Voltage divider calculator
Output voltage, current and per-leg power of a two-resistor divider.
Fill in the fields and the result will appear here automatically.
Calculate unloaded output voltage, common current and each leg dissipation for two resistors. R1 is the upper leg from input to output; R2 is the lower leg from output to reference. Their ratio sets the fraction, while absolute values affect current, heating and loading sensitivity. Signed or zero input voltage is accepted.
How it works
Formula and logic
Uout=Uin·R2/(R1+R2), I=Uin/(R1+R2), P1=I²R1, P2=I²R2. Enter V and Ω; output current is mA and powers mW. R1, R2>0. Negative Uin reverses Uout and I, but leaves powers unchanged. For AC use RMS if the elements remain purely resistive.
Example
12 V, 10000 Ω and 4700 Ω give 3.837 V, 0.8163 mA and 31.9728% of input. At −12 V output and current reverse sign while powers stay unchanged; at 0 V they are all zero. An attached load requires a different calculation.
Fields and units
- Input voltage — V
- Upper resistor R1 — Ω
- Lower resistor R2 — Ω
How to use
- — Enter V and Ω: 10 kΩ means 10000 Ω.
- — Check R1/R2 placement: output is across R2.
- — A measurement input is not automatically an insignificant load.
- — Check voltage, dissipation, rated power and tolerance against actual component data.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: common series current and parallel-branch voltage
- Limitation
- Two purely resistive legs without output loading; not a regulator or loaded/reactive network calculation.
FAQ
Can a divider power a load?
A load RL is parallel with R2. With Rth=R1 R2/(R1+R2), Uloaded/Uopen=1/(1+Rth/RL). Equal legs and RL=10 R2 give 4.7619% droop, which is not automatically an acceptable error.
Which resistor values should I pick for a given ratio?
Any pair with the same ratio gives the same unloaded fraction. Current, losses, output Rth, parasitic capacitance and tolerances still matter. There is no universal kilohm range or 0.25 W rating.
Why is power computed from the current rather than the voltage?
Both forms are equivalent, but the current is shared by both legs, which makes it the shorter route: each leg dissipates current squared times its own resistance.
Does this work on AC?
For a purely resistive divider yes, reading the voltage as RMS. Add capacitance or inductance and a frequency dependence appears that is not modelled here.