Power associated with inductor - Alternating Current Video Lecture - Class 12

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going ahead with pack 10 now let us talk
about power associated with inductor in
the similar way as we talked about in
case of the resistor so let us start
with my voltage associated with an
inductor which is V is equal to VM sine
Omega T the current in an inductor is I
M sine Omega t minus PI by 2 so let me
now calculate the instantaneous power so
I will follow the same approach as I did
in case of a resistive circuit so
instantaneous power which we denote with
a small P will be equal to the
instantaneous voltage multiplied by the
instantaneous current so that would be V
M I M sine Omega T into sine Omega t
minus PI by 2 right so this we can write
it as VM I M sine Omega t 2 sine theta
minus PI by 2 so this we can write it as
sine theta minus PI by 2 M is equal to
minus cos theta so let me write it like
that this would be minus cos Omega T
right now we know that 2 sine a cos B is
equal to 2 sine a cos a is equal to sine
2a or you can write it as theta 2 sine
theta cos theta is equal to sine 2 theta
this is all another trigonometric
identity so using that we can say that
sine theta cos theta can be written as
sine 2 theta so we can write it as V M I
I am sine 2 Omega T divided by 2 right
because sine theta cos theta would be
equal to sine 2 theta divided by 2 so
this is my instantaneous power but I
want to calculate the average power over
one complete cycle so how do I calculate
average power so let me try to do that
average power denoted by capital P so
since right now I am talking about an
inductive circuit let me use a subscript
L so average power would be equal to
integration of instantaneous power over
the entire time interval of one time
period divided by the time period so
this would be equal to 1 by T
integration of V M I M sine 2 Omega T DT
divided by 2 so this comes out to be V M
I M divided by 2 T integration of sine 2
Omega T DT from 0 to T so this can be
written as V M I M divided by 2 T into
cos 2 Omega T divided by 2 Omega from 0
to T so I am just doing simple
integrations so this would be V M I M
divided by 2 T into cos 2 Omega T
divided by 2 Omega minus cos 0 so this
can again be written as V M I M divided
by 2 T now what is this Omega into T we
know that Omega is equal
two pi by T so we can write Omega T as
two pi so this will be equal to cos four
pi divided by 2 Omega minus 1 now what
is the value of cos 4 pi the value of
cos 4 pi comes out to be 1 so we get I
am sorry I miss to take this to Omega
outside because it has nothing to do
with the upper and lower limits so this
value comes out to be 1 so this comes
out to be 0 therefore the value comes
out to be 0 so we get that the average
power associated with an inductor is
equal to 0 over one complete cycle so
this is very much unlike to what
happened in case of a resistive circuit
in case of a resistive circuit the
average power over one complete cycle
was non zero but here it is zero so we
see that the average power supplied to
an inductor over one complete cycle is
zero right so now let us look at the
phasor diagram for current voltage
relationship of an inductive circuit so
this was the graphical representation
right which was clear to us now let us
represent the voltage looking at the
graph corresponding to this graph I have
drawn a Frizzle V making an angle Omega
T 1 because I am considering this
instant of time right and the vertical
component represents the instantaneous
value that is VM sine Omega T 1 and the
length of the phasor represents the peak
value at this instant of time now what
about the current phaser how do I
represent the current Fiza the current
phasor here is lesser than the voltage
phasor
as far as peak value is concerned
therefore its length is shortened and we
also know that it lags behind the
voltage by PI by 2 that means instead of
drawing it overlapping on voltage I do
it PI by 2 that is 90 degrees before
voltage that means now voltage will keep
moving current will keep following
voltage right and the difference
between voltage and current at any
instant would be PI by two that I mean
in phasor diagram how do the phasers
move they just keep rotating like this
so if voltage will keep rotating like
this and current will keep following the
voltage because current lags behind
therefore we have drawn current PI by
two behind
voltage right so now you can see that
how is the phase diagram advantageous in
phase diagram you can very easily plot
the voltage and current relationship
because you just need to know the phase
difference okay you know that okay
current lags behind voltage so just draw
a vector for voltage and ninety degree
before that draw another vector for
current okay so things are becoming
simpler when we are considering
different types of AC circuits you will
see later that when we consider even
more complicated error AC circuits the
phasor diagram will be helping you or
not in determining phase relationships
between voltage between different
voltages and different currents right so
I hope that this phasor diagram is clear
to you okay so please I will repeat it
once again the phase led the way I have
drawn phasor diagram first you draw a
voltage in the graph and plot a
corresponding phasor on the phasor
diagram and then you look at the phase
relationship between voltage and current
if they are in phase then voltage and
current will overlap if they add out of
phase then the current will be behind
voltage or ahead of voltage depending
upon the phase relationship right so
since here the current was lacking
behind voltage so the current was drawn
after voltage I mean it was drawn behind
voltage and the phase difference was
five by two therefore the angle between
the voltage phasor and the current
phaser is taken as pi by two so here
also this the vertical component of
current represents the instantaneous
value of current that is I am sine Omega
t minus PI by 2 and the length of the
current phaser represents the peak value
of current at this instant of time thank
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