---
title: "Potentiometer loading effect and linearity"
subject: "Instrumentation I"
chapter: 3
key: "semester_4/instrumentation_i:potentiometer-loading-effect"
source: https://www.bctengineer.com/q/semester_4/instrumentation_i/potentiometer-loading-effect
site: BCT Engineering
---

# Potentiometer loading effect and linearity

Instrumentation I · Chapter 3: Transducer

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_i/potentiometer-loading-effect
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_i/chapter-3

## Question

Explain the loading effect on a potentiometer and explain the effect of load resistance on the linearity of the potentiometer. Also show that the error occurs at the midpoint of the potentiometer.

## Short answer

Loading from a meter of resistance Rm makes the output eo/Ei = x/[1 + r·x(1−x)] with r = Rp/Rm, introducing non-linearity. Linearity worsens as Rm decreases; maximum error occurs at x = 2/3, and at midpoint the error is about r/8, e.g. 10% when Rp = Rm.

## Answer

A resistive potentiometer is a displacement transducer. The wiper position x gives an output voltage proportional to the displacement, provided that no current is drawn from the wiper.

#### Loading effect
Let the total resistance be Rp, the wiper be at fraction x = Rx/Rp of its travel, the supply be Ei, and the load (meter) resistance be Rm.
```text
 Ei (+) o-----+
              |
             [ ] (Rp - Rx) = (1 - x)Rp
              |
              +<--- wiper ---+------------o  eo
              |              |
             [ ] Rx = xRp   [Rm]  load (meter)
              |              |
 Ei (-) o-----+--------------+------------o
```
Without load: eo = Ei·x (a straight line).

With load, Rm is in parallel with Rx. The parallel resistance is Rx·Rm/(Rx + Rm), and the total resistance across the supply is:

R = (Rp − Rx) + RxRm/(Rx + Rm)

eo = Ei · [RxRm/(Rx+Rm)] / R

Substituting Rx = xRp and r = Rp/Rm:

**eo/Ei = x / [1 + r·x(1 − x)]**

So the loaded output is smaller than the ideal value Ei·x. The meter draws current, and the potentiometer output becomes non-linear.

#### Effect of load resistance on linearity
- If Rm ≫ Rp (r → 0): eo/Ei → x, and the characteristic is nearly a straight line.
- As Rm decreases (r becomes large), the curve bends further below the ideal line, with the greatest deviation in the upper-middle region.
- Error = ideal − actual = Ei·x − eo:

**ε/Ei = r·x²(1 − x) / [1 + r·x(1 − x)]**

For small r the error ≈ r·x²(1−x). It is zero at x = 0 and x = 1, and the maximum occurs where d/dx[x²(1−x)] = 0, i.e. 2x − 3x² = 0, giving x = 2/3, with error ≈ 0.148 r.

#### Error at the midpoint
At the mid position x = 0.5:

eo/Ei = 0.5 / (1 + r/4)

ε/Ei = r·(0.25)(0.5)/(1 + r/4) = 0.125 r / (1 + 0.25 r) ≈ r/8

So the midpoint reading is lower than 0.5·Ei by about r/8 of the supply; that is, an error is present at the midpoint. Example: Rp = Rm (r = 1) gives eo = 0.4 Ei instead of 0.5 Ei, an error of 0.1 Ei = 10 % of full scale.

To keep this error low, use Rm ≥ 10 Rp or a high-impedance (buffered) meter.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does a load resistance Rm make a potentiometer non-linear?
   A. Rm is in parallel with Rx, so the meter draws current and the output falls below Ei·x
   B. Rm is in series with Rp
   C. Rm increases the supply voltage
   D. Rm makes the wiper stick
   Answer: A. Rm in parallel with the wiper section changes the divider ratio, giving eo/Ei = x/[1 + r x(1 - x)]. (Key point: Loading effect cause)

2. With r = Rp/Rm and x the wiper fraction, the loaded output ratio eo/Ei is:
   A. x(1 + r)
   B. x / [1 + r x (1 - x)]
   C. x [1 + r x (1 - x)]
   D. x / (1 + r)
   Answer: B. Derived from the parallel resistance Rx·Rm/(Rx+Rm) and the total resistance across the supply. (Key point: Loaded output formula)

3. What keeps the loading error small?
   A. Rm much smaller than Rp
   B. Rm equal to zero
   C. A wiper at the end only
   D. Rm much larger than Rp (r close to 0) or a buffered meter
   Answer: D. As r tends to 0 the output tends to Ei·x; use Rm of at least 10 Rp or a high-impedance meter. (Key point: Condition for linearity)

4. For Rp = Rm (r = 1), what is the output at the midpoint x = 0.5?
   A. 0.5 Ei
   B. 0.4 Ei
   C. 0.45 Ei
   D. 0.6 Ei
   Answer: B. eo/Ei = 0.5/(1 + 1/4) = 0.4, so the error is 0.1 Ei instead of zero. (Key point: Midpoint error)

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