Circuit design / passive networks

Voltage divider

Solve a two-resistor divider with the load taken into account, not just the textbook open-circuit ratio. The error introduced by the load and the Thévenin output impedance are reported so you can size the divider properly.

Current0.83 mADissipation9.9 mWLoadingModerate

Output voltage

3.7179 V

With the load connected

Unloaded output

3.8367 V

Open-circuit divider output

Divider ratio

31.97%

R₂ / (R₁ + R₂)

Load error

-3.10%

How much the load pulls the output down

Divider operating point

Unloaded and loaded output, the current drawn, and the power in each element.

32.0% ratio

Divider operating point
Quantity	Value	Basis
Input voltage	12.000 V	Source
Output, no load	3.8367 V	R₂/(R₁+R₂) = 0.3197
Output, loaded	3.7179 V	R₂‖R_L = 4489 Ω
Load error	-3.10%	Deviation from the unloaded value
Divider current	0.828 mA	Through R₁ and the R₂‖R_L pair
Power in R₁	6.86 mW	I²R₁
Power in R₂	3.08 mW	I²(R₂‖R_L)
Thévenin resistance	3197 Ω	R₁‖R₂ — the source impedance seen by the load

A stiff divider wastes current; a high-impedance divider is easily disturbed by the load. The Thévenin resistance is the figure of merit for that trade-off.

Output vs load resistance

As the load approaches the Thévenin resistance, the output collapses.

Transfer characteristic

Loaded output as R₂ sweeps across three decades.

Canonical equations

The exact relationships implemented by this engine.

V_out = V_in · R₂/(R₁+R₂) · exact only with no load

R₂‖R_L = R₂R_L/(R₂+R_L) · the load appears in parallel with R₂

I = V_in/(R₁ + R₂‖R_L) · shared by both branches

R_th = R₁‖R₂ · output impedance of the divider

Show model notation and computationOpen the symbolic form, the variable definitions, and the step-by-step derivation when you need the detail.

A clear path from inputs to results.

A voltage divider sets an output voltage as a fraction of the input, but only exactly when nothing draws current from it. This engine reports both the open-circuit ratio and the loaded output, so the error from whatever the divider drives is visible rather than assumed away.

Step-by-step methodWorked exampleAssumptions explained

How to use

Enter the input voltage.

Set R₁, the series (top) resistor.

Set R₂, the shunt (bottom) resistor.

Enter the load resistance. Use a very large value to model an open circuit.

Compare the unloaded and loaded outputs — the difference is the loading error.

Check the Thévenin resistance: keep it well below the load for an accurate divider.

The formula

V_{\text{out}} = V_{\text{in}}\frac{R_2}{R_1+R_2}, \qquad V_{\text{loaded}} = V_{\text{in}}\frac{R_2 \parallel R_L}{R_1 + R_2 \parallel R_L}, \qquad R_{th} = R_1 \parallel R_2

Reference: Thévenin equivalent of a resistive network; IEC 60038 for nominal low-voltage levels.

Worked example

12 V input, 10 kΩ and 4.7 kΩ, 100 kΩ load (default inputs)

Inputs

Input 12 V.
R₁ = 10 kΩ, R₂ = 4.7 kΩ.
Load 100 kΩ.

Calculation path

1
Divider ratio: 4,700 / 14,700 = 31.97%.
2
Unloaded output: 12 × 0.3197 = 3.837 V.
3
R₂ in parallel with 100 kΩ is 4,485 Ω, so the loaded output is 3.718 V.
4
Loading error: −3.10%.
5
Thévenin resistance: 10 kΩ ‖ 4.7 kΩ = 3,197 Ω.

Result

Unloaded 3.837 V · loaded 3.718 V · error −3.10% · output impedance 3.2 kΩ.

Interpreting the result

A 100 kΩ load pulls the output down by 3%. That is often acceptable, but if the load were 10 kΩ the error would be about 24% and the divider would be useless as a reference. The rule is to keep the divider current at least ten times the load current, which is equivalent to keeping R_th below a tenth of R_L.

Assumptions & limitations

Resistors are ideal and exactly at their nominal values. A 1% pair contributes up to 2% of ratio error before loading.

The source has zero output impedance. A real source resistance adds to R₁ and shifts the ratio.

The load is purely resistive. Capacitive or inductive loads make the divider frequency dependent.

Temperature coefficients and self-heating are ignored.

No current is drawn by whatever measures the output — a meter with finite impedance loads the divider too.

Frequently asked questions

Why is my measured output lower than calculated?

Almost always loading. The meter or circuit downstream has finite impedance in parallel with R₂, which lowers the effective shunt resistance. Compare the loaded and unloaded rows to quantify it.

What is the Thévenin resistance for?

It is the output impedance of the divider — the resistance the load sees looking back into the network. It sets how much the output sags under load, and it forms a low-pass filter with any cable capacitance.

How does a potentiometer relate to this?

A pot is a divider with a variable tap. The same loading equations apply, with R₁ and R₂ being the two segments either side of the wiper.

How do I choose the resistor values?

Pick the ratio first, then scale both resistors to set the current. Lower values give a stiffer divider but waste more power; higher values save power but are more easily disturbed by the load.

Can I use a divider to power something?

Only for very light loads. A divider is a reference, not a supply: it wastes current continuously and its output collapses as load current rises. Use a regulator or a buffer for real loads.

Does this work for AC?

Only at low frequency with resistive loads. At higher frequencies the parasitic capacitance across each resistor and the load reactance change the division ratio.