Question

Show that voltage leads current by $\frac{\pi }{2}$, when A.C. voltage applied to pure inductance.

Answer 1

Voltage $V=V_0\sin \omega t$ is applied to pure inductor (zero resistance) coil of inductance L
The current through the inductor varies and opposing induced emf is generated in it as per Lenz Law and
is given by $-L\frac{di}{dt}$.
Kirchoff's loop rule gives us ,
$V_0\sin \omega t-L\frac{di}{dt}=0$
$di=\frac{V_0\sin \omega tdt}{L}$
Integrating both sides we get,
$i=-\frac{V_0\cos \omega t}{\omega L}+C$
C-integration constant .
Since source has an emf which oscillates symmetrically about zero, the current it sustain also oscillates symmetrically about zero, so there is no time independent component of current that exists. Thus constant C=0
$\therefore i=-\frac{V_0\cos \omega t}{\omega L}=i_0\sin (\omega t-\frac{\pi }{2})$
where $i_0=\frac{V_0}{\omega L}$ is the peak value of current .
From instantaneous values of current and voltage we see that in pure inductive circuit the voltage leads current by a phase angle of $\frac{\pi }{2}$