Build and analyze a voltage divider: Visual walkthrough

We assume you know nothing except that electricity flows in loops. Every symbol is earned before use.

Step 1 — Two resistors, one loop, one road

WHAT. Draw a battery. From its top terminal a wire goes down through a first resistor, then through a second resistor, then back to the battery's bottom terminal. That is the whole circuit.

WHY. A resistor is a component that pushes back against flow — think of a narrow stretch of pipe. When two of them sit one-after-the-other on the same wire with no fork between them, we call that series. The key fact we will lean on the entire page: in series, the same amount of electricity passes through both, because there is only one road and nowhere to escape.

PICTURE. Look at the figure. The blue arrow is the single current path — it enters $R_{1}$ , leaves $R_{1}$ , enters $R_{2}$ , and there is no side-branch where any could leak away.

Figure — Build and analyze a voltage divider

Let me name the quantities, each one a plain physical thing:

This is Ohm's Law and Series and Parallel Resistors applied on one road.

Step 2 — Two roadblocks in a row = add their resistances

WHAT. Replace the two resistors, in our heads, with a single resistor whose value is $R_{1} + R_{2}$ .

WHY. If electricity must squeeze through a narrow pipe and then immediately through another narrow pipe, the total narrowness it feels is just the two added together. There is no shortcut. So the whole loop resists as much as one big resistor:

$R_{t o t a l} = first block R_{1} + second block R_{2}$

Every symbol here is already defined; the plus sign means "one after the other, so pile them up."

PICTURE. The figure shows the two stacked resistors on the left collapsing into one tall equivalent resistor on the right, labelled $R_{t o t a l}$ . Same total height = same total resistance.

Step 3 — Ask Ohm how much current flows

WHAT. Compute the current $I$ in the loop.

WHY. We know the push ( $V_{in}$ ) and we know the total resistance ( $R_{t o t a l}$ ). Ohm's Law is the tool that connects push, resistance, and flow. It says: flow equals push divided by resistance. We reach for this tool and not any other because it is precisely the rule that answers "given a voltage and a resistance, how much current?" — nothing else does that job.

$I = \frac{V _{in}}{R _{t o t a l}} = \frac{V _{in} push, volts}{total resistance, Ω R _{1} + R _{2}}$

Term by term: the numerator is the whole battery push; the denominator is everything resisting it. Big denominator ⇒ small current. Makes sense — more roadblocks, less flow.

PICTURE. The figure shows the same loop with the current arrow now thick or thin depending on the denominator: a mental slider where increasing $R_{1} + R_{2}$ thins the arrow.

Step 4 — That same current drops voltage across each resistor

WHAT. Turn the single current $I$ back into two voltage drops, one per resistor.

WHY. Voltage isn't used up by magic — it is spent pushing current through each resistor. Ohm's law read the other way says: the voltage a resistor "eats" equals its resistance times the current through it, $V = I \cdot R$ . Because $I$ is the same through both (Step 1), the resistor with more ohms eats more volts.

$V_{R_{1}} = I \cdot R_{1} V_{R_{2}} = I \cdot R_{2}$

$V_{R_{1}}$ — volts eaten by the top resistor.
$V_{R_{2}}$ — volts eaten by the bottom resistor. This is exactly $V_{o u t}$ , because $V_{o u t}$ is measured across $R_{2}$ .

PICTURE. The figure marks two voltage-drop brackets on the vertical wire: a tall bracket over $R_{1}$ , a shorter one over $R_{2}$ (or vice-versa), and the sum of the two brackets spans the full $V_{in}$ .

Step 5 — Substitute and clean up into the divider formula

WHAT. Put the current from Step 3 into the $V_{o u t} = I \cdot R_{2}$ from Step 4.

WHY. We want $V_{o u t}$ written in terms of things we set ( $V_{in}, R_{1}, R_{2}$ ), not the middle-man $I$ . So we substitute $I$ away.

$V_{o u t} = I \cdot R_{2} = that is I \frac{V _{in}}{R _{1} + R _{2}} \cdot R_{2} = V_{in} \cdot \frac{R _{2}}{R _{1} + R _{2}}$

Now read the final fraction as a share:

$V_{o u t} = V_{in} \cdot R_{2} ’s slice of the total \frac{R _{2}}{R _{1} + R _{2}}$

The fraction is always between $0$ and $1$ (a slice can't exceed the whole).
It is exactly the portion of total resistance that $R_{2}$ owns. Own half the ohms ⇒ get half the volts.

PICTURE. The figure shows a horizontal bar of total length $V_{in}$ , split into an $R_{1}$ -slice and an $R_{2}$ -slice, with the $R_{2}$ -slice's length labelled $V_{o u t}$ and its width proportional to $R_{2} / (R_{1} + R_{2})$ .

Step 6 — Every case, so you never hit a surprise

WHAT. Sweep the ratio $R_{2} / (R_{1} + R_{2})$ across all its possibilities and watch $V_{o u t}$ .

WHY. A formula you trust must be checked at its extremes, not just the comfortable middle.

PICTURE. The figure plots $V_{o u t}$ against $R_{2}$ (with $R_{1}$ fixed), marking the special points below.

Case	What $R_{2}$ is	Fraction	$V_{o u t}$	Meaning
Bottom shorted	$R_{2} = 0$	$0$	$0$ V	Nothing to drop across; output pinned to ground.
Equal split	$R_{2} = R_{1}$	$\frac{1}{2}$	$V_{in} /2$	The half-way tap.
Top shorted	$R_{1} = 0$	$1$	$V_{in}$	No first block, so full push reaches the tap.
Bottom huge	$R_{2} \to \infty$	$\to 1$	$\to V_{in}$	Output approaches $V_{in}$ but never fully with finite $R_{1}$ .

$R_{2} = 0$ : the fraction is $0/ (R_{1} + 0) = 0$ . The output node is wired straight to ground.
$R_{1} = 0$ : fraction is $R_{2} / R_{2} = 1$ . The wire from the battery reaches the tap with nothing in the way.
The curve is monotonic (always rising as $R_{2}$ grows) and never leaves $[0, V_{in}]$ — so no combination of positive resistors can ever give you more than you put in.

Step 7 — The degenerate case that bites people: loading

WHAT. Connect a real circuit (a load $R_{L}$ ) across $R_{2}$ and re-derive.

WHY. A load is just another resistor sitting in parallel with $R_{2}$ , offering a second road for current at the tap. The moment there are two roads, Step 1's "same current everywhere" no longer holds below the tap — so the divider must be re-analysed with $R_{2}$ replaced by the parallel combination.

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

Because the parallel value is smaller than $R_{2}$ alone, $R_{2}$ 's slice shrinks, so:

$V_{o u t}^{loaded} = V_{in} \cdot \frac{R _{2} ∥ R _{L}}{R _{1} + ( R _{2} ∥ R _{L} )} < V_{o u t}^{unloaded}$

PICTURE. The figure adds $R_{L}$ across $R_{2}$ , shows current forking at the tap node, and drops the $V_{o u t}$ bracket lower than its unloaded dashed level.

This is why a divider is a sensing/reference circuit, not a power supply — the fix is an Op-amp Buffer, and the clean way to think about the loaded output is the Thevenin Equivalent with $R_{t h} = R_{1} ∥ R_{2}$ . Two dividers compared side-by-side is a Wheatstone Bridge.

The one-picture summary

Everything on this page, compressed: the loop, the single current, the two drops that add to $V_{in}$ , and the slice-of-the-bar that is $V_{o u t}$ .

Recall Feynman: tell the whole walkthrough in plain words

Water pours down from a tank (that's $V_{in}$ ). It goes through one narrow pipe, then a second narrow pipe, back to the tank — one loop, so the same water flow passes both pipes (that's $I$ , found by dividing the tank's pressure by the total narrowness $R_{1} + R_{2}$ ). Each pipe "uses up" pressure equal to its own narrowness times the flow. If you dip a gauge in at the joint between the pipes, you read the pressure the second pipe still has to spend — that's $V_{o u t}$ , and it's just the second pipe's share of the total narrowness times the full pressure. Make the second pipe half the total narrowness and you read half the pressure. Zero it and you read nothing; make it the whole thing and you read everything (but never more). And the instant you attach a real machine at the joint, you open a second drain — the joint's pressure sags below what you predicted. That sag is "loading," and it's why a divider tells you a voltage but shouldn't try to feed one.

Active recall

Which fact from Step 1 makes the whole derivation possible?

The same current

I

flows through both resistors (series, one path).

Why is

R_{t o t a l} = R_{1} + R_{2}

Series roadblocks pile up; there is no shortcut, so resistances add.

Which tool turns "push and resistance" into current, and why that one?

Ohm's law

I = V / R

— it is exactly the rule linking voltage, resistance, and flow.

Why is

V_{o u t}

the

R_{2}

drop and not

R_{1}

The output is measured across

R_{2}

, so

V_{o u t} = I R_{2}

What does the fraction

R_{2} / (R_{1} + R_{2})

physically represent?

R_{2}

's share of the total resistance = its share of the voltage.

What is

V_{o u t}

when

R_{2} = 0

and when

R_{1} = 0

0

V (bottom shorted) and

V_{in}

(top shorted).

Can

V_{o u t}

ever exceed

V_{in}

with positive resistors?

No — the fraction saturates at

1

What happens to

V_{o u t}

when a load

R_{L}

is attached across

R_{2}

Replace

R_{2}

with

R_{2} ∥ R_{L}

(smaller), so

V_{o u t}

drops.

Connections

Build and analyze a voltage divider — the parent topic this page derives in pictures.
Ohm's Law — used in Steps 3 and 4.
Series and Parallel Resistors — Step 2 (series add), Step 7 (parallel load).
Kirchhoff's Voltage Law — Step 4 sanity check.
Thevenin Equivalent — clean model of the loaded output.
Op-amp Buffer — fixes the loading of Step 7.
Wheatstone Bridge — two dividers compared.