---
title: "Potentiometer transducer and loading error"
subject: "Instrumentation and Measurement"
chapter: 5
key: "semester_4/instrumentation_and_measurement:potentiometer-transducer-loading-error"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/potentiometer-transducer-loading-error
site: BCT Engineering
---

# Potentiometer transducer and loading error

Instrumentation and Measurement · Chapter 5: Sensors

- Asked in: year not given, 4+4 marks
- Video: [Transducer | Lecture 9 | Loading Error in Potentiometer Transducer |](https://www.youtube.com/watch?v=9C9-IEYYEw8) — Harishankar Swarnkar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/potentiometer-transducer-loading-error
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-5

## Question

(a) Explain the working of a sliding-contact (potentiometer) displacement transducer. State its advantages and limitations.

(b) A 10 k$\Omega$ linear potentiometer with a 10 V supply drives a 50 k$\Omega$ load. Find the output voltage and loading error at 50% and 75% of the travel, and state how the loading error can be reduced.

## Short answer

A potentiometer transducer gives output voltage e_o = E_s·x/L proportional to wiper position. With R_p/R_m = 0.2, at 50% travel e_o = 4.76 V (error 0.238 V); at 75% travel e_o = 7.23 V (error 0.271 V). Loading error arises from finite load resistance; reduce by making R_m ≫ R_p or using a buffer amplifier.

## Answer

#### (a) Potentiometer transducer
A resistance potentiometer (pot) has a resistive element (wire-wound, carbon or conductive plastic) with a sliding contact (wiper) mechanically linked to the moving member. A supply $E_s$ is applied across the full length $L$, and the output is taken between the wiper and one end. With no load, the output is linear:

$$
e_o = E_s\frac{x}{L}
$$

```text
           Es (supply)
   o-------+
           |
          [|]  resistive element, resistance Rp, length L
          [|]
          [|]<--- wiper, moved by displacement x ---o  e_o
          [|]
           |
   o-------+----------------------------------------o  common
```

**Advantages:** simple, cheap, high output, no amplifier needed, works on dc, available in linear or rotary form.

**Limitations:**
- Wear and friction of the sliding contact; limited life.
- Finite resolution for wire-wound types (a step of one wire turn).
- Loading error with a finite load resistance.
- Limited frequency response because of wiper inertia; noise at the contact.

#### (b) Loading error
With a load $R_m$ across the wiper output, and $x$ = fractional position:

$$
\frac{e_o}{E_s} = \frac{x}{1 + x(1-x)\dfrac{R_p}{R_m}}
$$

$R_p/R_m = 10/50 = 0.2$, $E_s = 10$ V.

| x | Ideal (V) | Actual $e_o$ (V) | Error (V) | Error % of $E_s$ |
|---|---|---|---|---|
| 0.50 | 5.000 | $\dfrac{5}{1 + 0.25(0.2)} = 4.762$ | 0.238 | 2.38 |
| 0.75 | 7.500 | $\dfrac{7.5}{1 + 0.1875(0.2)} = 7.229$ | 0.271 | 2.71 |

**Answer: $e_o = 4.76$ V at 50% (error 0.238 V) and $7.23$ V at 75% (error 0.271 V).**

The loading error is largest at about 2/3 of the travel. To reduce it:
- Make $R_m \gg R_p$ (use a high-input-impedance voltmeter or a buffer amplifier).
- Use a smaller $R_p$ (limited by power dissipation).
- Use a tapped (non-linear) pot to compensate.

## Check yourself

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

1. The unloaded output of a pot is
   A. Es x/L
   B. Es L/x
   C. Es (x/L)^2
   D. Es / x
   Answer: A. Output is linear with wiper displacement. (Key point: Unloaded output)

2. What does a finite load resistance Rm do to the pot output?
   A. Raises it above ideal
   B. Lowers it below ideal, giving a loading error
   C. Makes it exactly linear
   D. No effect
   Answer: B. The load shunts part of the pot, so actual output is less than ideal. (Key point: Loading effect)

3. With Rp/Rm = 0.2 and Es = 10 V, output at 50% travel is
   A. 5.00 V
   B. 4.76 V
   C. 4.50 V
   D. 5.24 V
   Answer: B. eo = 5/(1 + 0.25×0.2) = 4.762 V. (Key point: Output at 50%)

4. How can loading error be reduced?
   A. Make Rm much smaller than Rp
   B. Make Rm much larger than Rp, or use a buffer amplifier
   C. Increase Rp
   D. Lower the supply voltage only
   Answer: B. A high-impedance load or buffer and a smaller Rp reduce the loading error. (Key point: Reducing error)

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## Related in other subjects

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