1.

What is the value of inductance L for which the current is maximum in a series LCR circuit with C=10 mu F and omega =1000 s^(-1) ?

Answer»

1 mH
cannot be calculated unless R is known
10 mH
100 mH

Solution :In SERIES LCR, CURRENT is maximum at resonance.
`therefore` Resonant FREQUENCY `omega = (1)/(sqrt(LC))`
`therefore omega^(2)=(1)/(LC)` or, `L=(1)/(omega^(2)C)`
Given `omega = 1000 s^(-1)` and `C=10 MU F`
`therefore L=(1)/(1000xx1000xx10xx10^(-6))=0.1 H=100 mH`.


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