A `200 Omega` resistor and 1H inductor are joined in series with an ac source of emf `10 sqrt(2) sin (200t)V`. Calculate the phases difference between emf and current.

Explanation

`L=1H, R=200 Omega, E=10 sqrt(2) sin (200t)V` ltbr. `tan (phi)=(X_(C)-X_(L))/(R )=(-omega L)/(R )=-(200xx1)/(200)=-1`
`implies phi=tan^(-1)(-1)=-(pi)/(4)`
` phi` is negative, hence current legs behind voltage by `pi//4`.