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Guided experiment 01

What does a load do to a divider?

Connect a load, predict output and compare the drop with the unloaded value.

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Fixed 5 V supply; R1 = R2 = 1 kΩ. Observe the unloaded output first and predict what connecting a load will do.

Ω
For the current configuration, compared with 2.5 V, output will be…

Load5 VR1 · 1 kΩR2 · 1 kΩOutput: 2.5 V0 V

Load disconnected. No current flows in the right-hand branch.

No load2.5 V
Current configuration2.5 V
Load current0 mA
Drop from ideal0 %

Initial values are explained in the worked example.

One change to identify a cause

  1. With the load disconnected, record 2.5 V and select “Equal” to check that state.
  2. Before connecting it, predict the output change. Enable the 1 kΩ load and compare.
  3. Change only RL to 10 kΩ and then 100 Ω. Describe which load changes the voltage most.
  4. Disconnect it again. Output returns to 2.5 V because the additional current path is gone.

Worked solution: the load is part of the divider

R2 ∥ RL = 1 / (1/R2 + 1/RL)
With RL = 1 kΩ: parallel combination = 500 Ω
Vsal = 5 × 500 / (1000 + 500) = 1.6667 V

Load current is 1.6667 mA. It comes through R1 before splitting, not from nowhere. A 10 kΩ load gives 2.3810 V; 100 Ω gives 0.4167 V. Lower load resistance draws more current and changes this output more.

The comparison refers to the current configuration: disconnected means 2.5 V; any positive finite load resistance in range gives a lower output. Checking a prediction does not lock the controls.

What this experiment demonstrates

This demonstrates resistive DC loading with ideal sources and resistors. It excludes tolerances, capacitors and changing current demand. Do not choose a regulated supply based on an unloaded voltage alone.

Why does a 100 Ω load not make the output exactly 0 V?

Because its resistance is still positive and finite. Its parallel connection with R2 is not a short circuit; it still produces a calculable voltage drop.