Find the maximum value of current when an inductance of 2H is connected to 150 V, 50 cycle supply.

Explanation

Here `L = 2H , E_(rms) = 150 V, v = 50 Hz`
`X_(L) = L omega = L xx 2pi v`
`2 xx 2 xx 3.14 xx 50 = 628` ohm
RMS value of current through the inductor,
`I_(rms) = (E_(rms))/(X_(L)) = (150)/(628) = 0.24`
Maximum value (or peak value) of current is given by
`I_(rms) = (I_(0))/(sqrt(2))` or `I_(0) = sqrt(2)I_(rms)`
`= 1.414 xx 0.24 = 0.339 A`