Here, inductance, `L= 2 `henry
r.m.s. voltage, ` E_(v) = 150` volt
frequency of A.C. supply, `V = 50 c//s`.
`:.` Inductvie reactance, `X_(L) = omega L`
`= 2 pi v L = 2 (22)/(7) xx 50 xx 2 = (4400)/(7) ohm`
If `E_(0)` is the peak value of the alternating voltage, then maximum value of current `(I_(0))` is given by
`I_(0) = (E_(0))/(X_(L))`
or `I_(0) = (sqrt2 xx E_(V))/(X_(L)) = (sqrt 2 xx 150)/(4400//7) = 0.337 A`
