Question

show that in an circuit containing a pure inductor, the voltage leads the current π/2 in phase​

Answer 1

Voltage V=V0sinωt is applied to pure inductor (zero resistance) coil of inductance L

The current through the inductor varies and opposing induced emf is generated in it as per Lenz Law  and

 is given by −Ldtdi.

Kirchoff's loop rule gives us ,

V0sinωt−Ldtdi=0

di=LV0sinωtdt

Integrating both sides we get,

i=−ωLV0cosωt+C

C-integration constant .

Since source has an emf which oscillates symmetrically about zero, the current it sustain also oscillates symmetrically about zero, so there is no time independent component of current that exists. Thus constant C=0

∴i=−ωLV0cosωt=i0sin(ωt−2π)

where i0=ωLV0 is the peak value of current .

From instantaneous values of current and voltage  we see that in pure inductive circuit the voltage leads current by a phase angle of π/2.

Hope this helps you

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