Question:

Published on: 23 June, 2022

**A single phase kWhr meter makes 500 revolutions per kWhr. It is found on testing that it is making 40 revolutions in 58.1 seconds at 5kW load. Find out the percentage of error.**

Answer:

Energy consumed in 58.1 seconds=\(\frac{5\ast58.1}{3600}\) kWh=0.08069kWh

Energy consumption registered=\(\frac{Number\ of\ revolution\ made}{meter\ constant\ in\ revolution/kWh}\)

\( =\frac{40}{500}=0.08kWh\)

Percentage error =\(\frac{Actual\ registration-true\ energy\ consumption}{True\ energy\ consumption}\ast100\)

\(=\frac{0.08-0.08069}{0.08069}*100=-0.8555% \)

Subjects

Trending

**How can potentiometer be used for**

**Calibration of****ammeter****Calibration of****voltmeter**

View : 58

25 September, 2022

Random questions