Answer Posted / vignesh r

Under no load condition it is 20% of rated current of the
device.And in full load condition it is 120% of rated
current.For eg., consider the rated current of a motor is
19ampere.then under no load it is 20% of 19 =3.8ampere
=5amp(approx).for full load it is 120%of 19 = 22.8ampere=20
ampere(approx).

An impedance of 100 ohms resistance and an unknown inductance is connected across the capacitor. The resulting impedance is a pure resistance of 500 ohms ifω = 10^5 rad/sec. calculate the values of inductor and capacitor.

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