Electronics · Circuit theory · Beginner

Voltage dividers: calculate the output and load effect

Use two series resistors to find an output voltage, then see why connecting a load resistor changes the result.

By Amit Kumar Das About the author Published

Who this is for: Beginners who know Ohm’s law and how to combine resistors in series and parallel.

What you will learn: find an unloaded divider output, calculate the effect of a load connected across the output, and identify the assumptions behind the simple formulas.

Checked example: The results below were checked by direct arithmetic. The loaded example uses the same source and resistor values as the default circuit in the Thevenin calculator.

The basic voltage divider

A voltage divider uses two resistors in series across a source. The output is measured across the lower resistor, R2:

Unloaded divider · Vs = 12 V
Two-resistor voltage divider A 12 volt source feeds a 1 kilo-ohm resistor in series with a 2 kilo-ohm resistor. The output is measured across the lower resistor and is 8 volts with no load connected. R1 1 kΩ R2 2 kΩ Vs = 12 V Vout 8 V 0 V

With no other component connected to Vout, the same current flows through R1 and R2. The output is the fraction of the source voltage that appears across R2:

Vout = Vs × R2 / (R1 + R2)

This assumes an ideal voltage source and that the output is not significantly loaded by the next circuit.

Worked example: no load connected

Use a 12 V source, R1 = 1 kΩ, and R2 = 2 kΩ. The series resistance is 3 kΩ, so the divider current is 12 V / 3 kΩ = 4 mA. The output across R2 is:

Vout = 12 × 2 / (1 + 2)
     = 8 V

The source voltage divides in proportion to resistance: 4 V is across R1 and 8 V is across R2. Together they add to the 12 V supply.

What changes when you connect a load?

A real next stage draws some current. If a load resistor RL is connected from Vout to 0 V, it is in parallel with R2. The divider’s lower resistance becomes:

Rbottom = R2 × RL / (R2 + RL)
Vout    = Vs × Rbottom / (R1 + Rbottom)

For the same 12 V source and 1 kΩ / 2 kΩ divider, connect a 1 kΩ load:

  1. Combine R2 and RL in parallel: 2 kΩ × 1 kΩ / (2 kΩ + 1 kΩ) = 0.667 kΩ.
  2. Find the output: 12 V × 0.667 / (1 + 0.667) ≈ 4.8 V.
  3. The output falls from 8 V to about 4.8 V because the load makes the lower leg of the divider effectively smaller.

The loaded result can also be checked using the Thevenin equivalent of the divider. For this example, Vth = 8 V and Rth ≈ 0.667 kΩ, so a 1 kΩ load receives 8 × 1 / (0.667 + 1) ≈ 4.8 V.

Try it yourself

A 9 V ideal source is connected to R1 = 1 kΩ and R2 = 2 kΩ. First calculate Vout with no load. Then connect RL = 2 kΩ across the output. What is the loaded output?

Show the calculation

With no load: Vout = 9 × 2 / (1 + 2) = 6 V.

With RL = 2 kΩ: R2 and RL are 1 kΩ in parallel, so Vout = 9 × 1 / (1 + 1) = 4.5 V.

When a divider is—and is not—a good choice

  • A divider is useful for creating a reference or signal voltage when the following input draws little current compared with the divider current.
  • If the load is significant, include it in the calculation. Do not assume the unloaded formula still gives the loaded output.
  • For powering a load that needs a stable voltage, a resistor divider may be unsuitable because the output changes with load current and resistor tolerances. Consider an appropriate regulator or buffer for the application.
  • These calculations do not check resistor power ratings, source limits, temperature effects, or circuit safety. Verify component ratings and the complete circuit before building it.

Continue learning

Review how to read resistor color codes or decode physical bands with the resistor color code calculator. Then learn how a divider can be replaced by an equivalent source in the Thevenin theorem lesson. Try the Thevenin calculator with the example values; its model covers this one ideal loaded-divider topology.