AC Voltage Applied To An Inductor - Practice Questions & MCQ

AC voltage applied to an inductor

Voltage applied in the circuit is  V=V₀sinωt is applied to pure inductor  coil of inductance L. As the current through the inductor varies and opposing induced emf is generated in it and is given by  $-L \frac{d i}{d t}$

From Kirchhoff's loop rule: 

$
V_0 \sin \omega t-L \frac{d i}{d t}=0
$

or

$
d i=\frac{V_0}{L} \sin \omega t d t
$

Integrating both sides we get,

$
i=-\frac{V_0}{\omega L} \cos \omega t+C
$

Where C is the constant of integration. This integration constant has dimensions of current and is independent of time. 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 $\mathrm{C}=0$.

So we have,

$
\begin{aligned}
i & =\frac{-V_0}{\omega L} \cos \omega t \\
& =\frac{V_0}{\omega L} \sin \left(\omega t-\frac{\pi}{2}\right) \\
i & =i_0 \sin \left(\omega t-\frac{\pi}{2}\right)
\end{aligned}
$

Where $i_0=\frac{V_0}{\omega L}$ is called peak value of the current.

From instantaneous values of current and voltage we see that in pure inductive circuit the current lags behind emf by a phase angle of π/2.

This phase relationship is graphically shown below in the figure-

Since peak value of current in the coil is $\qquad$
Comparing it with the ohm's law we find product $\omega \mathrm{L}$ has dimension of resistance and it can be represented by

$
X_L=\omega L
$

where $X_L$ is known as reactance of the coil which represents the effective opposition of the coil to the flow of alternating current.

Phase difference (between voltage and current):

$
\phi=+\frac{\pi}{2}
$

Power factor:

$
\cos (\phi)=0
$

Time difference:

$
\mathrm{T} . \mathrm{D}=\frac{T}{4}
$