Obtain an expression for the current through an inductor when an AC emf is applied.

Explanation

Circuit consists of pure capacitor of capacitance C. Let an A.C emf `V = V_(m)sin omega t` is applied to it. Let I and q be the instantaneous values of current and charge.
Potential difference across capacitor `= - (q)/(C )`
Total emf `= V_(m)sin omega t - (q)/(C )`
According to Ohms law this must be equal to iR = 0
`V_(m)sin omega t-(q)/(C )= 0`

`(q)/(C )=V_(m)sin omega t` .....(1)
`q=V_(m)C sin omega t`
Differentiating w.r.to time
`(dq)/(dt)=V_(m)omega C cos omega t[because i=(dQ)/(dt)]` ...(2)
`i=(V_(m))/(1//omega C)sin(omegat+(pi)/(2))(because i_(0)=(V_(m))/(1//omega C))`
`i=i_(0)sin (omega t+(pi)/(2))` ....(3)
`i_(0)` is peak value of current. Here the current leads the applied e.m.f by `(pi)/(2)` radian (or) `90^(@)`.