Online Voltage Divider Calculator: Formula and Examples

An online voltage divider calculator works out the output voltage from a resistor divider when you enter the input voltage and the two resistor values. It can also show the output with a load attached, which matters because a real circuit may produce less voltage than the unloaded formula predicts.

Formula

The calculator models a source voltage, two series divider resistors, and an optional load connected across the lower resistor. It first combines the lower resistor and load as a parallel equivalent:

R_eq = R₂R_L / (R₂ + R_L)

Then the loaded output voltage is V_out = V_in × R_eq / (R₁ + R_eq). With no load, take R_L as infinite, so R_eq equals R₂ and the familiar formula is V_out = V_in × R₂ / (R₁ + R₂).

V_in is the source voltage in volts (V).
R₁ is the resistor between the source and output, in ohms (Ω).
R₂ is the resistor between output and ground, in ohms (Ω).
R_L is the load resistance across the output and ground, in ohms (Ω). Enter 0 to represent no load in this calculator.
R_eq is the parallel equivalent of R₂ and R_L, in ohms.

The other outputs are source current, I_in = V_in / (R₁ + R_eq), and the ideal resistor dissipations: P₁ = I_in²R₁ and P₂ = (V_out/R₂)²R₂. Current is in amperes (A) and power in watts (W). The P2 result is the power in R2 itself, not the power consumed by the load.

Worked Example

Using the defaults—12 V, R1 = 10,000 Ω, R2 = 10,000 Ω, and no load—the unloaded divider has R_eq = 10,000 Ω.

Output: 12 × 10,000 / (10,000 + 10,000) = 6.000 V.
Source current: 12 / 20,000 = 0.000600 A (0.6 mA).
R1 power: 0.000600² × 10,000 = 0.003600 W.
R2 power: (6 / 10,000)² × 10,000 = 0.003600 W.

The displayed results are rounded to the number of decimal places shown in the calculator block. The current and power calculations use the ideal circuit values, not resistor tolerances or temperature effects.

How to Use a Voltage Divider

Enter the supply voltage and the resistance of each divider resistor. Use ohms for both resistor entries; for example, 10 kΩ is 10,000 Ω. If a circuit has a load connected from the output node to ground, enter its resistance in RL. If there is no load, leave RL at 0. The output voltage is measured from the junction of R1 and R2 to ground.

For a first estimate of an unloaded output, use the two-resistor formula. Equal resistors produce half the input voltage.
For a loaded output, include the load resistance. A smaller load resistance draws more current and pulls the output down.
Use the current and power results as circuit estimates, then compare component stress with the resistor specifications and actual operating conditions.

Voltage dividers are commonly used to create a reference or scale a voltage for a high-impedance input, such as a measurement input. They are not usually a good substitute for a regulated power supply: a load that changes its current also changes the divider output. For a circuit that must hold a stable voltage while delivering meaningful current, consider an appropriate regulator or buffer instead.

Loading, Accuracy, and Resistor Choice

The simple formula assumes the output is not disturbed by anything connected to it. A measuring instrument, input pin, or other circuit behaves as a load with finite resistance. If its resistance is comparable to R2, it is effectively in parallel with R2, reducing the lower branch resistance and therefore reducing Vout. The loaded calculation captures this first-order effect.

As a practical check, compare the load resistance with R2. A load resistance much higher than R2 changes the divider less; a resistance close to R2 changes it substantially. If the input impedance is specified as a range, calculate the output at the least favorable expected value as well. Some electronic inputs are not simple resistors: protection diodes, bias currents, switching, and capacitance can make their behavior depend on voltage or time. In those cases the equivalent-resistance model may not describe the full circuit.

Resistors have tolerances, so the actual output can differ from the nominal calculation. For the unloaded divider, a low R1 paired with a high R2 raises the output, while a high R1 paired with a low R2 lowers it. For precision work, check the resistor tolerance, temperature coefficient, source variation, and input loading. The calculator uses nominal values and does not perform a tolerance or temperature analysis.

Common Mistakes and Limits

Mixing units: use ohms consistently. Entering 10 for a 10 kΩ resistor means 10 Ω, not 10 kΩ.
Reversing the resistors: R1 is above the output node; R2 is below it, between output and ground. Swapping their values changes the result.
Ignoring the load: an unloaded result may be too high once a real circuit is connected.
Assuming a divider regulates: the output depends on input voltage, resistor values, and load current; it is not inherently fixed.
Overlooking power: calculate each resistor’s dissipation and check it against the part’s rated power under the actual conditions. The calculator does not select components or account for derating.

The model assumes a steady DC source, ideal wiring, and linear resistors. It does not include resistor tolerance, self-heating, source resistance, wiring resistance, transient behavior, or nonlinear loads. The zero-ohm boundary for a resistor is mathematically allowed by the input fields but can make the circuit singular, such as a short from the output node to ground; treat such boundary cases as invalid circuit designs rather than useful divider values.

Frequently Asked Questions

What is the voltage divider formula? For an unloaded divider, Vout = Vin × R2 / (R1 + R2), where R1 is above the output node and R2 is below it to ground. A load across R2 must be included as a parallel resistance.

How do I calculate a voltage divider with a load? Find the parallel equivalent of R2 and the load: Req = R2RL/(R2 + RL). Then use Vout = Vin × Req/(R1 + Req). This assumes the load can be represented by a resistance.

Why is my measured divider voltage lower than the calculator result? The connected circuit may load the divider, or the supply and resistor values may differ from their nominal values. Check the input impedance, resistor tolerances, grounding, and whether the measurement connection is across the intended nodes.

Can a voltage divider reduce voltage for powering a device? It can reduce voltage for a light, predictable load, but the output changes with load current. For a device that needs a stable supply or draws varying current, a suitable regulator or buffer is generally more appropriate.

What resistor values should I use? Choose values based on the required output, acceptable divider current, load impedance, accuracy, and resistor power. Lower values make the divider less affected by a given load but draw more current; higher values save current but make loading and leakage more significant.

Formula used

in vin | Input voltage | V | 12 | 0
in r1 | Upper resistor R1 | ohm | 10000 | 0
in r2 | Lower resistor R2 | ohm | 10000 | 0
in rl | Load resistance RL (0 = no load) | ohm | 0
out vout | Output voltage | V | vin * (r2 * rl / (r2 + rl)) / (r1 + (r2 * rl / (r2 + rl))) | 3
out iin | Current from source | A | vin / (r1 + (r2 * rl / (r2 + rl))) | 6
out p1 | Power in R1 | W | iin ^ 2 * r1 | 6
out p2 | Power in R2 | (W) | (vout / r2) ^ 2 * r2 | 6

Each line is one input (in, sel, list) or one result (out: label, unit, formula, decimals). How the calculators work

Method: the standard formula for this calculation, rounded as each result states. See how the calculators work or report an error.

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