A 0.21 H inductor and a `12 Omega ` resistance are connected in series to a `220 V, 50 Hz ac source.
The current in the circuit is

Options

A. `(220)/(sqrt(4400))A`
B. `(22)/(3sqrt(5)) A`
C. `(220)/(sqrt(4500))A`
D. `(22)/(5sqrt(3)) A`

Answer

B

Explanation

here,
`X_(L)=omega L =2 pi L = 2pi xx 50 xx 0.21 = 21 pi Omega`
so, `Z=sqrt(R^(2)+X_(L)^(2))=sqrt(12^(2)+(21 pi)^(2))`
`=sqrt(12^(2)+(21 xx 21//7)^(2))=sqrt(4500))=30 sqrt(5) Omega`
so, (a) `I=(V)/(Z) =(220)/(3 sqrt(5))A`
and (b) `phi=tan^(_1)((X_L)/(R )) = tan^(-1) ((21 pi)/(12)) = tan^(-1)((7 pi)/(4))`
i.e. the current will lag the applied voltage.