Why did adding a load lower the output voltage?

The load is in parallel with R2, reducing the effective lower-leg resistance and therefore the divider ratio.

Is this enough to design an ADC input divider?

It is a useful DC check, but an ADC may also impose source-impedance, acquisition-time, leakage, clamp-current, filtering, and absolute-maximum constraints from its datasheet.

Electronics / LOCAL TOOL

Voltage Divider Calculator

Calculate loaded or unloaded divider voltage, currents, impedance, and power.

Runs locally — input is never uploaded

Divider valuesOptional load supported

Input voltage (V) R1, upper resistor (Ω) R2, lower resistor (Ω) Load to ground (Ω), optional

Plain values use the units in the labels; suffixes such as 12 V, 4.7 kΩ, and 100 kΩ are accepted too.

ResultReady when you are

V_{out} = V_{in} \frac{R_{2}}{R_{1} + R_{2}}

R_{lower} = R_{2} ∥ R_{load} V_{out} = V_{in} \frac{R_{lower}}{R_{1} + R_{lower}}

WHAT IT DOES

See what the divider delivers after the load is connected.

Calculate the output of a two-resistor DC divider with an optional load to ground. The result includes source and load current, Thevenin output impedance, and the individual resistor dissipation that a voltage-only answer hides.

Why use it?

Use it to estimate a divider's output and see why connecting a load changes the voltage you expected from the resistor ratio alone.

How to use this tool

Enter input voltage, upper resistor R1, and lower resistor R2; add the load resistance when one is connected to ground.
Calculate the output and individual dissipations. Two 10 kΩ resistors across 5 V give 2.5 V with no load.
Repeat with the intended load and compare the result. A divider is not a regulated supply, and a high-impedance calculation does not model every ADC or dynamic load.

Details and limitations

With a load present, the lower leg is calculated as R2 in parallel with the load before the divider ratio is applied.
Output impedance is reported with the ideal voltage source shorted, so it is R1 in parallel with R2 and does not include a stated load.
Real sources, ADC sampling capacitors, input leakage, resistor tolerance, noise, and bandwidth can change a practical divider's behaviour.

WORKED EXAMPLES

Show why a load pulls the output down

Calculate this 5 V divider with a 10 kΩ load, then clear only the load field and calculate again. The load is in parallel with the lower resistor.

Input voltage

5 V

Upper resistor

10 kΩ

Lower resistor

10 kΩ

Load

10 kΩ

Expected result

With the load, output is approximately 1.667 V. Without it, output is 2.5 V. The reported divider output impedance is 5 kΩ in both calculations; it excludes the external load.

COMMON QUESTIONS

Voltage Divider questions

Why did adding a load lower the output voltage?: The load is in parallel with R2, reducing the effective lower-leg resistance and therefore the divider ratio.
Is this enough to design an ADC input divider?: It is a useful DC check, but an ADC may also impose source-impedance, acquisition-time, leakage, clamp-current, filtering, and absolute-maximum constraints from its datasheet.