Question

Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.​

Answer 1

The following figure shows an AC source, generating a voltage

$\sf{e = e_0 sin wt,}$

Connected to a key K and a pure inductor of inductance L to form a closed circuit

AN AC SOURCE CONNECTED TO AN INDUCTOR

On closing the key K, an emf is induced in the in doctor as the magnetic flux linked with it changes with time This emf opposes that applied emf and according to the laws of electromagnetic induction by Faraday and Lenz we have,

$\sf{e' = -L\frac{di}{dt}------------(1)}$

Where e' is the induced emf and i is the current through the inductor To maintain the current, e and e' must be equal in magnitude and opposite in direction

According to Kirchhoff's Voltage law, as the resistance to the inductor is assumed to be zero, we have,

$\sf{-e' = -L\frac{di}{dt}----------(2)}$

$\therefore\sf{\frac{di}{dt}= \frac{e}{L}= \frac{e_0 sin\omega t}{L}}$

$\therefore\sf{\int di= \int \frac{e_0 sin\omega t }{L} dt}$

$\therefore\sf{i= -\frac{e_0}{\omega L}cos\omega t + C}$

Where C is constant of integration.

C must be time independent and have the dimension of current As e oscillates about zero and hence there cannot be any time independent emf component of current

$\therefore\sf{C=0}$

$\therefore\sf{i = -\frac{e_0}{\omega L}cos\omega t }$

$\sf{= -\frac{e_0}{\omega L} sin (\frac{π}{2}-\omega t})$

$\therefore\sf{i= \frac{e_0}{\omega L} sin (\omega t -\frac{π}{2} eq(3)}$

$as sin (-\theta )= - sin\theta$

From EQ(3),

$\sf{i_peak = i_0 =\frac{e_0}{\omega L}}$

$\therefore\sf{i = i_0 sin(\omega t -\frac{π}{2}) EQ(4)}$

Comparison of this equation with

$e = e_0 sin wt$

shows that e leads i by π/2 rad, ie, the voltage is ahead of current by π/2 rad in phase

HOPE IT HELPS

Answer 2

AC circuit containing Inductor only (Inductive circuit) : Consider a circuit having no ohmic resistance and only inductance L as shown in the figure. When an AC current is passed through the circuit a magnetic flux is set a which induces alternating e.m.f in the inductance which is L$\bigg(\dfrac{di}{dt}\bigg)$and opposes the variation of current through it every instànt. If the applied voltage is $\sf v_{max}\sin \omega t$, we have,

• $L\dfrac{di}{dt} = \sf v_{max}\sin \omega t$

• $\sf Ldi = v_{max}\sin\omega tdt$

• $\displaystyle \int\limits_{0}^{I_{max}}{L \, di}$ = $\displaystyle \int\limits_{0}^{t}{V_{max} \sin \omega t\, dt}$

By integrating the above relation we have,

• $\sf i = I_{max} \sin( \omega t - \dfrac{\pi}{2}$) where $\sf I_{max} = \dfrac{V_{max}}{\omega L}$

From the above relation it is clear that there is a phase difference of $\dfrac{\pi}{2}$ between voltage and current, that is, current lags voltage by $\dfrac{\pi}{2}$ radians.

It can also be done by this method.

• $\displaystyle \int\limits_{0}^{I}{dI} = \displaystyle \int\limits_{0}^{ \pi/2}{E_o/L \sin \omega t dt}$

• $\displaystyle I = \dfrac{E_o}{L} \int \sin t dt$

• $I = \dfrac{E_o}{L} \bigg(\dfrac{- cos \omega t dt}{ \omega} \bigg)$

• $I = -\dfrac{E_o}{ \omega L} \sin \bigg(\dfrac{ \pi}{2}- \omega t \bigg)$

• $I =\dfrac{E_o}{\omega L} \sin \bigg( \omega t-\dfrac{ \pi}{2} \bigg)$

• $I = -\dfrac{E_o}{X_L} \sin \bigg(\dfrac{ \pi}{2}- \omega t \bigg)$

The current will be maximum,

• I = $\sf I_o$, when $\sin\bigg(\dfrac{\pi}{2} - \omega t \bigg)$ = maximum = 1

• l = $\sf I_o$ $\sin\bigg(\dfrac{\pi}{2}- \omega t \bigg)$