Question

A 50-A, 230-V energy meter is a full-load test makes 61 revolutions in 37 seconds if the normal speed of the disc is 520 revolutions/kWh, compute the percentage error

Answer 1

Unity power-factor is assumed.

Energy consumed, in kWh, in 37 seconds

= (50 x 230)/1000 x 37/3600 = 0.1182kWh

Number of revolutions corresponding to this energy

= 520 × 0.1182 = 61.464

The meter makes 61 revolutions

∴ % Error = (61 - 61.464)/61.464 x 100% = - 0.755%