How to Build and Measure a Voltage Divider: Why the Output Drops Under Load

Series navigation: Episode 1: The Roadmap · Episode 2: Digital Multimeter and LED Circuit · Episode 3

What you will build: A 4.5 V voltage divider using two 10 kΩ resistors.
What you will observe: About 2.25 V with no load and about 1.50 V after adding a 10 kΩ load.
What you will understand: A connected device becomes part of the circuit; it does not merely receive the calculated voltage.

Reader guide

Indicator	Details
Article type	Guided circuit and measurement tutorial
Reading time	About 11 minutes
Difficulty	2/5 — Beginner
Hands-on time	25–40 minutes
Estimated cost	A few resistors if you already have the tools from the previous guide
Prerequisites	Basic digital multimeter operation and Ohm's law
Safety level	Extra-low-voltage DC only; maximum 4.5 V in this procedure
Reader outcome	Calculate and explain both unloaded and loaded voltage-divider output

The voltage was correct—until I connected something

I first met voltage dividers while trying to connect real components to an Arduino.

The idea looked simple. Two resistors could reduce a voltage to a level that a microcontroller could measure. I entered the values into a calculator, built the circuit on a breadboard, and checked the output with a digital multimeter.

The display showed almost exactly what I expected.

Then I connected another component.

The voltage dropped.

At first, I assumed that I had selected the wrong resistor, misunderstood the breadboard, or damaged a component. I checked the wiring and repeated the calculation. Everything still appeared correct, but the circuit no longer produced the voltage predicted by the familiar formula.

This is a common moment when learning electronics with an Arduino or Raspberry Pi. A voltage divider may show the expected value when measured by itself, but behave differently after it is connected to a sensor, module, LED, transistor, or analog-to-digital converter.

The formula was not wrong. My model of the circuit was incomplete.

The new component was not merely observing the voltage. It had become part of the circuit.

In this guide, we will build a simple voltage divider, predict its output, and measure it with a digital multimeter. Then we will add a known load and watch the expected output change. By the end, we will understand how to calculate loaded voltage and why a voltage divider should not normally be used as a power supply.

Safety boundary

This guide uses a holder containing three ordinary AA cells, approximately 4.5 V DC. Do not substitute mains power, an unprotected lithium cell, a high-current pack, or an unknown adapter.

Disconnect the battery before changing a resistor or jumper.
Measure resistance only with power disconnected.
Keep the meter's red lead in the V/Ω socket for every measurement in this guide.
Do not connect the divider to an Arduino, Raspberry Pi, ADC, or other board until you have checked that board's absolute input-voltage limit.
Never apply 5 V to a Raspberry Pi computer's 3.3 V GPIO input.
Stop if any component becomes warm, smells, swells, smokes, or behaves inconsistently.

The microcontroller examples later in the article explain design consequences. The experiment itself requires only a battery, resistors, breadboard, and multimeter.

Parts and tools

Item	Specification	Quantity	Purpose
Digital multimeter	DC voltage and resistance modes	1	Measures input, output, and resistor values
Battery holder	3 × AA, approximately 4.5 V	1	Extra-low-voltage source
R1	10 kΩ, ¼ W, preferably 1% or 5%	1	Upper divider resistor
R2	10 kΩ, ¼ W, preferably 1% or 5%	1	Lower divider resistor
RL	10 kΩ, ¼ W, preferably 1% or 5%	1	Known removable load
Breadboard	Known internal connection pattern	1	Reversible construction
Jumper wires	Intact	4–6	Connects the nodes

The three resistors may look identical, so measure and label them with power disconnected before building.

The unloaded voltage-divider formula

A voltage divider contains two series resistors. R1 connects from the source to the output node, and R2 connects from the output node to ground.

Vin ── R1 ──┬── Vout
            │
            R2
            │
Ground ─────┴── 0 V

With nothing else connected to Vout, the ideal output is:

Vout = Vin × R2 / (R1 + R2)

For a 4.5 V source and two equal 10 kΩ resistors:

Vout = 4.5 V × 10 kΩ / (10 kΩ + 10 kΩ)
     = 2.25 V

Equal resistors produce half the input voltage. The divider current is:

I = 4.5 V / (10 kΩ + 10 kΩ)
  = 0.225 mA

This calculation contains a hidden assumption: no meaningful current leaves the output node.

What changes when a load is connected

Now connect another 10 kΩ resistor, RL, from Vout to ground. It is in parallel with R2.

Comparison of an unloaded and loaded voltage divider

Two equal 10 kΩ resistors in parallel are equivalent to 5 kΩ:

Rlower = R2 || RL
       = (R2 × RL) / (R2 + RL)
       = (10 kΩ × 10 kΩ) / (10 kΩ + 10 kΩ)
       = 5 kΩ

The original formula still works, but its lower resistance is no longer 10 kΩ:

Vout_loaded = 4.5 V × 5 kΩ / (10 kΩ + 5 kΩ)
            = 1.50 V

The output did not mysteriously fail. We changed the circuit from a 10 kΩ lower leg to a 5 kΩ lower leg.

This is the key beginner insight:

A load connected to a voltage divider changes the divider unless its resistance is much larger than the divider's output resistance.

Breadboard wiring

Breadboard wiring for the voltage divider and removable load

The exact row numbers do not matter. The electrical nodes do.

Node	Connections
`Vin`	Battery positive and top of R1
`Vout`	Bottom of R1, top of R2, optional top of RL, red meter probe
Ground	Battery negative, bottom of R2, optional bottom of RL, black meter probe

R2 and RL are parallel because both ends connect to the same two nodes: Vout and ground.

Build procedure

Disconnect the battery holder.
Measure all three resistors and identify R1, R2, and RL.
Connect R1 from the positive rail to an unused Vout row.
Connect R2 from that same Vout row to the ground rail.
Connect battery negative to ground and battery positive to the positive rail.
Leave RL disconnected for the first reading.
Check that positive and ground are not accidentally shorted.
Connect the battery.

For every voltage reading, keep the black probe on ground. Place the red probe on battery positive to measure Vin, then on the center node to measure Vout.

Measurement 1: no external load

With RL disconnected:

measure Vin between battery positive and ground;
measure Vout between the center node and ground;
calculate the expected output using the actual Vin, R1, and R2 values.

If the source measures 4.5 V and the resistors are close to 10 kΩ, expect approximately 2.25 V.

The digital multimeter is technically a load too. A common meter may have an input resistance around 10 MΩ, although the actual value must come from its manual. Ten megohms in parallel with 10 kΩ changes the lower resistance only slightly:

10 kΩ || 10 MΩ ≈ 9.99 kΩ

That is why the meter appears to “just observe” this particular divider.

Measurement 2: add the 10 kΩ load

Disconnect the battery.
Connect RL from Vout to ground, in parallel with R2.
Verify that RL is 10 kΩ and not a near-zero jumper.
Reconnect the battery.
Measure Vin and Vout again.

Expect Vout to be near 1.50 V rather than 2.25 V.

Circuit state	Effective lower resistance	Ideal output at 4.5 V
No external load	10 kΩ	2.25 V
10 kΩ load attached	5 kΩ	1.50 V

The load and R2 each carry approximately:

Ibranch = 1.50 V / 10 kΩ = 0.15 mA

Those two branch currents add to 0.30 mA through R1. Kirchhoff's current law is not a separate trick; it is the accounting rule that explains where the current goes.

The meter can cause the same problem

The multimeter's loading was negligible with 10 kΩ resistors because its input resistance was roughly one thousand times larger than R2.

Now imagine using two 10 MΩ divider resistors and a meter with a 10 MΩ input resistance. The meter appears in parallel with the lower 10 MΩ resistor:

10 MΩ || 10 MΩ = 5 MΩ

The measured output becomes one third of the input instead of one half:

Vout = Vin × 5 MΩ / (10 MΩ + 5 MΩ)
     = Vin / 3

The act of measuring changed the quantity being measured.

This matters with high-impedance sensors, tiny currents, long wires, and oscilloscope probes. Every instrument has an input characteristic. “High impedance” means the effect is small in a particular circuit, not that the instrument is invisible.

A faster model: Thevenin equivalent

From the load's point of view, the unloaded divider can be replaced by one ideal voltage source and one series resistance.

For the equal 10 kΩ divider:

Vthevenin = 2.25 V
Rthevenin = R1 || R2 = 5 kΩ

When a 10 kΩ load is attached, those values form another divider:

Vout = 2.25 V × 10 kΩ / (5 kΩ + 10 kΩ)
     = 1.50 V

Thevenin's model makes the design rule easy to see: if the load resistance is not much larger than the source's Thevenin resistance, the output will sag.

A common rough target is to make the load resistance at least ten times the divider's Thevenin resistance when only modest error is acceptable. Precision work requires calculating the actual error rather than relying on a rule of thumb.

What this means for Arduino

An Arduino analog input is not a perfect, permanently infinite resistance. Its ADC includes a sample-and-hold capacitor that must charge from the signal source.

For the ATmega328P used by classic boards such as the Arduino Uno R3, the manufacturer states that the ADC is optimized for analog sources with an output impedance of approximately 10 kΩ or less. A higher-impedance divider may appear correct on a slowly responding multimeter but give inaccurate or slow-settling ADC readings.

Practical responses include:

reduce the divider resistor values while checking current and power;
add a capacitor at the ADC input when appropriate for a slowly changing signal;
allow acquisition time before conversion;
buffer the divider with an op-amp designed for the voltage range;
follow the datasheet for the exact microcontroller, not a generic Arduino assumption.

Also confirm the ADC reference voltage and the pin's absolute maximum rating before connecting the divider.

What this means for Raspberry Pi

A Raspberry Pi computer's standard GPIO pins are 3.3 V digital inputs and outputs. They are not general-purpose analog inputs. To measure an analog divider output, use an external ADC such as an appropriate SPI or I²C converter.

The external ADC then becomes the load whose input impedance and sampling behavior must be checked.

Do not confuse Raspberry Pi computers with Raspberry Pi Pico microcontroller boards. Pico-class boards include ADC inputs on specified pins, while the 40-pin header on a Raspberry Pi computer exposes digital GPIO. In both cases, consult the exact board documentation and never exceed the allowed input voltage.

Why a voltage divider is not a power supply

A divider is excellent for scaling a signal when the receiving input draws very little and predictable current. It is usually a poor way to power a sensor, module, motor, radio, or LED strip.

Real loads change current over time. A radio may draw short bursts. A sensor may consume more current while measuring. A microcontroller draws different current while sleeping and executing. Every change alters the divider output.

A proper power source uses active regulation and is designed to maintain voltage across a stated load range. Depending on the application, use:

an LDO regulator;
a buck converter;
a boost converter;
a buck-boost converter;
a buffer amplifier for a signal rather than a power rail;
a logic-level shifter for digital communication.

Choose the circuit based on whether you are scaling a measurement, translating a digital signal, or delivering power. Those are different jobs.

Troubleshooting

Symptom	Likely cause	Safe check	Correction
Vout is almost Vin	R2 is disconnected	Power off and continuity-check the ground path	Reconnect R2 to ground
Vout is near zero	R1 is disconnected or output is shorted	Power off and measure resistance between nodes	Correct the wiring before repowering
Unloaded output is not near half	Resistor values differ or rails are split	Measure Vin, R1, and R2 separately	Use the measured values in the formula
Output does not change with RL	RL is not connected across Vout and ground	Power off and trace both ends of RL	Place RL in parallel with R2
Output falls much lower than 1.5 V	Wrong load value or another hidden load	Disconnect loads and measure each resistance	Identify every path from Vout to ground
Reading changes when touching the circuit	Floating node or high impedance	Check for missing ground and loose contacts	Repair the reference and connections

Completion checklist

[ ] I can identify Vin, Vout, and ground as electrical nodes.
[ ] I can calculate the unloaded divider output.
[ ] I can recognize that a load is parallel with R2.
[ ] I can calculate R2 || RL and the loaded output.
[ ] I know that a multimeter and ADC are also loads.
[ ] I can explain why a divider is suitable for some signals but not for powering a module.
[ ] I know that Raspberry Pi computer GPIO is 3.3 V digital I/O and requires an external ADC for analog measurements.

Next experiment

The divider looked static on a multimeter because the voltage changed slowly. The next step is to watch voltage change over time and discover what a multimeter display cannot show.

That means meeting the oscilloscope.

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