Question

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

Answer 1

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 $\dfrac{{dI}}{{dt}}$.Now,
$V = L\dfrac{{dI}}{{dt}} \\\ \Rightarrow \dfrac{{dI}}{{dt}} = \dfrac{V}{L}$
In an inductor $V = {V_m}\sin \omega t$,
$\dfrac{{dI}}{{dt}} = \dfrac{{{V_m}\sin \omega t}}{L}$

Integrating the equation we get,
$\int {dI = \dfrac{{{V_m}}}{L}} \int {\sin \omega tdt}$
$\Rightarrow I = - \dfrac{{{V_m}}}{{\omega L}}\cos \omega t + c - - - - - - \left( 1 \right)$
$c$ 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 $\cos x = \sin \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2} \pm x} \right]$ in equation $\left( 1 \right)$ and we get,
$I = \dfrac{{{V_m}}}{{\omega L}}\sin \left( {\omega t - \dfrac{\pi }{2}} \right) - - - - - \left( 2 \right)$ , $c = 0$.

We must also keep an eye on the sign by using the quadrant concept. The given $\cos$ value here is negative and hence $\sin \left( {\omega t - \dfrac{\pi }{2}} \right)$ falls into IV quadrant which is also negative. Hence, there is no alternation in magnitude or sign.In terms of reactance we get,
${I_m} = \dfrac{{{V_m}}}{{\omega L}} = \dfrac{{{V_m}}}{{{X_L}}}$
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 $\dfrac{\pi }{2}$.

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.