Voltage divider formula
Vout = Vin × R2 / (R1 + R2)
Example: 12 V, R1 = 10 kΩ, R2 = 4.7 kΩ → Vout = 12 × 4.7 / 14.7 ≈ 3.84 V.
What a load does to the output
Anything connected to the output sits in parallel with R2, so the effective bottom resistor becomes R2 ∥ RL = R2·RL / (R2 + RL) and the output drops. With RL = 10 kΩ in the example above, the output falls from 3.84 V to about 2.9 V. Rule of thumb: keep the divider resistors at least ten times smaller than the load, or buffer the output with an op-amp follower. A divider is fine for reading a voltage with an ADC; it is not a power supply.
Full explanation: why a voltage divider's output drops under load.
Why simulate it?
A simulation puts a voltmeter on the real node, so you can change the load and watch the output move. Press Simulate this circuit to open the divider in CircuPilot with these values.
FAQ
How do I get 5 V from 12 V with a voltage divider?
Choose R2 / (R1 + R2) = 5/12, for example R1 = 7 kΩ and R2 = 5 kΩ, or standard values R1 = 6.8 kΩ, R2 = 4.7 kΩ (≈4.9 V). This only works for a high-impedance input such as an ADC pin — to power something, use a regulator.
Why is my voltage divider output lower than calculated?
Because something is loading it. The load resistance is in parallel with R2 and pulls the output down. Enter the load in the calculator to see by how much.
What resistor values should I use?
Low enough that the load (including an ADC input) barely matters, high enough not to waste current. For reading a sensor with a microcontroller, a total of 10 kΩ to 100 kΩ is typical.