An inductor has a resistance `R` inductance `L`. It is connected to an `A.C` source of e.m.f. `E_(V)` and angular frequency `omega` then the current `I_(v)` in the circuit is

Options

A. `(E_(V))/(omega L)`
B. `(E_(V))/(R )`
C. `(E_(V))/(sqrt(R^(2) + omega^(2) L^(2)))`
D. `sqrt(((E_(V))/(R ))^(2) + ((E_(V))/(omega L))^(2))`

Answer

3

Explanation

`i= (E_(0))/(sqrt(R^(2) + X_(L)^(2))), X_(L) = L omega`