There is a `5 Omega` resistance in an `AC`, circuit. Impedence of `0.1 H` is connected with it in series. If equation of `AC` emf is `5 sin 50 t` then the phase difference between current and e.m.f. is

Options

A. `(pi)/2`
B. `(pi)/6`
C. `(pi)/4`
D. `0`

Answer

C

Explanation

`cos varphi=R/Z=R/(sqrt(R^(2)+omega^(2)L^(2)))=5/(sqrt(25+(50)^(2)xx(0.1)^(2)))`
`=5/(sqrt(25+25))=1/(sqrt(2)) implies varphi=pi//4`