Electrical · Calculator

Voltage divider.

Voltage divider: Calculate the output voltage of a resistor voltage divider given Vin, R1, and R2.

Two-resistor voltage divider — output voltage, current draw, and power dissipation in each resistor. Plus the closest E12 and E24 standard resistor values.

How the math works

Vout = Vin × R2 / (R1 + R2)

Current  I = Vin / (R1 + R2)
Power R1   P1 = I² × R1
Power R2   P2 = I² × R2

The formula assumes no load on the output — i.e., the downstream circuit draws negligible current compared to what flows through the divider. If the load draws current comparable to I, the actual Vout will sag (it forms a Thévenin source with output impedance R1∥R2).

Rule of thumb: load resistance ≫ R1∥R2

For the voltage divider to work as designed, the load resistance should be at least 10× higher than the parallel combination of R1 and R2. If your load is 10 kΩ, your divider's R1∥R2 should be ≤ 1 kΩ.

This trades off against power dissipation: lower R values mean more current flows continuously, which means more wasted power and more heat. The classic compromise is to make R1+R2 around 10-50× the load impedance.

Common pitfalls

Common questions

What is a voltage divider?

Two resistors in series across a voltage source. The output, tapped between them, is a fixed fraction of the input: Vout = Vin × R2 ÷ (R1 + R2).

Can I power a circuit from a voltage divider?

Generally no. A divider only holds its calculated voltage when almost no current is drawn from the tap. As soon as a load pulls current, the output sags. Use dividers for reference or sense voltages — for supplying power, use a voltage regulator or buck converter.

How do I choose R1 and R2 values?

Set the ratio for your target voltage, then scale both resistors for your current budget. Smaller values waste more power but tolerate heavier loads; larger values save power but are more sensitive to loading and noise. A combined 10k–100k Ω is a common compromise.

How much power do the resistors dissipate?

A standing current of I = Vin ÷ (R1 + R2) flows through the pair continuously, dissipating P = Vin² ÷ (R1 + R2). With low-value resistors this can be significant — check each resistor's wattage rating against its I²R.

Why is my measured output different from the calculation?

Loading (your meter or the downstream circuit draws current), resistor tolerance (±5% or ±1%), and temperature all shift it. A 10 MΩ multimeter barely loads a 10k divider but noticeably loads a 1 MΩ one.

What are the E12 and E24 resistor series?

They are the standard manufactured resistor values: E12 has 12 values per decade (±10% parts), E24 has 24 (±5% parts). The calculator suggests the nearest standard pair because arbitrary resistance values aren't stocked.

Sources

Disclaimer. Voltage dividers are appropriate for level-shifting and sensing applications. For power delivery, use a voltage regulator (LDO, switching regulator) instead.

See also