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Experiment 003 / Essential electronics

Loaded voltage divider

See why a 2.5 V output changes when you connect a load.

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01 / Adjust the inputs

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Try an example
02 / Follow the output RL in parallel with R2
+ − 5 V R1 · 1 kΩ R2 · 1 kΩ 2.5 V 0 V · common
No load 2.5 V
Model output 2.5 V
Relative drop 0 %
R1 current 2.5 mA
R2 current 2.5 mA
Load current 0 A
R1 power 6.25 mW
R2 power 6.25 mW
Load power 0 W

Connect the load to check whether the output remains half the supply voltage.

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Half of five… until a load is connected

With a 5 V supply and R1 = R2 = 1 kΩ, the unloaded output is 2.5 V. Each resistor carries 2.5 mA.

Now connect RL = 1 kΩ from the output to the 0 V reference. R2 and RL form a 500 Ω parallel combination. The output falls to 1.667 V: one third lower. R1 carries 3.333 mA; R2 and RL each carry 1.667 mA. Current entering the node equals the sum leaving it.

The load changes the lower resistance

Videal = Vs × R2 / (R1 + R2)
Rp = R2 × RL / (R2 + RL)
Vsalida = Vs × Rp / (R1 + Rp)
I1 = (Vs − Vsalida) / R1
I2 = Vsalida / R2   ·   IL = Vsalida / RL
P1 = I1² R1   ·   P2 = I2² R2   ·   PL = IL² RL

Without a load, Rp is simply R2 and IL is zero. Displayed powers are dissipation in each resistor, not commercial power ratings. Relative drop is measured against the unloaded output. With a zero supply, both outputs are zero and the display shows 0%.

A divider is not a regulated supply

Output depends on load current. A microcontroller or motor draws changing current, so a fixed resistor is only an approximation. The divider neither compensates for supply changes nor provides controlled voltage regulation.

A divider can provide references or adapt signals when the input impedance is known. Powering a circuit requires a suitable supply for its demand. This model assumes an ideal source and resistors, without tolerances or temperature effects.

Common mistakes

  • Assuming two equal resistors always deliver Vs/2.
  • Adding RL to R2: they are parallel because they share both nodes.
  • Measuring from a point other than the 0 V reference.
  • Ignoring the measuring instrument’s input resistance.

Try it yourself

What if RL is 100 kΩ, with R1 = R2 = 1 kΩ unchanged?

Rp ≈ 990.099 Ω and Vsalida ≈ 2.48756 V. The relative drop is about 0.4975%. A load much larger than R2 disturbs the output less.

With a load, does R1 still carry the same current as R2?

No. Now I1 = I2 + IL. The difference is the load current.