Voltage Divider Calculator

Solve any unknown in a two-resistor voltage divider: Vout, Vin, R1, or R2.

Solve for

Vin — Input Voltage

R1 — Top Resistor

R2 — Bottom Resistor

Vout — Output Voltage

Vin ──[R1]──┬──[R2]── GND │ Vout

Vout

8.0000 V

Current through divider: 4.000 mA (Vin / (R1 + R2) = 12V / 3000Ω)

Voltage Divider Calculator

A voltage divider is a fundamental passive circuit that uses two resistors in series to produce an output voltage that is a fraction of the input. The formula is Vout = Vin × R2 / (R1 + R2). This tool lets you solve for any one of the four variables by entering the other three. It also shows the quiescent current flowing through the divider, which is useful for power budget calculations. All math runs locally in your browser.

Built and maintained by Meet Shah · Last updated 19 July 2026

What this tool is used for

Choosing resistor values to produce a target output voltage.
Working out the output of an existing divider.
Solving for one resistor given the other and the target.
Checking a divider's current draw before choosing values.
Confirming a measured output against the design.

Frequently Asked Questions

Why does the output voltage sag when I connect a load?: Because the load sits in parallel with R2 and lowers the effective resistance, which shifts the ratio. The usual rule of thumb is to make the divider current at least ten times the load current — otherwise the circuit you attach becomes part of the divider and the maths above stops describing it.
What is the quiescent current for?: It is the current flowing through the divider itself, Vin / (R1 + R2), wasted as heat whether anything is connected or not. On a battery it matters a great deal: a 12 V supply across 3 kΩ burns 4 mA continuously, which is far more than a sleeping microcontroller draws.
How do I choose the resistor values themselves?: The ratio sets the voltage; the magnitudes trade current waste against noise. Low values (hundreds of ohms) waste power but are stiff and immune to leakage; high values (megohms) sip current but pick up noise and are loaded by the input impedance of whatever reads them.
Can a divider power a circuit?: No — it is a reference, not a regulator. It cannot supply meaningful current, and its output moves with the input, so a sagging battery drags Vout down with it. Use it for level-shifting a signal or reading a sensor; use an LDO or buck converter to actually power something.
Why does solving for R1 or R2 sometimes give a negative answer?: Because a passive divider can only attenuate. Asking for Vout above Vin has no solution with positive resistors, and the algebra dutifully returns a negative value, which is rejected here rather than displayed. Raising a voltage needs a boost converter or a charge pump.

Common errors and gotchas

Ignoring the load, which draws current and pulls the output below the unloaded calculation.
Choosing very high resistances, which makes the output sensitive to noise and load.
Choosing very low resistances, which wastes power continuously.
Using a divider to power something rather than to provide a reference, which it cannot do.
Forgetting resistor tolerance, which shifts the output from the calculated value.