An inductor `(L = 200 mH)` is connected to an `AC` source of peak emf `210 V` and frequency `50 Hz`. Calculate the peak current. What is the instantaneous voltage of the source when the current is at its peak value?

Explanation

Here, `L = 200 mH = 200 xx 10^(-3) H, E_(0) = 210 V`,
`v = 50 Hz, I_(0) = ?`
`I_(0) = (E_(0))/(X_(L)) = (E_(0))/(omega L) = (E_(0))/(2 pi v L) = (210)/(2 xx 3.14 xx 50 xx 0.2)`
`= 3.3 A`
Through L. current lags behind the voltage by a phase angle of `90^(@)`. Therefore,when current is at its peak value voltage must be zero.