Voltage Divider Calculator

Two resistors turn one voltage into a smaller one. Set the input voltage and the two resistors R1 and R2, and the calculator gives the tapped output from Vout = Vin · R2/(R1+R2) — the output is the same fraction of the input as R2 is of the total. Add a load and watch it pull the output down as R2 and the load combine in parallel, with the loaded and unloaded values shown side by side, plus the current the divider draws and the power it burns. A live schematic tracks it all, on your device.

You are in the Physics lab.

The output is the input scaled by R2/(R1+R2) — a load across R2 drags it lower.
Output voltage Vout
Output ratio
Source current
Total power
The output is a fraction, set by the ratio. Because the same current runs through both resistors, the input voltage splits between them in proportion to their sizes — so the tap across R2 gives Vout = Vin · R2/(R1+R2). It depends only on the ratio of the resistors, not their absolute size, which is why a 1 kΩ / 1 kΩ pair and a 1 MΩ / 1 MΩ pair both give half. The absolute values decide something else: how much current the divider wastes, and how hard a load can drag the output down.

Reading the calculator

What the tap voltage, the load droop and the fraction bar tell you.

1

The tap and the fraction bar

R1 sits on top, R2 below, and the output is tapped at the junction between them. The bar on the right shows Vout as a percentage of Vin — raise R2 relative to R1 and the bar climbs toward 100%, lower it and the bar drops. It is a direct picture of the divider ratio R2/(R1+R2).

2

Add a load and watch it droop

Connect a load and a third resistor appears across R2. Because it sits in parallel with R2, it lowers the effective bottom resistance and pulls the output down. The readout shows both the loaded and the unloaded value so you can see exactly how much the load costs you — a load much larger than R2 barely matters; one comparable to R2 halves the effect.

3

Current and wasted power

The divider draws a source current of Vin/(R1+R2) even with nothing connected, and burns it as heat — that is the total power readout. Big resistor values sip almost nothing but are easily disturbed by a load; small values hold steady but waste current. The art of a divider is picking values that are stiff enough for the load yet frugal enough to ignore.

How It Works

Series resistors, a shared current, and the loading rule.

1

Vout = Vin · R2/(R1+R2)

R1 and R2 in series carry one current, I = Vin/(R1+R2). The output across R2 is that current times R2, which simplifies to Vin·R2/(R1+R2). The output is therefore the same fraction of the input as R2 is of the total resistance — pure proportion, straight from Ohm's law.

2

Only the ratio matters (until you load it)

Scale both resistors by the same factor and the ratio — and the output — do not change. What does change is the current: doubling both halves the current and the wasted power, but also makes the output twice as sensitive to a load. That trade-off is the whole design decision behind choosing divider values.

3

A load parallels R2

Any load connected across the output sits in parallel with R2, combining to R2' = R2·RL/(R2+RL), which is smaller than either — so the output falls to Vin·R2'/(R1+R2'). Keep the load resistance well above R2 (ten times or more) and the droop is negligible; let it approach R2 and the divider sags badly. Our Ohm's law calculator and circuit simulator explore the same resistors further.

What is the voltage divider formula?
Vout = Vin · R2/(R1+R2). Two series resistors carry the same current, so the voltage splits in proportion to their resistances — the tap across R2 is the same fraction of Vin as R2 is of the total. Equal resistors give half; a larger R2 raises the output toward Vin, a smaller one lowers it toward zero.
How does a load affect a voltage divider?
A load across the output sits in parallel with R2, lowering the effective bottom resistance to R2' = R2·RL/(R2+RL) and pulling the output down. The effect is small when the load is much larger than R2, large when it is comparable. The tool shows the loaded and unloaded output side by side.
How do I get a specific output voltage?
Set the ratio: R2/(R1+R2) = Vout/Vin. For 3.3 V from 5 V that ratio is 0.66, so R2 ≈ twice R1 (such as R1 = 1.7 kΩ, R2 = 3.3 kΩ). Then pick values large enough not to waste current but small enough that the load doesn't disturb them. Adjust R1 and R2 here until the output reads what you need.
Can a voltage divider power a circuit?
Only a very light one. A divider is a reference, not a supply — any current the load draws shifts the output, and stiffening it with small resistors wastes power. For anything drawing real current use a proper regulator, and keep the divider for setting a reference voltage or scaling a signal.

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