Voltage Divider Calculator
Find the output voltage of a two-resistor divider, plus its ratio and current.
Divider ratio R2 ÷ (R1 + R2) = 0.5 · Divider current I = Vin ÷ (R1 + R2) = 250 µA.
A voltage divider outputs Vout = Vin × R2 ÷ (R1 + R2), tapped between two series resistors. With Vin = 5 V and R1 = R2 = 10 kΩ, Vout = 5 × 10000 ÷ 20000 = 2.5 V. Swap to 12 V with 1 kΩ and 2 kΩ and Vout = 12 × 2000 ÷ 3000 = 8 V.
How a voltage divider works
Two resistors in series carry the same current, so the input voltage splits between them in proportion to their resistance. The output is taken at the node between R1 (the top resistor, tied to Vin) and R2 (the bottom resistor, tied to ground). Because R2 gets the fraction R2 ÷ (R1 + R2) of the total, that same fraction of Vin appears at the tap. It is the workhorse behind sensor scaling, reference voltages, and reading a signal that is too high for a microcontroller pin. When you already know the current, reach for Ohm’s law instead.
Vin = input voltage (V), R1 = top resistor (Ω), R2 = bottom resistor (Ω), Vout = output at the tap (V)
Worked example
Drop a 12 V supply to a lower rail using R1 = 1 kΩ and R2 = 2 kΩ:
- 1 Add the two resistors. R1 + R2 = 1000 + 2000 = 3000 Ω. This sets the total resistance the supply sees.
- 2 Find the divider ratio. R2 ÷ (R1 + R2) = 2000 ÷ 3000 = 0.667 — the fraction of Vin that reaches the output.
- 3 Multiply by the input voltage. Vout = Vin × ratio = 12 × 0.667 = 8 V.
- 4 Check the current (optional). I = Vin ÷ (R1 + R2) = 12 ÷ 3000 = 4 mA flows through both resistors.
Common divider ratios
Vout as a fraction of Vin for a few R1 : R2 pairs (no load).
| R1 : R2 | Ratio R2 ÷ (R1 + R2) | Vout fraction of Vin |
|---|---|---|
| 1 : 1 | 0.500 | Vin ÷ 2 |
| 2 : 1 | 0.333 | Vin ÷ 3 |
| 1 : 2 | 0.667 | 2 × Vin ÷ 3 |
| 9 : 1 | 0.100 | Vin ÷ 10 |
| 1 : 3 | 0.750 | 3 × Vin ÷ 4 |
The loading effect — and how to avoid it
A load changes the answer. The formula assumes nothing draws current at the tap. Connect a load, and it sits in parallel with R2, lowering the effective bottom resistance and pulling Vout below the ideal figure. The heavier the load (the lower its resistance), the bigger the sag.
Keep the divider current well above the load current. A common rule of thumb is to size R1 and R2 so the current through the divider is at least ten times what the load draws; then the load barely perturbs the ratio. That is why dividers are poor power sources — use them for sensing and reference levels, not to feed a hungry circuit. For a stiff, load-independent output, use a voltage regulator instead.