Potentiometer & Rheostat Calculator

Wiper output voltage of a pot (with taper and load), plus resistance and current for a rheostat.
Potentiometer (Divider)
Rheostat (Series)

Potentiometer as a Voltage Divider

Vout = Vin × Rlower/(Rupper+Rlower)  •  with load: Rlower → Rlower ∥ RL
5V, 10k, 50%, no load
50% with 10k load
Log taper, 50%
V
kΩ
%
kΩ
Enter values and press Calculate.

Rheostat (Variable Series Resistor)

R = Rtotal × position  •  I = V / (R + Rload)  •  Prheo = I²R
12V, 1kΩ, 50%, 10Ω load
5V, 10kΩ, 25%
V
kΩ
%
Enter values and press Calculate.

Potentiometer vs Rheostat

A potentiometer uses all three terminals as an adjustable voltage divider — the wiper taps off a fraction of the input voltage. A rheostat uses just two terminals as a variable resistor in series with a load to control current. The taper (how resistance varies with rotation) is linear for most control and sensing, or logarithmic for audio volume, which matches how we hear loudness.

QuantityFormula
Ideal wiper voltageVout = Vin × fraction
Loaded wiper voltageVout = Vin × (Rlower∥RL)/(Rupper+Rlower∥RL)
Rheostat resistanceR = Rtotal × fraction
Rheostat currentI = V / (R + Rload)

Watch the loading effect: if the load resistance isn't much larger than the pot, it pulls the wiper voltage down and makes the response non-linear. Keep RL >> Rtotal for a faithful divider.

Real-World Applications & Examples

Worked examples

1. Half rotation, no load. 5 V, 10 kΩ linear at 50%: Vout=5×0.5=2.5 V.
2. Loading effect. Add a 10 kΩ load at 50%: the lower half (5 kΩ) in parallel with 10 kΩ is 3.33 kΩ, so Vout=5×3.33/(5+3.33)=2.0 V — the load pulled it down.
3. Log taper. A log pot at 50% delivers only ~24% of the input (~1.2 V), matching how our ears perceive volume.
4. Rheostat current. 12 V, 1 kΩ at 50% (=500 Ω) with a 10 Ω load: I=12/(500+10)=23.5 mA.
5. Full brightness. Turning the rheostat to 0% leaves just the 10 Ω load, so I=12/10=1.2 A — maximum current.
6. Power in the wiper. A rheostat carrying current dissipates I²R — make sure the pot\'s power rating covers it, especially near the low-resistance end.

Frequently Asked Questions

What is the difference between a potentiometer and a rheostat?

A potentiometer uses all three terminals as a voltage divider to tap off a fraction of a voltage; a rheostat uses two terminals as a variable resistor to control current in series with a load.

What is taper?

How the resistance changes as you turn the pot. Linear taper changes evenly; logarithmic (audio) taper changes slowly then quickly, matching how we perceive sound level.

When should I use a log (audio) taper?

For volume and other controls perceived logarithmically. A linear pot would make most of the useful volume change happen in a small part of the rotation.

What is the loading effect?

When the load resistance is not much bigger than the pot, it draws current from the wiper, lowering the output voltage and bending the once-linear response into a curve.

How do I avoid the loading effect?

Make the load resistance at least ten times the pot's total resistance, or buffer the wiper with a high-impedance op-amp follower.

What value potentiometer should I use?

A value about ten times smaller than the load it drives, so loading is negligible, but large enough to keep current and power low. 10 kΩ is a common general-purpose choice.

Can a potentiometer be used as a rheostat?

Yes — connect the wiper and one end terminal (leave the third unconnected) and it acts as a two-terminal variable resistor.

Does the wiper position map linearly to voltage?

For a linear pot with no load, yes. With a log taper or a significant load, the wiper voltage is a non-linear function of position.

How much power can a potentiometer handle?

Small pots are rated a fraction of a watt. As a rheostat carrying current, check I²R against the rating — the low-resistance end can get hot.

What is a trimmer potentiometer?

A small, screwdriver-adjusted pot for one-time calibration or trimming inside a circuit, not for frequent user adjustment.

Why does my volume control feel uneven?

It probably uses a linear pot where a log taper is needed, or the load is changing its response. A log-taper pot fixes the perceived unevenness.

What happens at the extreme wiper positions?

At one end the output is 0 V (or full resistance), at the other it is the full input (or zero resistance). Real pots may have a small end resistance rather than exactly zero.

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