1.

A 100 muF capacitor in series with a 40 omega resister connected to a 110 V, 60 Hz supply What is the maximum current in the circuit ?

Answer»

SOLUTION :`X_c = sqrt(R^2 + (1/(OMEGAC))^2`
`= sqrt ((40)^2 + (1/(2PI xx 60 xx 100 xx 10^(-6)))^20mega` = 48 W


Discussion

No Comment Found

Related InterviewSolutions