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.

Explanation

`"Here" X_(L)=omegaL=2pif L=2pixx50xx(9)/(100pi)=9Omega`
`"So", Z=sqrt(R^(2)+X_(L)^(2))=sqrt(12^(@)+9^(2))=15Omega`
`"So" (a)I=(V)/(Z)=(225)/(15)=15A`
`and (b) phi=tan^(-1)((X_(L))/(R))=tan^(-1)((9)/(12))=tan^(-1)3//4=37^(@)`
i.e. the current will lag the applied voltage by `37^(@)` in phase.