Calculators
Voltage Divider Calculator
Calculate voltage-divider output or solve R2 for a target voltage.
Calculate the unloaded output of a two-resistor voltage divider, or solve the ideal R2 value for a target output. Resistance inputs are in ohms.
Output voltage
6 V
Divider current0.006 A
R1 power0.036 W
R2 power0.036 W
Vout = Vin × R2 ÷ (R1 + R2)
This is an ideal unloaded divider model. A connected load changes the effective lower resistance and therefore the output voltage.
About This Tool
A two-resistor voltage divider produces a fraction of an input voltage by placing R1 and R2 in series and taking the output across R2. This calculator finds the ideal unloaded output voltage, current through the divider, and power dissipated by each resistor. It can also rearrange the divider equation to estimate R2 for a desired output when Vin and R1 are known. The calculation runs locally in your browser and is intended for learning, prototyping, and checking ideal circuit calculations.
How To Use It
- Choose Calculate output to enter Vin, R1, and R2, or Find R2 to enter Vin, R1, and a target output voltage.
- Enter voltage in volts and resistance in ohms. R1 and R2 must be greater than zero for the output calculation.
- Review the output voltage and, when calculating a divider, the ideal divider current and resistor power values.
- For a real circuit, account for the connected load, resistor tolerances, source limits, component ratings, and electrical safety before selecting parts.
Examples
Half the input voltage
With Vin = 12 V, R1 = 1 kΩ, and R2 = 1 kΩ, Vout = 12 × 1000 / (1000 + 1000) = 6 V.
Find R2 for 5 V from 12 V
With Vin = 12 V and R1 = 1 kΩ, the ideal R2 for 5 V is about 714.29 Ω. A standard resistor value will produce a slightly different actual output.
Divider current and power
A 12 V divider using two 1 kΩ resistors carries 6 mA ideally. Each resistor dissipates 0.036 W before any external load is connected.
Useful Notes
Voltage divider formula
For an unloaded two-resistor divider with the output measured across R2, Vout = Vin × R2 / (R1 + R2). The same current flows through both series resistors: I = Vin / (R1 + R2).
Solving for R2
Rearranging the ideal divider equation gives R2 = Vout × R1 / (Vin − Vout). A passive divider cannot produce an unloaded output greater than or equal to its positive input voltage with finite positive resistors.
Loading changes the output
The basic equation assumes the output is unloaded or measured by an effectively infinite input resistance. A real load appears in parallel with R2, reducing the effective lower resistance and usually lowering Vout. For accurate design, include the load resistance or use a buffer/regulator where appropriate.
Resistor power
The calculator reports ideal resistor dissipation using P = I²R. Component selection should include appropriate margin and account for tolerance, temperature, environment, pulse conditions, and manufacturer ratings rather than choosing a resistor exactly at the calculated dissipation.
Divider versus voltage regulator
A resistor divider is useful for references, sensing, biasing, and other light-load signals. It is generally not a substitute for a regulated power supply when the load current changes, because the output depends on the load and continuously draws divider current.
Electrical safety
A correct divider calculation does not make a circuit safe. High voltage, mains power, batteries capable of high fault current, and unsuitable component ratings can cause shock, fire, or equipment damage. Use appropriately rated components and qualified guidance for hazardous circuits.
FAQ
Why is the output across R2?
This calculator uses the common convention where R1 connects from Vin to the output node and R2 connects from the output node to the reference or ground. Vout is therefore the voltage across R2.
Can a voltage divider power a device?
It can supply very small, predictable loads in some designs, but changing load current changes the output voltage. Regulators or buffered circuits are usually more appropriate for powering devices.
Can I enter resistor values in kΩ?
Inputs are labeled in ohms, so enter 1 kΩ as 1000. Because the divider ratio depends on resistance ratios, using consistent resistance units gives the same voltage ratio, but the current and power calculations require the actual ohm values.
Why is my measured output different?
Common causes include load resistance, resistor tolerance, source resistance, meter input resistance, wiring, temperature, and a Vin value that differs from the assumed input.
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