1.

A coil of inductance 0.50 H and resistance 100omegais connected to a 240 V, 50 Hz a.c. supply What is maximum current in the coil ?

Answer»

SOLUTION :Impedance Z = `sqrt(R^2 + omega^2 L^2)`
`=sqrt((1000)^2 - 1 (2PI XX 50 xx 0.5)^2 omega)`= 186.21 `omega`


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