- Uout = Uin · R₂ / (R₁ + R₂)
The same current flows through both resistors, so each takes a share of the input voltage in proportion to its resistance. The output is taken across R₂, the one connected to ground. It follows from Ohm’s law and the rule for resistors in series.
Holds when nothing draws current from the output. A load across R₂ is a resistor in parallel with it, which lowers the effective R₂ and the output with it.
A worked example
12 V across 2,000 Ω and 1,000 Ω: Uout = 12 · 1,000 / 3,000 = 4 V. The current through the pair is 12 / 3,000 = 4 mA, and the divider dissipates 48 mW.
The load
Connect a 1,000 Ω load across the output. R₂ and the load in parallel make 500 Ω, and the output falls to 12 · 500 / 2,500 = 2.4 V. The rule of thumb is to make the divider’s current at least ten times the load’s, which means small resistors, which means more power wasted in the divider. A divider feeds a measurement input or a reference, not a motor.
Choosing the resistors
The ratio sets the output; the size sets the current. 20 kΩ and 10 kΩ give the same 4 V as the example at a tenth of the current and the power, and suit an input that draws almost nothing. The card above solves for either resistor from the other three.
Common mistakes
- Taking the output across R₁. That gives Uin − Uout, the other share.
- Using a divider to power something. The output sags under load and the resistors heat.
- Expecting an output above the input. A divider only lowers a voltage, which is the condition in the card.
Frequently asked questions
What is a potentiometer?
A divider whose tap slides along one resistor, so R₁ and R₂ change together with a constant sum.
Can I divide an AC voltage?
With resistors, yes, in the same ratio. With capacitors or inductors the division depends on frequency.
Which values do I actually buy?
The nearest preferred values in the E12 or E24 series; the
resistor colour code calculator names them and their bands.