Let, \(V=V_0 \sin \omega t\) --------- (1) is the applied alternating voltage.
And \(V^1=-L.\frac{dI}{dt}\)
Applying KVL for the circuit, we get
\(V+V^1=0\,\,\,\,(\therefore R=0,IR=0)\)
\(V=L.\frac{dI}{dt}\)
\(V_0\sin\omega t=L.\frac{dI}{dt}\)
\(dI=\frac{V_0}{L}\sin \omega t.dt\)
Integrating we get,
\(I=\frac{V_0}{\omega L}.\sin(\omega t-\pi/2)\) \(-\cos \omega t=\sin(\omega t-\pi/2)\)
Current is maximum when \(\sin(\omega t-\pi/2)=\pm1\),
that current is called as 'peak value of current and it is denoted as \(I_0\).
\(I_0=V_0/\omega L\) When \(\sin(\omega t- \pi/2)=\pm1\)
\(I=I_0\sin(\omega t- \pi/2)\) --------(2)
This is the expression for current in inductance circuit.
