Voltage Divider Calculator
Calculate the output voltage of a two-resistor voltage divider, solve for the resistor or supply voltage needed to hit a target output, see how a real load resistor pulls that output down, or explore how a potentiometer's output changes as the wiper sweeps and how loading distorts it.
Background
A voltage divider splits an input voltage between two resistors in series: Vout = Vin × R2/(R1+R2). It's one of the most common building blocks in electronics — used for logic-level shifting, biasing, and sensor reference voltages. The catch is that this formula assumes nothing else is connected to the output. The moment a real load draws current, the divider behaves a little differently, and potentiometers (which are really just adjustable dividers) show this effect especially clearly.
How to use this calculator
- Choose Basic Divider to solve the plain, unloaded divider equation for Vout, Vin, R1, or R2.
- Choose Loaded Divider to see how a real load resistor connected at the output actually pulls Vout down from the ideal value.
- Choose Potentiometer Sweep to see the output voltage across the full range of wiper travel, with or without a load, and how loading bends the once-linear response.
- Click Calculate to see the diagram, the full step-by-step math, and a callout explaining what the result actually means.
How voltage dividers work
Two resistors in series split an input voltage in proportion to their resistances: Vout = Vin × R2/(R1+R2). The resistor closer to the output (R2) gets the larger share when it's the larger resistor.
This formula assumes nothing else draws current from the output node — an "unloaded" or "open-circuit" divider. It's an idealization that real circuits only approximate.
Every divider has its own output (Thevenin) resistance, R_th = R1×R2/(R1+R2) — this is the resistance the outside world "feels" looking into the Vout node. It's the key number for predicting loading effects.
When a real load RL is connected, the loaded output can be written elegantly as Vout_loaded = Vout_unloaded × RL/(RL+R_th) — the load and the divider's own output resistance form a second divider.
A common rule of thumb: keep RL at least about 10× the divider's R_th to hold loading error under roughly 10%. Dividers built from large resistors (hundreds of kΩ or more) are much more sensitive to loading than the same ratio built from small resistors.
A potentiometer is just a divider where R1 and R2 are two parts of one resistive track, split by a sliding wiper. With no load, Vout scales perfectly linearly with wiper position.
Add a load at the wiper and the response bows away from that straight line — the effect is zero at both ends of travel (0% and 100%) and largest near the middle, since that's where the pot's own output resistance is highest.
A divider isn't free: current I = Vin/(R1+R2) flows continuously through it, whether or not anything uses Vout, dissipating power P = I×Vin as heat. In battery-powered designs, that constant drain is a real design cost worth checking.
Formulas & Equations Used
Basic divider: Vout = Vin × R2 / (R1 + R2)
Solve for Vin: Vin = Vout × (R1+R2) / R2
Solve for R1: R1 = R2 × (Vin−Vout) / Vout
Solve for R2: R2 = Vout × R1 / (Vin−Vout)
Current drawn & power dissipated: I = Vin / (R1+R2), P_total = I × Vin = I²×R1 + I²×R2
Divider output resistance: R_th = R1×R2 / (R1+R2)
Loaded output: R2∥RL = R2×RL/(R2+RL), Vout_loaded = Vin×(R2∥RL)/(R1+R2∥RL) = Vout_unloaded × RL/(RL+R_th)
Potentiometer: R_bottom = x×Rpot, R_top = (1−x)×Rpot, Vout_ideal = Vin×x
Loaded potentiometer: R_bottom∥RL = (R_bottom×RL)/(R_bottom+RL), Vout_loaded = Vin×(R_bottom∥RL)/(R_top+R_bottom∥RL)
Example Problems & Step-by-Step Solutions
Example 1 — Basic divider, solve for Vout
A 5V rail feeds a divider with R1 = 10 kΩ, R2 = 20 kΩ.
Step: Vout = 5 × 20,000/(10,000+20,000) = 5 × 0.6667.
Result: Vout ≈ 3.33V — close enough to 3.3V for many logic-level shifting circuits. This divider also draws I = 5/30,000 ≈ 0.167 mA continuously, dissipating about 0.83 mW.
Example 2 — Solve for R2 at an exact midpoint
You want exactly 2.5V from a 5V rail, using R1 = 10 kΩ.
Step: R2 = 2.5 × 10,000/(5−2.5) = 25,000/2.5.
Result: R2 = 10 kΩ — equal resistors always give exactly half the input voltage.
Example 3 — Solve for R1 to regulate a battery
A 9V battery needs to look like 5V through R2 = 10 kΩ.
Step: R1 = 10,000 × (9−5)/5 = 10,000 × 0.8.
Result: R1 = 8 kΩ.
Example 4 — A load that crushes the output
Vin = 5V, R1 = R2 = 1 kΩ, loaded by RL = 1 kΩ.
Step: R_th = 1,000×1,000/2,000 = 500Ω. Vout_loaded = 2.5 × 1,000/(1,000+500).
Result: Vout drops from 2.5V unloaded to 1.667V loaded — a 33% error, because the load is only 2× R_th, far below the "10×" comfort margin.
Example 5 — A divider that barely notices its load
Same ratio as Example 4, but scaled up: R1 = R2 = 10 kΩ, loaded by a 100 kΩ multimeter.
Step: R_th = 5 kΩ. Vout_loaded = 2.5 × 100,000/(100,000+5,000).
Result: Vout only drops to 2.381V — a 4.8% error, because 100 kΩ is 20× R_th, safely above the rule-of-thumb margin.
Example 6 — A volume knob pulled down at the middle
A 10 kΩ potentiometer at Vin = 5V, wiper at 50%, feeding a 10 kΩ amplifier input.
Step: R_bottom = R_top = 5 kΩ. R_bottom∥RL = 5,000×10,000/15,000 = 3,333Ω. Vout_loaded = 5 × 3,333/(5,000+3,333).
Result: Instead of the expected 2.5V at the halfway mark, the load pulls it down to 2.0V — the pot "feels" quieter than the dial suggests.
Example 7 — The same pot with nothing attached
A 10 kΩ pot at Vin = 9V, wiper at 50%, with no load connected.
Step: Vout_ideal = 9 × 0.5.
Result: Vout = 4.5V — exactly half of Vin, exactly what the dial position promises, since there's no load to distort it.
Example 8 — Why R1 = 0 can't give a lower output
You try to solve for R2 with Vin = 5V, Vout = 2V, but R1 = 0.
Step: With R1 = 0, Vout = Vin×R2/(0+R2) = Vin for any R2 > 0 — the output is pinned to Vin no matter what R2 is chosen.
Result: There is no valid R2 that produces 2V here; the calculator flags this instead of returning the misleading "R2 = 0Ω."
Frequently Asked Questions
What does the "no-load" assumption actually mean?
It means the basic formula assumes no current flows out of the Vout node besides the current already flowing through R1 and R2. Any real device connected to Vout draws some current, which is exactly what the Loaded Divider mode accounts for.
What is the divider's "output resistance" and why does it matter?
R_th = R1×R2/(R1+R2) is the resistance a load effectively sees looking back into the Vout node. The load and R_th form a second divider, so comparing your load resistance to R_th tells you instantly whether loading will be a problem.
Why can't Vout exceed Vin in a basic divider?
Vout = Vin × R2/(R1+R2), and R2/(R1+R2) can never exceed 1 for positive resistors, so Vout is always a fraction of Vin — never more.
Why are large-resistance dividers more sensitive to loading?
A 1 MΩ/1 MΩ divider has R_th = 500 kΩ, so even a "high impedance" 1 MΩ load is only 2× R_th and causes serious error. The same 1:1 ratio built from 100Ω resistors has R_th = 50Ω, so almost any realistic load is enormous by comparison.
What's the "10× rule" for choosing a load?
Keeping RL at least 10 times R_th keeps the loading error under about 10%, since error ≈ R_th/(RL+R_th). It's a quick sanity check, not a hard limit — smaller ratios just mean progressively larger errors.
Why does a potentiometer's loading error peak in the middle?
The pot's own output resistance, R_top∥R_bottom-style, is largest when the wiper is near the middle (splitting the track roughly evenly) and shrinks to zero at both ends of travel, so that's where a fixed load distorts the output the most.
Can R1 or R2 be zero?
R2 = 0 is valid and just means the tap is grounded directly (Vout = 0). R1 = 0 is also valid and means Vout always equals Vin — but that also means R1 = 0 can never be part of a solution where Vout is supposed to be lower than Vin, which the calculator now checks for directly.
Is a potentiometer really just two resistors?
Electrically, yes — the resistive track above and below the wiper act exactly like R1 and R2 in a fixed divider, except their values change continuously and in lockstep as the wiper moves, always adding up to the pot's total resistance.
Why does a divider draw current even with nothing connected to Vout?
Current always flows from Vin, through R1 and R2, to ground, purely to establish the voltage split — that's the mechanism the divider relies on. This constant draw (I = Vin/(R1+R2)) is a real, ongoing power cost, which matters most in battery-powered or energy-constrained designs.
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