What is the voltage divider output?
Type the input voltage and the two resistors to get the output voltage of a voltage divider. Or type any three of Vin, R1, R2 and Vout to find the fourth, for example the R1 that gives 3.3 V from 12 V. The page also gives the current and power.
- Output voltage Vout
- 1 V
With 9 V across 80 Ω and 10 Ω, the output is 1 V.
- Current
- 0.1 A
- Vout ÷ Vin
- 0.111111
- Power used
- 0.9 W
Output voltage Vout: 1 V. With 9 V across 80 Ω and 10 Ω, the output is 1 V.
How to calculate
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.
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.
Formula: Vout = Vin × R2 ÷ (R1 + R2), because the same current I = Vin ÷ (R1 + R2) flows through both resistors and Vout = I × R2.
- 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
Each example is checked against the calculator on every build.
- Input voltage Vin 9 V, R1 (top) 80 Ω, R2 (bottom) 10 Ω gives Output voltage Vout 1 V, Current 0.1 A, Vout ÷ Vin 0.111111, Power used 0.9 W.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)
- Input voltage Vin 5 V, R1 (top) 10,000 Ω, R2 (bottom) 10,000 Ω gives Output voltage Vout 2.5 V, Current 0.00025 A, Vout ÷ Vin 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)
- Input voltage Vin 12 V, Output voltage Vout 3.3 V, R2 (bottom) 10,000 Ω gives R1 (top) 26,364 Ω.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)
- Output voltage Vout 2 V, R1 (top) 30 Ω, R2 (bottom) 60 Ω gives Input voltage Vin 3 V.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)
How it works
R1 connects the input to the output; R2 connects the output to ground. With nothing connected across the output:
- Current: I = Vin ÷ (R1 + R2)
- Vout = I × R2 = Vin × R2 ÷ (R1 + R2)
- Solved for the others: Vin = Vout × (R1 + R2) ÷ R2; R1 = R2 × (Vin − Vout) ÷ Vout; R2 = R1 × Vout ÷ (Vin − Vout)
- Divider ratio: Vout ÷ Vin = R2 ÷ (R1 + R2)
- Power in both resistors: P = Vin × I
Type any three of Vin, R1, R2 and Vout; the fourth is worked out. Values run in volts, ohms and amperes (base units), in double precision.
Rules
- Every value is more than 0. If R1 + R2 is beyond the largest double-precision number, the current, ratio and power are left out. A worked-out value that is 0 or less (Vout of Vin or more when solving for a resistor) gives no answer.
- No load on the output; ideal resistors; DC, or RMS values in a resistive AC circuit.
Output format. Voltages, current and power in the unit picked (V, A and W unless you pick another unit); the ratio as a plain number.
Worked examples by hand
9 V, R1 = 80 Ω, R2 = 10 Ω. I = 9 ÷ 90 = 0.1 A. Vout = 0.1 × 10 = 1 V. Ratio 10 ÷ 90 = 0.1111. Power 9 × 0.1 = 0.9 W.
5 V, R1 = R2 = 10 kΩ. Vout = 5 × 10,000 ÷ 20,000 = 2.5 V; I = 5 ÷ 20,000 = 0.25 mA.
R1 for 3.3 V from 12 V, R2 = 10 kΩ. R1 = 10,000 × (12 − 3.3) ÷ 3.3 = 26,363.6 Ω (26.36 kΩ).
Vin from Vout = 2 V, R1 = 30 Ω, R2 = 60 Ω. Vin = 2 × 90 ÷ 60 = 3 V.
Other questions people ask
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.