Question

A 100V voltmeter has an accuracy of \( \pm \)2% of full-scale deflection (FSD). If it reads 50V, what is the limiting error?

• A. \( \pm \)2%
• B. \( \pm \)1%
• C. \( \pm \)4%
• D. \( \pm \)5%

Hint:

Limiting error is inversely proportional to the scale reading:
\[ \% E_L = \frac{V_{\text{FSD}}}{V_{\text{reading}}} \times (\text{\% Full Scale Error}) \] For this question: \( \frac{100}{50} \times \pm 2\% = \pm 4\% \).
This quick formula saves valuable time in competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to calculate the limiting error (relative error) when a voltmeter with a full-scale deflection (FSD) of 100 V and an accuracy of \( \pm 2\% \) of FSD is used to measure a voltage of 50 V.
The accuracy of an instrument is usually specified at its full scale, but the error increases in percentage terms when measuring smaller values on the same scale.

Step 2: Key Formula or Approach:
The absolute error (\( \delta V \)) of the instrument is constant over the entire scale and is computed as:
\[ \delta V = \pm (\text{Accuracy \%}) \times V_{\text{FSD}} \] The relative limiting error (\( E_L \)) at any measured reading (\( V \)) is given by:
\[ E_L = \pm \frac{\delta V}{V} \times 100\% \]

Step 3: Detailed Explanation:
Let us perform the calculations step-by-step:
Given data:
Full-scale deflection, \( V_{\text{FSD}} = 100\text{ V} \)
Guaranteed accuracy = \( \pm 2\% \) of FSD
Measured voltage, \( V = 50\text{ V} \)

•

Step 1: Compute the absolute constant error (\( \delta V \)):
\[ \delta V = \pm 2\% \text{ of } 100\text{ V} = \pm \frac{2}{100} \times 100 = \pm 2\text{ V} \] This means any reading taken on this scale has an inherent uncertainty of \( \pm 2\text{ V} \).

•

Step 2: Compute the relative limiting error at the 50 V reading:
Using the limiting error formula:
\[ E_L = \pm \frac{\delta V}{V} \times 100\% = \pm \frac{2\text{ V}}{50\text{ V}} \times 100\% \] \[ E_L = \pm \frac{1}{25} \times 100\% = \pm 4\% \] This calculation demonstrates that as the measured value drops to half of the full-scale value, the relative limiting error doubles from \( \pm 2\% \) to \( \pm 4\% \).

Step 4: Final Answer:
The limiting error of the voltmeter when reading 50 V is \( \pm 4\% \).