A voltage divider under load

Why does the output fall when a load is connected?

Predict the output. Then change the circuit.

Ideal 5 V source

Upper R₁ Lower R₂ Connect a 1 kΩ load

2.50 VEqual resistors share the supply voltage equally.

Vout = 5 × Rlower / (R₁ + Rlower). With a load, Rlower = R₂ ∥ Rload.

Ideal DC source and resistors. No tolerances, temperature effects, source impedance or transient response are modelled. Current is shown in mA.

Teaching notes · 30 minutes

Calculate an ideal divider output and explain how a parallel load changes the effective lower resistance.

Ohm’s law, series circuits, parallel resistance, and the meaning of voltage relative to ground.

Two equal resistors divide 5 V into 2.5 V. Will the output stay at 2.5 V when another 1 kΩ resistor is connected across the lower resistor?

A misconception to listen for

A divider does not maintain a fixed voltage for every connected load. The load becomes part of the circuit.

Student task

Find R₁ and R₂ values that give 1 V without a load. Then explain why adding a load reduces that voltage.

Exit question & expected reasoning

With R₁ = R₂ = Rload = 1 kΩ, why is the output about 1.67 V?

R₂ and the load are in parallel, giving 0.5 kΩ. The output is 5 × 0.5 / (1 + 0.5), approximately 1.67 V.