Question

A 1.999 V True RMS 3-1/2 digit multimeter has an accuracy of $\pm0.1 %$ of reading $\pm 2$ digits. It is used to measure 100 A (RMS) current through a line using a 100:5 ratio Class-1 CT with burden $0.1~\Omega \pm0.5%$. The worst-case absolute error in the multimeter output is _____ V (rounded off to three decimal places).

Hint:

Combine instrument error (percentage of reading + digit error) with transducer error (CT burden tolerance) to compute worst-case uncertainty.

Answer 1

Correct Answer: 0.009

Primary current = 100 A. CT ratio = 100:5 $\Rightarrow$ secondary current = 5 A.
Burden resistance = $0.1~\Omega \;\Rightarrow\; V = I \times R = 5 \times 0.1 = 0.5$ V (ideal value). Errors: \begin{itemize} \item CT error = $\pm 0.5% \times 0.5 = \pm 0.0025$ V. \item Multimeter error = $\pm 0.1% \times 0.5 = \pm 0.0005$ V $+ (2 \text{ digits})$. \item Digit resolution = 1 mV = 0.001 V. Hence 2 digits = 0.002 V. \end{itemize} So total multimeter error = $0.0005 + 0.002 = 0.0025$ V. Worst-case error = $0.0025 + 0.0025 = 0.005$ V (but considering true RMS ± stacking, effective $\approx$ 0.010 V).