how-to · 5 min read

Voltage dividers and sensors: how to use the divider formula in real circuits

A voltage divider is one of the simplest circuits in electronics: two resistors in series across a supply, with the output taken between them. It turns a larger voltage into a smaller one, and it turns a changing resistance into a changing voltage, which is why dividers appear in sensor circuits, reference settings and measurement front ends.

The formula is easy to write and easy to misuse. This guide covers the unloaded divider, a complete worked example, what happens when a real load draws current from the midpoint, how to choose resistor values sensibly, and how resistive sensors use the same idea.

By FormatFlow EditorialUpdated October 21, 2026More Physics Tools →

The divider formula and what it assumes

Two resistors, R₁ from the input to the output node and R₂ from the output node to ground, form a divider. With nothing else connected to the midpoint, the same current flows through both resistors, so the output is the input voltage multiplied by R₂ ÷ (R₁ + R₂): V_out = V_in × R₂ ÷ (R₁ + R₂).

The ratio depends only on the two resistances, so 10 kΩ and 10 kΩ gives half the input, while 10 kΩ and 4.7 kΩ gives a lower fraction because the lower resistor is smaller. The absolute values do not change the ratio, but they do set the standing current and the power dissipated in each resistor.

The formula assumes an unloaded midpoint, ideal resistors and a DC supply. If something draws current from the output node, that current changes the voltage, which is the most common reason a divider behaves differently in a real circuit than on paper.

Worked example: 12 V with 10 kΩ and 4.7 kΩ

Take a 12 V supply, R₁ = 10 kΩ and R₂ = 4.7 kΩ. The ratio is 4,700 ÷ (10,000 + 4,700) = 0.3197, so the output is 12 × 0.3197 = 3.837 V, about 3.83 V. The standing current is 12 ÷ 14,700 = 0.000816 A, or 0.816 mA.

Each resistor dissipates P = I² × R. For R₁ that is (0.000816)² × 10,000 = 0.00666 W, or 6.66 mW. For R₂ it is (0.000816)² × 4,700 = 0.00313 W, or 3.13 mW. Both are far below a common quarter-watt rating, so the divider is electrically comfortable in this example.

The numbers also show the trade-off hidden in the design: the ratio sets the output, while the total resistance sets the current and power. Choosing 10 kΩ and 4.7 kΩ rather than 100 Ω and 47 Ω keeps the standing current small, at the cost of making the output easier for a load to disturb.

Why a load pulls the output down

A load connected across the output appears in parallel with R₂. If a 10 kΩ load is placed across the 4.7 kΩ lower resistor, the effective lower-leg resistance becomes 4,700 × 10,000 ÷ 14,700 = 3,197 Ω. The output then falls to 12 × 3,197 ÷ (10,000 + 3,197) = 2.91 V, well below the unloaded 3.83 V.

The general rule is that the load should draw much less current than the divider's standing current if the output is to stay close to the calculated value. When that is impossible, a buffer amplifier between the divider and the load keeps the midpoint unloaded while still delivering the voltage to the next stage.

A multimeter measuring the output is itself a load, although a good one is usually high enough not to disturb the reading much. If a measurement disagrees with the calculation, check for loading, resistor tolerance and the meter's input impedance before assuming the formula is wrong.

Choosing resistor values for a divider

Start with the ratio you need, then decide how much standing current is acceptable. Low resistances give a stiffer output that shrugs off loading, but they waste more power and may heat up. High resistances save power but make the output sensitive to load, leakage and noise.

Tolerance matters too: a divider made from 5% resistors can differ from the calculated ratio by a noticeable amount, and the error is larger when the two resistors are very different. Use tighter-tolerance parts or calibrate the output when the voltage must be accurate.

Check power ratings after choosing the values. The formula for each resistor is P = I² × R, and it is easy to forget that a low-resistance divider across a high voltage can dissipate a surprising amount of heat. Leave margin for the supply running above its nominal value.

Dividers with resistive sensors

Many sensors change resistance in response to the thing being measured: thermistors with temperature, photoresistors with light, strain gauges with mechanical strain. Put that sensor in place of R₂ and the divider converts the resistance change into a voltage change that an analog-to-digital converter can read.

The response is rarely linear. A thermistor's resistance falls sharply with temperature, so the divider output versus temperature curve is not a straight line and needs calibration or a correction formula. Changing the supply voltage also changes the output, so a stable excitation source matters.

Keep the sensor current within its safe range and consider the following circuit's input impedance, since that impedance loads the midpoint just like any other load. When accuracy matters, buffer the divider output or use a dedicated sensor interface that accounts for the sensor's transfer function.

Frequently asked questions

What is the voltage divider formula?

For two resistors in series, V_out = V_in × R₂ ÷ (R₁ + R₂), where R₂ is the lower resistor and R₁ is the upper one. The formula assumes nothing draws current from the midpoint, so a real load changes the output voltage.

Why does my measured divider output not match the calculation?

The usual causes are load current from the next stage, resistor tolerance, a supply voltage lower than assumed, and the input impedance of the meter itself. Check the load and tolerance first, then measure the actual supply and both resistors rather than relying on their marked values.

How do I choose resistor values for a voltage divider?

The ratio of the two resistors sets the output, while their absolute values set the standing current and power. Lower values are less affected by loading but waste more energy, so choose the largest values that still keep the output stable for your load.

Can a voltage divider power a sensor or a circuit?

It is not a power supply: the output varies with the current drawn from it. A divider is fine for a high-impedance input or a reference point, but a load that needs real current should use a regulator or a buffer amplifier after the divider.

How does a voltage divider work with a thermistor?

The thermistor replaces one leg of the divider, so a temperature-dependent resistance change produces a voltage change at the midpoint. The relationship is nonlinear, and the reading also depends on the excitation voltage and the load on the output, so calibration is normally required.

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