Question

In the case of an inductor
A. voltage lags the current by ${\displaystyle \frac{\pi }{2}}$.
B. voltage leads the current by ${\displaystyle \frac{\pi }{2}}$.
C. voltage leads the current by ${\displaystyle \frac{\pi }{3}}$.
D. voltage leads the current by ${\displaystyle \frac{\pi }{4}}$.

Answer 1

Hint: An inductor is a passive two terminal electrical device which stores energy in the magnetic field when electric current flows through it. The voltage applied across an inductor is equal to the self-induced emf across the inductor itself.

Complete step by step answer:
Inductor is a device which is meant to store energy when electric current flows through it.
According to the known principle that,
Voltage across an inductor$=$ Self-induced emf across inductor
Let the voltage be $V$, self-induced emf be $L$ and change in current with time be ${\displaystyle \frac{dI}{dt}}$.Now,
$$\begin{gathered} V=L{\displaystyle \frac{dI}{dt}} \\ \Rightarrow {\displaystyle \frac{dI}{dt}}={\displaystyle \frac{V}{L}} \end{gathered}$$
In an inductor $V=V_m\sin \omega t$,
${\displaystyle \frac{dI}{dt}}={\displaystyle \frac{V_m\sin \omega t}{L}}$

Integrating the equation we get,
$\int dI={\displaystyle \frac{V_m}{L}}\int \sin \omega tdt$
$\Rightarrow I=-{\displaystyle \frac{V_m}{\omega L}}\cos \omega t+c------\left(1\right)$
$c$ is the constant of integration and it is $0$.
As the $\cos$ component is in negative value so it is either in II or III quadrant.
Converting $\cos$ in terms of $\sin$ component by trigonometric property of $\cos x=\sin \left[\left(2n+1\right){\displaystyle \frac{\pi }{2}}\pm x\right]$ in equation $\left(1\right)$ and we get,
$I={\displaystyle \frac{V_m}{\omega L}}\sin \left(\omega t-{\displaystyle \frac{\pi }{2}}\right)-----\left(2\right)$ , $c=0$.

We must also keep an eye on the sign by using the quadrant concept. The given $\cos$ value here is negative and hence $\sin \left(\omega t-{\displaystyle \frac{\pi }{2}}\right)$ falls into IV quadrant which is also negative. Hence, there is no alternation in magnitude or sign.In terms of reactance we get,
$I_m={\displaystyle \frac{V_m}{\omega L}}={\displaystyle \frac{V_m}{X_L}}$
where $X_L$ is the resistance offered by the inductor in the flow of A.C. and hence is known as inductive reactance. Thus, it is clear from equation $\left(2\right)$ that current through an inductor lags behind the voltage by ${\displaystyle \frac{\pi }{2}}$.

So, the correct option is A.

Note: It must be noted here that the constant of integration here is $0$ as the logic behind it is that current that flows in the circuit actually oscillates continuously and has no constant term. An ideal inductor has always zero resistance.