AC through Pure Inductance – AC Circuits – Basic Electrical Engineering – First Year Engineering

hello friends in this video we are going to see if a/c supplies pass through pure inductance how it will behave so pure inductor practically not possible but still in

order to understand the behavior of inductor under the influence of AC supply we are considering a pure inductor so

this is circuit AC supply is given to a pure inductor the AC supply we have considered as a VM sine omega-t having a phase angle 0 in turns it will become a reference quantity now this voltage will give rise to a flux phi and this change in flux

Phi with respect to time linked with the inductor and we are having Faraday's law of electromagnetic induction which tells EMF induced is minus D Phi by DT and Phi is given by L into I so ultimately I will get equation EMF induced which is a self EMF equal

to E equal to minus L di by DT as per the Lenz's law if it will oppose the cause effect is self induced EMF E and cos is voltage V so I can say voltage applied is minus e so that is minus bracket minus L di by DT

equal to L di by DT what I want what is the current I so for that I will consider di on one side and remaining term VM by L sine Omega T DT on other side I want i+ integrate so after integral I will get integral V M

by L sine Omega T DT V M by L is a constant I take it out so in bracket if I integrate sine Omega T I will get minus cos Omega T upon Omega that equal to VM by Omega L what I've done over here minus cos Omega

T I consider as sine Omega t minus PI by 2 because we know minus cos theta can be represented as sine theta minus PI by 2 why I have done this because after comparing with a standard equation I of T equal to I am sine Omega T plus

Phi I will get I am as VM by Omega L which I can say VM by Excel Excel is nothing but inductive reactance and Phi I will get as minus PI by 2 Radian or minus 90 degree so by considering voltage as a reference having phase angles is

zero if I pass this voltage to a pure inductor I will get a current passing through it having a phase angle minus 90 degree so the conclusion is in purely inductive circuit the current lags the voltage applied by PI by 2 radians or 90 degree so the phase

angle of current flowing through inductor is having a phase angle of 90 degree and that is minus 90 degree means it is lagging so see the waveform and the phasor diagram this is the reference here taken that is real and current is delayed this reference by angle PI

by 2 Radian or 90 degree corresponding phasor diagram have shown like this voltage is a reference and current lags if I take an anti-clockwise direction current lags voltage by 90 degree now one quantity we have introduced that is XL so XL is nothing but Omega L and that

is given by 2 pi FL so what is the XL XL is nothing but an inductive reactance is defined as the opposition of what by inductance of circuit to the flow of alternating sinusoidal current so it is just like a resistance but it is a opposition offered by

inductor circuit to an alternating quantity and so obviously as I said earlier it is like a resistance its measurement can be done in ohms now lets come to the power so instantaneous power for this circuit is instantaneous voltage multiplied by instantaneous current so instance voltage is VM sin

Omega T and instantaneous current I am getting I am sine Omega t minus Phi by 2 so if I solve I will get minus VM I am into sine Omega T into cos Omega T because sine Omega t minus Phi by 2 is nothing but cos Omega T

so if I simply substitute it minus cos Omega T if I am multiplied by 2 and divide by 2 this term 2 sine Omega T into cos Omega T is sine 2 Omega T so ultimately instantaneous power is minus VM I M by 2 sine 2 Omega T

so it is a perfectly sinusoidal quantity I am getting so I represented like this now average power of this circuit will be P average equal to 0 to 2 pi vmim by 2 sine 2 Omega TD Omega T now what I have done simply I have taken average

of this waveform over a complete cycle so you know whatever positive values are present for AC waveform same negative values are also present so that sums up to a 0 so average of this power over a complete cycle is 0 so what conclusion we can draw is whenever

you are giving a AC supply to a pure inductor inductor never consumes a power so power taken by inductor is always zero but remember this concept you are learning with respect to pure inductor thank you

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