Formula
Unloaded, the output is the input times R2 over R1 + R2. A load in parallel with R2 replaces R2 with that parallel value. The current is the input divided by R1 plus the bottom resistance.
Vout = Vin × R2 / (R1 + R2)
Rbottom = R2 ∥ RL
The nearest pair searches 100 Ω to 1 MΩ in the series you choose. R1 is the top resistor. Ties go to the pair whose geometric mean is closer to 10 kΩ. For 12 V and a 3.3 V target, E24 returns 24 kΩ over 9.1 kΩ, which is 3.299 V. That pair is the result of this search. It is not a value IEC 60063 tells you to use. The string current is an Ohm’s law current. If it is a branch load, the wire size and voltage drop pages take it from here. Supply watts over time are the amp calculator, the battery runtime, and the kWh calculator.
Worked examples
12 V, R1 = R2 = 10 kΩ, no load. 6 V. String current is 0.6 mA.
The same divider with a 10 kΩ load. Bottom resistance is 5 kΩ. Output is 4 V.
12 V, target 3.3 V, E24. 24 kΩ on top, 9.1 kΩ on the bottom, 3.299 V. The error is about 0.027%.
Reference
E12, E24, and E96 are the IEC 60063 preferred numbers. The search range is 100 Ω to 1 MΩ.
Questions
What is the output of two 10 kΩ resistors across 12 V?
6 V, unloaded. The bottom resistor is half the string.
What if a 10 kΩ load sits on the output?
The bottom 10 kΩ in parallel with 10 kΩ is 5 kΩ. The output is 12 × 5/15 = 4 V.
How does the E-series pair get chosen?
The search tries every pair from 100 Ω to 1 MΩ in the series you pick and keeps the closest output. If two pairs tie, it keeps the one whose geometric mean is closer to 10 kΩ. That is this page’s search, not an IEC recommendation of a particular pair.
Sources
- Voltage divider Vout = Vin × R2 / (R1 + R2), with R2 replaced by R2 ∥ RL when a load is entered.
- IEC 60063 preferred numbers for the pair search.