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Free Voltage Divider Calculator

Compute the output voltage of a two-resistor voltage divider from your input voltage and resistor values, plus the current flowing through the divider chain in amps and milliamps.

Vout is measured across R2 (the lower resistor to ground).

Results
Output voltage Vout8 V
Current I0.004 A
Current I4 mA

Vout = Vin · R2 / (R1 + R2) · I = Vin / (R1 + R2). Assumes an unloaded divider; a load across the output lowers Vout.

Quick answer

A resistive voltage divider produces an output of Vout = Vin × R2 / (R1 + R2), where R2 is the resistor across which Vout is measured. The current through the unloaded divider is I = Vin / (R1 + R2). For example, with Vin = 12 V, R1 = 1000 Ω, and R2 = 2000 Ω, Vout = 12 × 2000 / 3000 = 8 V and I = 12 / 3000 = 4 mA.

Formula & method

Output voltage

Vout = Vin · R2 / (R1 + R2)
  • Vout — Output voltage measured across R2 (volts)
  • Vin — Input (source) voltage applied across the R1+R2 chain (volts)
  • R1 — Upper resistor, between Vin and the output node (ohms)
  • R2 — Lower resistor, between the output node and ground (ohms)

Vout is taken across R2 (the lower resistor to ground). Valid for an unloaded divider — any load resistance in parallel with R2 lowers Vout.

Divider current

I = Vin / (R1 + R2)

Series current through both resistors when nothing is connected to the output. Multiply by 1000 to express in milliamps (mA).

Examples

Example 1: Default 12 V divider (2:1 ratio)
Input
Vin = 12 V, R1 = 1000 Ω, R2 = 2000 Ω
Result
Vout = 8 V, I = 0.004 A (4 mA)
Why
Vout = 12 × 2000 / (1000 + 2000) = 12 × 2000 / 3000 = 24000 / 3000 = 8 V. The current is I = 12 / 3000 = 0.004 A, which is 4 mA. Because R2 is twice R1, the output keeps two-thirds of the input.
Example 2: Equal resistors halve the voltage
Input
Vin = 9 V, R1 = 4700 Ω, R2 = 4700 Ω
Result
Vout = 4.5 V, I ≈ 0.000957 A (≈ 0.957 mA)
Why
With equal resistors, Vout = 9 × 4700 / (4700 + 4700) = 9 × 4700 / 9400 = 9 / 2 = 4.5 V — exactly half of Vin. The current is I = 9 / 9400 ≈ 0.0009574 A ≈ 0.957 mA.
Example 3: Scaling 5 V down for a 3.3 V ADC input
Input
Vin = 5 V, R1 = 10000 Ω, R2 = 3300 Ω
Result
Vout ≈ 1.241 V, I ≈ 0.000376 A (≈ 0.376 mA)
Why
Vout = 5 × 3300 / (10000 + 3300) = 16500 / 13300 ≈ 1.241 V. The current is I = 5 / 13300 ≈ 0.0003759 A ≈ 0.376 mA. Larger resistor values reduce wasted current but make the output more sensitive to loading.
Example 4: 3.3 V reference from common E12 resistors
Input
Vin = 3.3 V, R1 = 1000 Ω, R2 = 2200 Ω
Result
Vout ≈ 2.269 V, I ≈ 0.001031 A (≈ 1.031 mA)
Why
Vout = 3.3 × 2200 / (1000 + 2200) = 7260 / 3200 = 2.26875 ≈ 2.269 V. The current is I = 3.3 / 3200 = 0.00103125 A ≈ 1.031 mA.

When to use this tool

  • Scaling a higher voltage down to a level a microcontroller ADC or logic input can safely read (for example 5 V or 12 V down to a 3.3 V range).
  • Setting a reference or bias voltage at the input of a high-impedance op-amp, comparator, or transistor base where the divider supplies almost no current.
  • Estimating the standing current and power dissipated by a resistor pair so you can choose resistor values that don't waste battery or overheat.
  • Quickly sizing R1 and R2 for a target output ratio during breadboarding or PCB design before refining values under load.

Common mistakes

  • Swapping R1 and R2. Vout is the fraction R2/(R1+R2), where R2 is the resistor between the output node and ground. Putting the values in the wrong slots gives the complementary voltage (across R1 instead).
  • Ignoring the load. The Vout = Vin·R2/(R1+R2) formula assumes nothing draws current from the output. Connecting a load whose resistance is comparable to R2 puts it in parallel with R2 and pulls Vout lower than predicted.
  • Treating a divider as a power supply. A divider sets a voltage, not a regulated rail — Vout sags as soon as appreciable current is drawn, so it is unsuitable for powering anything but high-impedance inputs.
  • Mixing units. Keep both resistors in the same unit (both ohms or both kΩ) before adding them; the ratio is unitless, but the current depends on the absolute resistance, so 1 kΩ must be entered as 1000 Ω.

Frequently asked questions

What is the voltage divider formula?

Vout = Vin × R2 / (R1 + R2). Vin is the input voltage applied across the series pair, R1 is the upper resistor between Vin and the output node, and R2 is the lower resistor between the output and ground. The output equals the input scaled by the fraction R2/(R1+R2).

How does a load resistor change the output?

A load connected across the output sits in parallel with R2, lowering the effective R2 and therefore Vout. To keep the error small, choose a load resistance at least 10 times R2 (or make R2 at least 10 times smaller than the load). For an exact result, replace R2 with the parallel combination R2·RL/(R2+RL) in the formula.

How do I pick R1 and R2 for a target Vout?

First fix the ratio: R2/(R1+R2) = Vout/Vin, which rearranges to R1 = R2 × (Vin − Vout) / Vout. Pick R2 from standard values, then compute R1. Smaller resistors give a stiffer (less load-sensitive) output but draw more current; larger resistors save power but are easily disturbed by loading and noise.

How much current does a voltage divider draw?

With no load, the series current is I = Vin / (R1 + R2). In the default example that is 12 / 3000 = 0.004 A = 4 mA. The power wasted as heat is P = Vin × I = Vin² / (R1 + R2), so raising the resistor values lowers both current and dissipated power.

Can I use a voltage divider to power a device?

No. A divider only sets a voltage for a high-impedance input. As soon as a real load draws current the output voltage drops, and the resistors waste power as heat. To power a device at a fixed lower voltage, use a voltage regulator or a DC-DC converter instead.

Why is my measured output lower than the calculated value?

The most common cause is loading: your meter or the next stage is drawing current and pulling Vout down. High-value dividers are also affected by a multimeter's input impedance (often 1–10 MΩ). Resistor tolerance (typically ±1% to ±5%) adds a smaller error. Lower the resistor values or account for the load in the formula to match the calculation.

Sources & references

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