cdot 0.36 = 1.8 V$ . ✓ And $R_{1} > R_{2}$ matched the forecast.

Worked example Ex 6 — Case H: loading effect

$V_{in} = 9 V$ , $R_{1} = R_{2} = 4.7 k Ω$ (unloaded $V_{o u t} = 4.5$ V). Now hang a load $R_{L} = 4.7 k Ω$ across the output. Find the new $V_{o u t}$ .

Forecast: the load sits in parallel with $R_{2}$ , and two equal resistors in parallel give half their value → effective bottom resistor drops to $2.35 k Ω$ . Smaller bottom → smaller share → $V_{o u t}$ falls below 4.5 V. Guess around 3 V.

Combine $R_{2}$ with the load: $R_{2} ∥ R_{L} = \frac{R _{2} R _{L}}{R _{2} + R _{L}} = \frac{4.7 \cdot 4.7}{4.7 + 4.7} = 2.35 k Ω$ . Why this step? A load across the tap is electrically parallel to $R_{2}$ (Series and Parallel Resistors).
Re-run the divider with this effective bottom resistor: $V_{o u t} = 9 \cdot \frac{2.35}{4.7 + 2.35} = 9 \cdot \frac{2.35}{7.05}$ . Why this step? Same divider formula, just with $R_{2} \to R_{2} ∥ R_{L}$ .
$V_{o u t} = 9 \cdot 0.3333 = 3.0 V$ . Why this step? Arithmetic; the load "stole" current and dragged the output down.

Verify: new output $3.0$ V $<$ unloaded $4.5$ V — loading always lowers $V_{o u t}$ . ✓ A Op-amp Buffer between tap and load would fix this by drawing ~no current. (See also the Thevenin Equivalent view: source $4.5$ V, $R_{t h} = R_{1} ∥ R_{2} = 2.35 k Ω$ .)

Worked example Ex 7 — Case I: real-world word problem (potentiometer / sensor)

A 3-terminal potentiometer (a variable divider) has total resistance $10 k Ω$ across $V_{in} = 3.3 V$ . The wiper is turned to $30%$ from the ground end. What voltage does a microcontroller read at the wiper?

Forecast: the wiper splits the total resistance into "below the wiper" ( $R_{2}$ ) and "above" ( $R_{1}$ ). At 30% from ground, $R_{2}$ is 30% of the track → $V_{o u t}$ should be 30% of 3.3 V ≈ 1 V.

$R_{2} = 30% \times 10 k = 3 k Ω$ (below wiper); $R_{1} = 7 k Ω$ (above). Why this step? A pot is literally one resistor split by the wiper into top and bottom pieces.
$V_{o u t} = 3.3 \cdot \frac{3}{7 + 3} = 3.3 \cdot 0.3$ . Why this step? Divider formula with the wiper-defined $R_{1}, R_{2}$ .
$V_{o u t} = 0.99 V$ . Why this step? 30% of the track → 30% of the voltage.

Verify: at 0% wiper → 0 V, at 100% → 3.3 V; 30% landing near 1 V is consistent. ✓ (Assumes the microcontroller input draws negligible current — a high-impedance ADC, so no loading. Otherwise use Case H.)

Worked example Ex 8 — Case J: exam twist (three resistors, tap the middle)

A chain from $V_{in} = 12 V$ to ground: $R_{1} = 1 k Ω$ , then $R_{2} = 2 k Ω$ , then $R_{3} = 3 k Ω$ , all in series. What is the voltage at the node between $R_{2}$ and $R_{3}$ (measured to ground)?

Forecast: the node above $R_{3}$ has everything below it ( $R_{3}$ alone) as its " $R_{2}$ ", and everything above it ( $R_{1} + R_{2}$ ) as its " $R_{1}$ ". So the effective bottom is $3 k$ out of $6 k$ total → half → guess 6 V.

Total resistance $= 1 + 2 + 3 = 6 k Ω$ ; same current everywhere (series). Why this step? One loop → one current, resistances add.
The measured node sits above $R_{3}$ , so treat $R_{3}$ as the bottom resistor: effective $R_{2}^{*} = R_{3} = 3 k Ω$ . Why this step? " $V_{o u t}$ = voltage across everything below the tap point."
$V_{n o d e} = 12 \cdot \frac{R _{3}}{R _{1} + R _{2} + R _{3}} = 12 \cdot \frac{3}{6} = 6 V$ . Why this step? Generalised divider — the numerator is the resistance below the tap.

Verify (KVL): drops are $R_{1} : 12 \cdot \frac{1}{6} = 2$ V, $R_{2} : 12 \cdot \frac{2}{6} = 4$ V, $R_{3} : 12 \cdot \frac{3}{6} = 6$ V; sum $= 12$ V ✓. Node voltage = drop across $R_{3}$ (bottom) = 6 V, matching the forecast. (Kirchhoff's Voltage Law.)

Active recall

Equal resistors, what fraction of

V_{in}

appears at the tap?

Exactly one half.

R_{2} = 0

, what is

V_{o u t}

0

V — the tap sits on ground through a wire.

R_{1} = 0

, what is

V_{o u t}

The full

V_{in}

— the tap sits directly on the input.

R_{2} \to \infty

with

R_{1}

fixed,

V_{o u t}

approaches?

V_{in}

(fraction → 1) but never exceeds it.

What replaces

R_{2}

when a load

R_{L}

is attached across the output?

The parallel combination

R_{2} ∥ R_{L} = \frac{R _{2} R _{L}}{R _{2} + R _{L}}

In a multi-resistor chain, what goes in the numerator of the divider?

All resistance below the tap.

Can a passive divider output more than

V_{in}

No — output is always strictly between

0

and

V_{in}

Connections

Build and analyze a voltage divider — parent topic; this page stress-tests its formula.
Ohm's Law — every step is Ohm's law twice.
Series and Parallel Resistors — series adds (chain), parallel appears under loading.
Kirchhoff's Voltage Law — the drops-sum-to- $V_{in}$ checks.
Thevenin Equivalent — the loading example is a Thevenin source in disguise.
Wheatstone Bridge — two of these dividers compared.
Op-amp Buffer — the cure for the loading case (Ex 6).