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Potential divider with a potentiometer

Curricula: GCSE, A-level, IB B.5

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Aim

Obtain an adjustable voltage from a potentiometer and see how a load changes the output of the divider.

The bench

A 10 V source is connected to the end terminals of a 1 kΩ potentiometer; the knob is in the middle position. The voltmeter is between the wiper and the negative terminal. A 1 kΩ load is connected to the output through a switch, which is open.

Procedure

  1. Note the voltmeter reading.
  2. Turn the potentiometer knob all the way in one direction and then in the other.
  3. Return the knob to the middle position and close the load switch.
  4. Note the voltmeter reading under load.
Expected result

In the middle position the output is about 5.00 V. Turning the knob changes it from 0 to 10 V: clockwise, the wiper moves towards the terminal connected to the positive terminal of the source, and the voltage rises. Under the 1 kΩ load the output drops to about 4.0 V: the load shunts the lower half of the potentiometer.

Questions

  1. Why does the divider voltage decrease under load?
  2. What must the load be for the drop to be less than 1%?
Answers
  1. The load is connected in parallel with the lower half of the potentiometer, and 500 Ω in parallel with 1 kΩ make about 333 Ω. The lower part of the divider now has less resistance than the upper 500 Ω and gets a smaller share of the 10 V: 10 V × 333 / 833 ≈ 4.0 V.
  2. Seen from the output, the divider in the middle position acts as a 5 V source with an output resistance \(R_{out}\) = 250 Ω, the two 500 Ω halves in parallel. The output falls by the fraction \(R_{out} / (R_{out} + R_L)\), so for a drop below 1% the load \(R_L\) must be more than 99 × 250 Ω ≈ 25 kΩ, 25 times the resistance of the potentiometer.