Question
Explain AC circuit with only an inductor.

Answer

→ As shown in fig., an AC source is connected to pure inductor. (A pure inductor means an inductor having negligibly small resistance.)
→ Let the voltage across the source be
$
v=v_m \sin \omega t
...(1)$
→ Using the kirchhoff's loop rule for the AC circuit shown in the fig.,
$
v-L \frac{d i}{d t}=0
...(2)$
→ From equation (2),
$
\begin{array}{l}
\therefore \quad v=L \frac{d i}{d t} \\
\therefore \quad v_m \sin \omega t=L \frac{d i}{d t} \\
\therefore \quad \frac{d i}{d t}=\frac{v_m}{L} \sin \omega t
...(3)\end{array}
$
→ To obtain the current, we integrate equation (3)
$
\begin{array}{l}
\therefore \quad \int \frac{d i}{d t} d t=\int \frac{v_m}{L} \sin (\omega t) d t \\
\therefore \quad i=-\frac{v_m}{\omega L} \cos (\omega t)+\text { constant }
\end{array}
$
→ The integration constant has the dimension of current and is time-independent. Since the source has an emf which oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero.
$
\therefore \quad i=-\frac{v_m}{\omega L} \cos \omega t
$
$
\therefore \quad i=\frac{v_m}{\omega L} \sin \left(\omega t-\frac{\pi}{2}\right)
$
$\therefore \quad i=i_m \sin \left(\omega t-\frac{\pi}{2}\right)...(4)$
$i_m=\frac{ U _m}{\omega L}$ Amplitude of the current
→ The quantity $\omega L$ is analogous to the resistance and is called inductive reactance, denoted by $X _{ L }$ :
$
\therefore \quad X_L=\omega L
$
Unit of $X _{ L }$ is ohm ( $\Omega$ ).
→ From eq. (1) and (4), it can be said that current lags the voltage by $\frac{\pi}{2} rad$. [or onequarter $\frac{1}{4}$ cycle].

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