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

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Correct Answer - `15A ; 37^(@)`
Here `(X_L)=2 pi fL= 2 pi L =2 pi xx50xx(9)/(100 pi) =9 Omega`
So, `Z=sqrt(R^(2_+X_(L)^(2)) =sqrt(12^(2)+9^(2))=15 Omega`
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^(@)`.

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