Answer to Question #123820 in Mechanical Engineering for sreeju

Question

A 220V,10A dc energy meter is tested at its marked ratings. The resistance of
pressure coil is 9900ohm and that of current coil is 0.5ohm .Calculate the power consumed
when testing the meter with
(i).Direct loading arrangements
(ii).Phantom loading with current circuit excited by a 6V battery

Answer 1

Given,

220v, 10A dc energy meter

Pressure coil resistance 9900 ohm

Current coil rating 0.5 ohm

i) Direct loading arrangements

by pressure coil in parallel

P_{pressure coil} = v²/R = 220²/9940 = 11 watts

Power consumed by current coil (in series)

220*10 = 2200 w

Total power (direct loading)

2200 + 11 = 2211 w

ii) Phantom loading

In this current coil is excited separately and pressure coil also Thus, power consumed by pressure coil = v²/R = 220²/9940 = 11 w

But power consumed by current coil = v*i = 6*10= 60w

Total = 60 + 11= 71w

Therefore, phantom loading consume less power than direct loading