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Lesson 3 of 6 · 15 minutes · Free

Voltage Dividers Under Load

Calculate loaded output voltage instead of assuming an unloaded divider ratio.

Try the experiment ↓Practice challenge ↓

See the idea

A voltage divider places two resistors in series and takes an output from their junction relative to the return node. With no load, the lower resistor determines the output fraction.

An attached load is in parallel with the lower resistor. That lowers the effective lower-leg resistance and pulls the output down. The divider is not an ideal voltage source.

The load switch on the bench adds a finite resistance across R2. A sensor input with very high resistance may cause little error; a low-resistance load can cause substantial error.

Rlower = R2 ‖ RL Vout = Vs × Rlower / (R1 + Rlower)

A divider ratio is only valid for the load condition used to calculate it.

Work through an example

  1. With Vs = 12 V and R1 = R2 = 1,000 Ω, unloaded output is 6 V.
  2. Attach RL = 1,000 Ω: Rlower = 1,000 ‖ 1,000 = 500 Ω.
  3. Vout = 12 × 500/1,500 = 4 V. Source current is 8 mA; 4 mA flows in R2 and 4 mA in the load.

Watch for this

Do not use a resistor divider as a general substitute for a regulated supply. Load changes alter both voltage and resistor dissipation.

Predict, then test

  1. Toggle the load at its default 1,000 Ω. Observe output move between 6 V and 4 V.
  2. Raise load resistance to 10,000 Ω. Compare output with the unloaded case.

Your virtual bench

Change a setting. See the circuit respond.

Swipe the diagram sideways to explore the connections.

±12 VR1 1000 ΩR21000 ΩReturn reference: 0 V · Vout at R1/R2 junction
Source current6 mA
Supply power0.072 W
R1 voltage6 V
Load / R2 voltage6 V

R2 current: 6 mA · RL current: 0 mA.

Model: Ideal low-voltage DC source and linear resistors; zero wire resistance except the selected connection fault. Values are calculated, not equipment measurements. Browser simulation only—do not connect live equipment.

Check your understanding

Connecting a finite load across the lower resistor of an ideal divider makes its output:

Put it to work

With 12 V, R1 = R2 = 1,000 Ω and RL = 10,000 Ω, predict output to two decimals.

Show the worked solution

Rlower = 909.09 Ω; Vout = 12 × 909.09/1,909.09 = 5.71 V, about 4.76% below the unloaded 6 V.

Try the experiment and answer the check, then record your completion.

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