What is the voltage divider formula?

For two resistors R1 and R2 in series across a source Vin, with the output taken across R2, Vout = Vin × R2 / (R1 + R2). R1 sits between the input and the output node, R2 sits between the output node and ground.

Why does the output voltage depend on R2 and not R1?

Because the output is measured across R2. The same current flows through both resistors, and Ohm’s law says the voltage across any resistor in that loop is that current times its own resistance — so the fraction of Vin that appears at the output is R2 divided by the total resistance R1 + R2.

How do I find R1 or R2 for a target output voltage?

Pick a ratio k = Vout / Vin, choose one resistor as a starting value, then solve the other from the rearranged formula: R2 = R1 × k / (1 − k), or R1 = R2 × (1 − k) / k. Start from a standard E-series value for whichever resistor you fix first.

Does a voltage divider work under load?

Only approximately. The formula assumes nothing draws current from the output node besides R2. Once you connect a real load in parallel with R2, its resistance combines with R2 and pulls the output down. A divider only holds its calculated voltage accurately when the load impedance is much larger than R1 and R2 combined.

What is the difference between a voltage divider and a series resistor?

A series resistor alone drops voltage across itself but has no defined second node to tap a fraction of the source from — the remaining voltage depends entirely on whatever is downstream. A voltage divider adds a second resistor to ground specifically to fix that output voltage independent of what a high-impedance load does next.

Can a voltage divider supply significant current?

Not efficiently. Because the divider itself must always draw Vin / (R1 + R2) through both resistors regardless of the load, using it to power something that needs real current wastes power in R1 and R2 and sags badly under load. For anything beyond a reference voltage or a signal into a high-impedance input, use a regulator instead.

Voltage Divider Formula & Derivation

Where Vout = Vin × R2 / (R1 + R2) comes from, and how to use it in both directions.

What Is a Voltage Divider?

A voltage divider is two resistors wired in series across a source voltage, with the output taken from the node where they meet. Because that node sits at a fixed fraction of the source, a divider is the simplest way to turn one voltage into a smaller, predictable one without any active components — no regulator, no op-amp, just two resistors and Ohm's law.

Dividers show up constantly in real circuits: scaling a sensor's output into a microcontroller's ADC range, shifting a 5 V logic signal down to 3.3 V, generating a reference or bias voltage, or forming the feedback network that sets a regulator's output. In every case the same two-resistor relationship is doing the work, and it is worth deriving once so you can apply it in either direction — given the resistors, find the voltage, or given the voltage, find the resistors.

Deriving the Formula from Ohm's Law

Take a source voltage Vin connected across two resistors, R1 and R2, wired end to end. R1 runs from the input down to a middle node; R2 continues from that same node down to ground. The output, Vout, is measured at the middle node relative to ground.

Because R1 and R2 sit in a single loop with nothing else branching off between them, the same current Iflows through both — that is what “series” means. Ohm's law lets you find that current directly from the total resistance the source sees:

I = \dfrac{V_{in}}{R_1 + R_2}

Vout is nothing more than the voltage that current produces across R2, so a second application of Ohm's law — this time isolated to R2 — gives:

V_{out} = I \cdot R_2

Substituting the expression for I from the first step replaces the current with quantities you actually know (the source voltage and the two resistor values), which gives the voltage divider formula:

V_{out} = V_{in} \cdot \dfrac{R_2}{R_1+R_2}

Read it as a ratio: Vout is whatever fraction of Vin that R2 represents of the total series resistance. Make R2 much larger than R1 and Vout approaches Vin; make R2 much smaller than R1 and Vout approaches zero. Set R1 = R2 and Vout is exactly half of Vin, regardless of the absolute resistance chosen.

Solving for R1 or R2

Designing a divider usually starts from the other direction: you know Vin, you know the Vout you want, and you need resistor values. Rearranging the formula to isolate each resistor gives two equivalent forms — solve for R2 if you have already picked R1, or solve for R1 if you have already picked R2:

R_2 = R_1 \cdot \dfrac{V_{out}}{V_{in}-V_{out}} \qquad\qquad R_1 = R_2 \cdot \dfrac{V_{in}-V_{out}}{V_{out}}

In practice, pick a standard E-series value for whichever resistor you fix first, solve for the other, then round that result to the nearest standard value your series offers. That rounding step is exactly why the two worked examples below land close to their target voltage rather than hitting it exactly — real resistors only come in discrete values, and the Voltage Divider Finder automates that search across the resistors you actually have on hand.

Worked Example: 12 V Down to 5 V

Suppose you have a 12 V supply and need roughly 5 V for a downstream circuit. Using two common E24 standard values, R1 = 4.7 kΩ and R2 = 3.3 kΩ, the total series resistance is 8.0 kΩ. The current through the loop is:

I = \dfrac{12\text{ V}}{4.7\text{ k}\Omega + 3.3\text{ k}\Omega} = \dfrac{12\text{ V}}{8.0\text{ k}\Omega} = 1.5\text{ mA}

That current across R2 gives the output voltage:

V_{out} = 1.5\text{ mA} \times 3.3\text{ k}\Omega = 4.95\text{ V}

4.95 V from two off-the-shelf resistor values — well within tolerance for most 5 V rails, and reached without any non-standard part.

Worked Example: 5 V Down to 3.3 V

The same approach handles the common 5 V-to-3.3 V logic level shift. With R1 = 1.0 kΩ and R2 = 2.0 kΩ (both standard E24 values), the total resistance is 3.0 kΩ:

I = \dfrac{5\text{ V}}{1.0\text{ k}\Omega + 2.0\text{ k}\Omega} = \dfrac{5\text{ V}}{3.0\text{ k}\Omega} \approx 1.667\text{ mA}

V_{out} = 1.667\text{ mA} \times 2.0\text{ k}\Omega \approx 3.33\text{ V}

3.33 V is close enough to 3.3 V for a signal feeding a high-impedance logic input, and both resistor values are common, cheap, and easy to source.

Loading Effects — Why a Real Divider Sags

Every derivation above assumes the only path for current out of the middle node is through R2. That holds as long as whatever is connected to Vout draws essentially no current — a high-impedance input like an ADC pin or an op-amp buffer. Connect a real load resistor RL across the output instead, and it sits electrically in parallel with R2, so the effective resistance the formula should use is R2 in parallel with RL, not R2 alone. Because a parallel combination is always smaller than either resistor on its own, the effective “R2” shrinks and Vout sags below the value you calculated.

The rule of thumb: a divider holds its designed output accurately only when the load impedance is much larger — commonly cited as at least 10 times — than R1 in parallel with R2. If a load needs to draw meaningful current without the output sagging, a plain resistive divider is the wrong tool; use a voltage regulator or a buffered reference instead, and reserve dividers for signal conditioning, feedback networks, and other high-impedance measurement points.

Explore the Tool

Voltage Divider Finder

Search for the 2- or 3-resistor divider that lands closest to a target output voltage.

Voltage Divider Builder

Already know your resistors and topology? Get the exact output voltage instantly.

Related Guides

Series vs Parallel Resistors — When to Use Which

A voltage divider is a specific application of series resistors — this guide covers how series and parallel networks behave more generally.

Frequently Asked Questions

What is the voltage divider formula?: For two resistors R1 and R2 in series across a source Vin, with the output taken across R2, Vout = Vin × R2 / (R1 + R2). R1 sits between the input and the output node, R2 sits between the output node and ground.
Why does the output voltage depend on R2 and not R1?: Because the output is measured across R2. The same current flows through both resistors, and Ohm’s law says the voltage across any resistor in that loop is that current times its own resistance — so the fraction of Vin that appears at the output is R2 divided by the total resistance R1 + R2.
How do I find R1 or R2 for a target output voltage?: Pick a ratio k = Vout / Vin, choose one resistor as a starting value, then solve the other from the rearranged formula: R2 = R1 × k / (1 − k), or R1 = R2 × (1 − k) / k. Start from a standard E-series value for whichever resistor you fix first.
Does a voltage divider work under load?: Only approximately. The formula assumes nothing draws current from the output node besides R2. Once you connect a real load in parallel with R2, its resistance combines with R2 and pulls the output down. A divider only holds its calculated voltage accurately when the load impedance is much larger than R1 and R2 combined.
What is the difference between a voltage divider and a series resistor?: A series resistor alone drops voltage across itself but has no defined second node to tap a fraction of the source from — the remaining voltage depends entirely on whatever is downstream. A voltage divider adds a second resistor to ground specifically to fix that output voltage independent of what a high-impedance load does next.
Can a voltage divider supply significant current?: Not efficiently. Because the divider itself must always draw Vin / (R1 + R2) through both resistors regardless of the load, using it to power something that needs real current wastes power in R1 and R2 and sags badly under load. For anything beyond a reference voltage or a signal into a high-impedance input, use a regulator instead.