---
title: "Limiting error in calculated power"
subject: "Instrumentation and Measurement"
chapter: 1
key: "semester_4/instrumentation_and_measurement:power-error-limiting-error"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/power-error-limiting-error
site: BCT Engineering
---

# Limiting error in calculated power

Instrumentation and Measurement · Chapter 1: Fundamentals of Measurement

- Asked in: year not given, 3+2 marks
- Video: [Combination of Quantities With Limiting Errors (SSM College Of Engineering)](https://www.youtube.com/watch?v=y7VDoWbTdg4) — Shah Anayat Ullah
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/power-error-limiting-error
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-1

## Question

(a) The power dissipated in a resistor is calculated from $P = V^2/R$. The voltage is measured as 120 V with a limiting error of $\pm 0.5\%$ and the resistance is 50 $\Omega$ with a guaranteed accuracy of $\pm 1\%$. Calculate the power and its limiting error in watt and in percent.

(b) Differentiate between systematic error and random error, giving one cause of each.

## Short answer

Power P = 120²/50 = 288 W. Limiting error: relative error = 2×0.5% + 1% = 2%, absolute error = ±5.76 W, so P = 288 ± 5.76 W (2%). Systematic error is a constant bias (e.g., zero error); random error varies unpredictably (e.g., noise) and is reduced by averaging.

## Answer

#### Power and its limiting error
Given: $V = 120$ V ($\pm 0.5\%$), $R = 50\ \Omega$ ($\pm 1\%$).

$$
P = \frac{V^2}{R} = \frac{120^2}{50} = 288\ \text{W}
$$

For a quantity $P = V^2 R^{-1}$, limiting (worst-case) relative errors add, each multiplied by its power:

$$
\frac{\delta P}{P} = 2\,\frac{\delta V}{V} + \frac{\delta R}{R} = 2(0.5) + 1 = 2\%
$$

$$
\delta P = 0.02 \times 288 = 5.76\ \text{W}
$$

**Answer: $P = 288 \pm 5.76$ W (limiting error $\pm 2\%$).**

If the errors are independent, the probable error is $\sqrt{(2\times 0.5)^2 + 1^2} = 1.41\%$, about $\pm 4.1$ W.

#### Systematic and random errors

| Point | Systematic error | Random error |
|---|---|---|
| Nature | Constant or varies in a known way, same sign | Varies unpredictably, positive or negative |
| Cause example | Zero error, wrong calibration, loading effect | Electrical noise, friction, vibration, observer variation |
| Effect | Shifts all readings (affects accuracy) | Scatters readings (affects precision) |
| Reduction | Calibration, correction, better method | Averaging many readings, statistics |

## Check yourself

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

1. What is the power P = V^2/R for V = 120 V and R = 50 Ω?
   A. 144 W
   B. 240 W
   C. 288 W
   D. 2.4 W
   Answer: C. P = 120^2/50 = 14400/50 = 288 W. (Key point: Power value)

2. For P = V^2/R, how is the limiting error in P found?
   A. 2(error in V) + (error in R) in percent
   B. (error in V) + (error in R) in percent
   C. (error in V) - (error in R) in percent
   D. 2(error in V) - (error in R) in percent
   Answer: A. Limiting (worst-case) errors add, each multiplied by its power; V is squared so its error counts twice. (Key point: Limiting error rule)

3. With V at ±0.5% and R at ±1%, what is the limiting error of P in watts?
   A. ±2.88 W
   B. ±5.76 W
   C. ±1.44 W
   D. ±4.1 W
   Answer: B. Error = 2(0.5)+1 = 2%, and 2% of 288 W = 5.76 W. (Key point: Limiting error result)

4. Which statement about systematic and random errors is correct?
   A. Systematic errors scatter readings and affect precision
   B. Random errors are reduced by calibration
   C. Systematic errors shift readings and affect accuracy; random errors are reduced by averaging
   D. Both are reduced only by averaging
   Answer: C. Systematic errors (e.g. zero error) shift all readings and are fixed by calibration; random errors scatter readings and are reduced by averaging. (Key point: Systematic vs random)

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