Voltage Divider
Calculates Vin, R1, R2 or Vout of a resistive divider from three known quantities.
V
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V
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Ω
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Ω
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Circuit diagram
Why the formula works
R1 and R2 in series form a single path for current — the same current I flows through both (Ohm's Law, see Ohm's Law & Power Calculator: I = Vin/(R1+R2)). Vout is the voltage measured only across R2, so applying V = I×R again with that same current: Vout = I×R2 = Vin×R2/(R1+R2) — the divider formula comes straight from Ohm's Law applied twice, not a new rule to memorize.
Voltage dividers under load
This calculator's math assumes nothing besides the multimeter/ADC is "pulling" current from the tap point (Vout) — a high-impedance input, exactly the case of a microcontroller's ADC pin. If something with low resistance is connected there (a motor, an LED with no resistor of its own, another low-impedance circuit), that load forms a third current path in parallel with R2, lowering the effective resistance the tap sees — the real Vout sags below the value calculated here. The lower R1/R2 are relative to the load's resistance, the smaller this effect; that's why a voltage divider should never be used as a power source for anything that actually draws current.
Frequently asked questions
Vin, R1, R2 and Vout are tied together by a single equation (Vout = Vin×R2/(R1+R2)) — one equation with four unknowns becomes fully determined as soon as three of them become known values. All four possible combinations (leaving out Vin, R1, R2, or Vout) solve the missing quantity uniquely, always isolating it directly in the equation.
Because something connected to the tap point is drawing current — see "Voltage dividers under load", above. This calculator's formula assumes an infinite-impedance load (nothing "pulling" current from the tap); in practice, every real instrument/circuit has some finite impedance, so there's always some sag, larger the closer the real load's resistance gets to R1/R2.
No — an LED or a motor draws real current (tens of mA to several A), very different from the negligible current an ADC pin/multimeter draws. Wired straight to the tap, that draw pulls Vout far below the calculated value (sometimes close to zero), and constantly wastes energy in both resistors besides. To power a current-drawing load, use a voltage regulator (linear or switching) — a component designed to hold the output voltage steady regardless of load, which a resistive divider never does.
10 kΩ/20 kΩ is the combination most cited in Arduino/ESP32 tutorials for this exact problem (stepping 5 V down for the ADC), using round, easy-to-find commercial values — the result (~3.33 V) sits within the ESP32 ADC's safe margin (3.3 V max, with some tolerance). Finding the exact pair for precisely 3.300000 V would require a non-standard resistance, with no real practical benefit.
A classic technique for analog circuits running off a single supply (no negative voltage available): using two equal-value resistors (R1 = R2) to create a reference point at Vin/2, serving as a "virtual zero" for AC signals that swing around it — common in audio preamps and reading AC sensors with an ADC that only accepts positive voltage.
Yes — type the number followed by an International System prefix letter (m = milli, k = kilo, M = mega, among others) and the value is interpreted automatically in the correct unit, same behavior as the Ohm's Law & Power Calculator. Calculated fields already display the result formatted the same way.
Nexinon Principles
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