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?

Options

A. `3.5` A, `(1/240)` s
B. `3.5` A, `(1/360)` s
C. `35` A, `(1/240)` s
D. `4.0` A, `(1/240)` s

Answer

A

Explanation

The rms current is
`i_(rms) = i_0/sqrt(2) = 3.5 A`.
The time period is
`T = 1/v = 1/60 s`.
The current takes one fourth of the time period to reach the peak value starting from zero. Thus, the time required is
`t = T/4 = 1/240 s`.