# What is the voltage divider output?

Finds the output voltage of a two-resistor voltage divider, Vout = Vin × R2 ÷ (R1 + R2), or the input voltage or either resistor from the other three, with the current and power.

- Page: https://www.acalculator.org/physics/voltage-divider-calculator
- JSON spec: https://www.acalculator.org/physics/voltage-divider-calculator.json
- Version: 9ecd4061d2a1

## Default answer

Example with the default inputs (Input voltage Vin 9 V, R1 (top) 80 Ω, R2 (bottom) 10 Ω): With 9 V across 80 Ω and 10 Ω, the output is 1 V.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| vin | Input voltage Vin | The voltage across both resistors, from the source. |
| r1 | R1 (top) | The resistor between the input and the output. |
| r2 | R2 (bottom) | The resistor between the output and ground; Vout is the voltage across it. |
| vout | Output voltage Vout | The voltage across R2, less than Vin. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| vin | Input voltage Vin | The voltage across both resistors, from the source. |
| r1 | R1 (top) | The resistor between the input and the output. |
| r2 | R2 (bottom) | The resistor between the output and ground; Vout is the voltage across it. |
| vout | Output voltage Vout | The voltage across R2. |
| current | Current | The current through both resistors: Vin ÷ (R1 + R2). |
| ratio | Vout ÷ Vin | The divider ratio R2 ÷ (R1 + R2). |
| power | Power used | The power both resistors turn into heat: Vin × current. |

## Method

Vout = Vin × R2 ÷ (R1 + R2), because the same current I = Vin ÷ (R1 + R2) flows through both resistors and Vout = I × R2.

## Assumptions

- Nothing is connected across the output (no load), so all the current flows through both resistors.
- Ideal resistors and a steady (DC) voltage, or RMS values in a purely resistive AC circuit.
- All values are more than 0; Vout must be less than Vin.

## Worked examples

1. vin = 9, r1 = 80, r2 = 10 gives vout = 1, current = 0.1, ratio = 0.111111, power = 0.9. Source: OpenStax, University Physics Volume 2, §10.2 Resistors in Series and Parallel (series resistance R_S = R₁ + R₂ + …, the same current I = V ÷ R_S through each, and the drop across each V_i = I R_i; Example 10.2: 9 V across 90 Ω gives 0.1 A, 2 V across each 20 Ω and 1 V across the 10 Ω), https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel (retrieved 2026-10-02).
2. vin = 5, r1 = 10,000, r2 = 10,000 gives vout = 2.5, current = 0.00025, ratio = 0.5. Source: OpenStax, University Physics Volume 2, §10.2 Resistors in Series and Parallel (series resistance R_S = R₁ + R₂ + …, the same current I = V ÷ R_S through each, and the drop across each V_i = I R_i; Example 10.2: 9 V across 90 Ω gives 0.1 A, 2 V across each 20 Ω and 1 V across the 10 Ω), https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel (retrieved 2026-10-02).
3. vin = 12, vout = 3.3, r2 = 10,000 gives r1 = 26,363.636364. Source: OpenStax, University Physics Volume 2, §10.2 Resistors in Series and Parallel (series resistance R_S = R₁ + R₂ + …, the same current I = V ÷ R_S through each, and the drop across each V_i = I R_i; Example 10.2: 9 V across 90 Ω gives 0.1 A, 2 V across each 20 Ω and 1 V across the 10 Ω), https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel (retrieved 2026-10-02).
4. vout = 2, r1 = 30, r2 = 60 gives vin = 3. Source: OpenStax, University Physics Volume 2, §10.2 Resistors in Series and Parallel (series resistance R_S = R₁ + R₂ + …, the same current I = V ÷ R_S through each, and the drop across each V_i = I R_i; Example 10.2: 9 V across 90 Ω gives 0.1 A, 2 V across each 20 Ω and 1 V across the 10 Ω), https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel (retrieved 2026-10-02).

## FAQ

### What is a voltage divider?

Two resistors in series across a voltage. The voltage at the point between them is a fixed share of the input: Vout = Vin × R2 ÷ (R1 + R2), where R2 is the resistor between the output and ground.

### Why does the formula work?

In series, the same current I = Vin ÷ (R1 + R2) flows through both resistors. The voltage across R2 is I × R2 (Ohm’s law), which gives Vin × R2 ÷ (R1 + R2).

### What is the output of 9 V across 80 Ω and 10 Ω?

Vout = 9 × 10 ÷ 90 = 1 V, with 0.1 A flowing. This matches OpenStax Example 10.2, where 9 V across 90 Ω of resistors in series drops 1 V across the 10 Ω one.

### How do I choose R1 for a given output?

Solve the formula for R1: R1 = R2 × (Vin − Vout) ÷ Vout. For 3.3 V from 12 V with R2 = 10 kΩ, R1 = 10,000 × 8.7 ÷ 3.3 ≈ 26.36 kΩ. Leave R1 empty and type the other three.

### What happens when I connect a load?

A load across the output is in parallel with R2, so the bottom resistance drops and Vout falls. Use R2 in parallel with the load as the new R2, or keep the load much larger than R2. This page assumes no load.

### Can the output be higher than the input?

No. A resistor divider only lowers a voltage: Vout is always less than Vin. If you type Vout of Vin or more, there is no answer.

### Should I use a voltage divider as a power supply?

Usually not. The output changes with the load and the resistors waste power all the time. Dividers suit reference and sensing inputs; a regulator suits powering a circuit.

## Sources

- OpenStax, University Physics Volume 2, §10.2 Resistors in Series and Parallel (series resistance R_S = R₁ + R₂ + …, the same current I = V ÷ R_S through each, and the drop across each V_i = I R_i; Example 10.2: 9 V across 90 Ω gives 0.1 A, 2 V across each 20 Ω and 1 V across the 10 Ω). https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel (retrieved 2026-10-02)