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A 9//(100 pi) H inductor and a 12 Omega ...

A `9//(100 pi) H` inductor and a `12 Omega` resistance are connected in series to a `225 V`, `50 Hz` ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

Text Solution

Verified by Experts

Here `X_(L)=omegaL=2pilL=2pixx50xx9/(100pi)=9Omega`
So,`Z=sqrt(R^(2)+X_(L)^(2))=sqrt(12^(2)+9^(2))=15Omega`
So (a)`I=V/Z=225/15=15A`
and (b)`phi=tan^(-1)(X/L)=tan^(-1)(9/12)`
i.e., the current will lag the applied voltage by `37^(@)`in phase.
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