Case 1
- Input voltage (V)
- 12V
- Resistance R₁ (Ω)
- 1000Ω
- Resistance R₂ (Ω)
- 2000Ω
8.0000V
Open with these valuesVoltage Divider Calculator
8.0000V
Result: 8.0000 VThe output is the input voltage times the share R₂ holds of the total resistance: Vout = Vin × R₂ ÷ (R₁ + R₂). R₁ is the top resistor, R₂ the bottom one between the output node and ground — swap them and the answer is wrong but still looks plausible. The figure is the unloaded output.
Worked examples
Case 2
- Input voltage (V)
- 9V
- Resistance R₁ (Ω)
- 1000Ω
- Resistance R₂ (Ω)
- 1000Ω
4.5000V
Open with these valuesCase 3
- Input voltage (V)
- 5V
- Resistance R₁ (Ω)
- 330Ω
- Resistance R₂ (Ω)
- 470Ω
2.9375V
Open with these valuesHow it's calculated
Vout = Vin × R₂ ÷ (R₁ + R₂)
- StepAdd both resistances to get the total the input voltage sits across.
- StepDivide R₂ by that total — this is the share the output keeps.
- ResultMultiply the input voltage by that share.
What this number means
This is the unloaded output voltage
The formula assumes nothing but R₂ draws current from the output node. Anything you connect sits in parallel with R₂ and pulls the output lower; the effect is small only when the load resistance is much larger than R₂.
R₂ is the bottom resistor, R₁ the top one
R₁ sits between the input and the output node, R₂ between the output node and ground, and the output is the voltage across R₂. Swap them and 12 V with 1 kΩ and 2 kΩ reads 4 V instead of 8 V — wrong, but still plausible.
Only the ratio of the two resistors sets the share
R₂ ÷ (R₁ + R₂) is 2 ÷ 3 for 1 kΩ and 2 kΩ just as it is for 10 Ω and 20 Ω, so both give 8 V from 12 V. The absolute values still decide how hard a connected load pulls the output down.
Commonly misread
Vout = Vin × R₂ ÷ R₁, so 12 V with 1 kΩ and 2 kΩ gives 24 V.
The denominator is the sum of both resistors, not R₁ alone: 12 × 2000 ÷ 3000 = 8 V. A divider can never output more than it is fed.
I put the 2 kΩ on top, which still counts as R₂.
R₂ is whichever resistor sits between the output node and ground. With the 2 kΩ on top the answer is 4 V, not 8 V.
I measured 8 V here, so my circuit will see 8 V.
8 V is the unloaded figure. Whatever you hang on the output parallels R₂ and lowers the voltage.
Reference table
| Vin, R₁, R₂ | Share R₂ holds | Vout |
|---|---|---|
| 12, 1000, 2000 | 2 ÷ 3 | 8 |
| 9, 1000, 1000 | 1 ÷ 2 | 4.5 |
| 5, 330, 470 | 0.5875 | 2.9375 |
| 10, 100, 900 | 9 ÷ 10 | 9 |
| 3.3, 1000, 2000 | 2 ÷ 3 | 2.2 |
Questions
How do I calculate a voltage divider?
Multiply the input voltage by R₂ and divide by the sum of both resistors: Vout = Vin × R₂ ÷ (R₁ + R₂). For 12 V with R₁ = 1 kΩ and R₂ = 2 kΩ that is 8 V.
Which resistor is R₁ and which is R₂?
R₁ is the top resistor, between the input voltage and the output node; R₂ is the bottom one, between the output node and ground. The output voltage is the voltage across R₂. Swapping the two changes the result.
What happens if both resistors are equal?
Equal resistors split the voltage exactly in half, because R₂ ÷ (R₁ + R₂) becomes one half. Nine volts across two 1 kΩ resistors gives 4.5 V.
Why is this the unloaded output voltage?
The formula assumes nothing but R₂ draws current from the output node. Whatever you connect sits in parallel with R₂ and pulls the output lower. The effect is small when the load resistance is much larger than R₂.
Where are voltage dividers used?
To scale a higher voltage down to a usable level: reading a battery with a microcontroller's analog input, setting a reference voltage, biasing a transistor. They also read sensors such as potentiometers and thermistors.
Sources and last check
Information, not professional advice.