Two resistors in series turn one voltage into a smaller one at the node between them — the cheapest voltage converter in electronics, and the input stage of half the sensor circuits ever built. This voltage divider calculator computes the output voltage from Vin, R1 and R2, together with the current the divider draws and the power it wastes as heat, because those last two numbers decide whether the elegant little circuit survives contact with a real load.
The Divider Equation
Series resistors share one current, so the voltage splits in proportion to resistance: Vout = Vin × R2 ÷ (R1 + R2). The page’s default — 12 V across 1 kΩ over 2 kΩ — gives 12 × 2000/3000 = 8.00 V at the middle node, with 4 mA flowing and 48 mW warming the pair. Intuition to keep: Vout is the fraction of Vin that R2 represents of the total; equal resistors halve the voltage; making R2 nine times R1 passes 90 percent; swapping the pair inverts the fraction. The formula knows nothing about absolute values — 1 Ω and 2 Ω divide identically to 1 MΩ and 2 MΩ — which is precisely where the design judgement enters, because current, power and loading all scale with the absolute choice.
Choosing the Absolute Values
The ratio sets the voltage; the magnitude sets everything else. Too small (ohms) and the divider drinks current and cooks — fine for a power-rail sense line, ruinous for a battery node. Too large (megohms) and the node becomes a high-impedance antenna: leakage, bias currents and the input impedance of whatever reads it all bend the ratio. The working ranges of practice: kilohms for general signal work, tens of kilohms for battery-powered sensing, and the rule that the divider’s own current should be at least ten times any current the output node will be asked to supply. The calculator prints current and power beside the voltage so the trade is made with eyes open, not discovered warm.
Loading: The Divider’s Original Sin
The formula describes an unloaded divider. Connect a load R_L to the output and it parallels R2, shrinking the effective lower resistance to R2‖R_L and dragging Vout down — a 2 kΩ R2 feeding a 2 kΩ load loses a quarter of its intended voltage, not a negligible few percent. This is why dividers feed high-impedance inputs (op-amp buffers, ADC pins, MOSFET gates) happily and power loads never: a divider is a signal device. The stiff-divider rule (load ≥ 10 × R2) keeps the error near or under 10 percent of the ratio step; precision work buffers the node with a follower and lets the op-amp’s near-infinite input impedance make the formula true.
Where Dividers Work for a Living
Level-shifting a 5 V sensor into a 3.3 V microcontroller pin (the 1k/2k of this page, give or take). Scaling a battery’s 12–14 V down into an ADC’s 0–3.3 V window for voltage monitoring — the divider ratio is the calibration constant of every DIY power meter. Setting op-amp gain and reference voltages. Reading resistive sensors — thermistors, photoresistors, force-sensitive resistors — where the sensor is one leg of the divider and the output voltage becomes the measurement. In that last role the design question inverts: choose the fixed resistor near the sensor’s mid-range resistance to maximise sensitivity where the readings will live.
Potentiometers: The Adjustable Divider
A potentiometer is this circuit with R1 and R2 trading share along one resistive track — Vout sweeps from 0 to Vin as the wiper travels, which is why volume knobs and trim pots are dividers wearing mechanical clothes. The same loading law applies (a pot feeding a low-impedance input bows its taper), and the same ratio logic explains the audio taper: human hearing is logarithmic, so audio pots are wound to make the ratio progress logarithmically with rotation. When a schematic shows a pot with its wiper tied to one end, you are looking at a variable rheostat — one leg of a divider whose other leg lives elsewhere in the circuit.
Common Mistakes
Taking Vout across the wrong resistor (the formula’s fraction belongs to R2, the ground-side leg; across R1 the output is Vin × R1/(R1+R2) measured from the top). Ignoring loading, then debugging a “wrong resistor” that was right unloaded. Using a divider as a power supply for anything hungrier than microamps — efficiency caps at the ratio itself and regulation is nonexistent. And precision theatre: 0.1 percent resistors feeding an ADC whose reference wanders by 2 percent — the divider is rarely the weakest link it is blamed for being.
Related Tools
The electronics shelf continues in the Physics Calculators hub — the Gear Ratio Calculator is this circuit’s mechanical twin (a ratio trading one quantity for another), and the Tension Calculator covers the statics bench.
Frequently Asked Questions
How do you calculate voltage divider output?
Vout = Vin × R2/(R1 + R2). With Vin = 12 V, R1 = 1 kΩ and R2 = 2 kΩ, Vout = 12 × 2/3 = 8.00 V. The fraction is set by R2’s share of the total resistance.
What resistors divide 12 V down to 5 V?
Any pair with R2/(R1+R2) = 5/12 ≈ 0.4167 — for example R1 = 7 kΩ and R2 = 5 kΩ, or R1 = 1.4 kΩ and R2 = 1 kΩ. Then check the current (12 V across 12 kΩ is 1 mA) suits your power and loading needs.
Why does my divider read low when connected?
Loading: your load sits in parallel with R2, reducing the effective lower resistance and the output with it. Keep the load at least ten times R2, or buffer the node with an op-amp follower.
Can a voltage divider power a device?
Essentially no — it is a signal circuit. Its output sags under load, its efficiency is capped by its ratio, and it has no regulation. Use a regulator or converter for power; use a divider to measure or signal.
What current does a divider draw?
I = Vin/(R1 + R2), flowing constantly whether the output is used or not. The defaults here draw 4 mA (48 mW) — trivial on a bench supply, a battery budget item in a sensor node.
How does a potentiometer relate to a divider?
It is one: the track is R1 + R2 and the wiper is the moving output node, sweeping Vout from 0 to Vin. Volume controls, trimmers and joystick sensors are all potentiometers dividing a supply.