Question

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

Answer 1

Question :

Show that in an AC circuit containing a pure

inductor, the voltage is ahead of current by π/2 in

phase.

Solution :

The following figure shows an AC source,

generating a voltage -

$\sf{ \purple { e = e_0 \: \sin(wt) }}$

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,

$\sf{ \blue { e^{/} = - L \dfrac{di}{dt} }}$

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

So we have,

$\sf{ \orange { e^{/} = - L \dfrac{di}{dt} }} \\ \\ \sf{ \purple { \therefore \dfrac{di}{dt} = \dfrac{e}{L} = \dfrac{ e_0 \: \sin(wt) }{L} }} \\ \\ \sf{ \bold { Transposing \: dt \: and \: integrating \: - }} \\ \\ \sf{ \red { \therefore \int di = \int \dfrac{ e_0 \: \sin(wt) }{L} dt }} \\ \\ \sf{ \blue { \therefore i = \dfrac{ - e_0 }{wl} \cos{wt} + c }}$

Here C is constant of integration.

C must be time independent and have the same dimension of current.

As e oscillates about zero and hence there cannot be any time independent emf component of current.

So, C = 0 .

$\sf{ \green { \therefore i = \dfrac{ - e_0 }{wl} \cos{wt} }} \\ \\ \sf{ \red { \therefore i = \dfrac{ e_0}{wl} \sin(wt - \dfrac{ \pi }{2}) }}$

$\sf{ \bold { At \: the \: peak \: voltage \: - }} \\ \\ \sf{ \orange { \therefore i = \dfrac{ i_0}{wl} \sin(wt - \dfrac{ \pi }{2}) }}$

$\sf{ \bold { Comparing \: , we \: get \: - }} \\ \\ \sf{ \purple { e = e_0 \sin(wt) }}$

This shows that e leads i by r/2 rad, i.e., the voltage is

ahead of current by n i/2 rad in phase.

Note : Here , w has been used in place of $\omega$

Image of the circuit is attached herewith !

Answer 2

Explanation:

AC voltage applied to an inductor

Source, v=v

m

sinωt

Using Kirchhoff's loop rule, ∑ε(t)=0

v−L

dt

di

=0

dt

di

=

L

v

=

L

v

m

sinωt

Integrating

dt

di

with respect to time,

∫

dt

di

dt=

L

v

m

∫sin(ω)dt

i=−

ω

v

m

Lcos(ωt)+constant

−cosωt=sin(ωt−

2

π

)

∴i=i

m

sin(ωt−

2

π

)

∵ Where, i

m

=

ωL

v

m

is the amplitude of current.

hope you get answers..