Power Associated with Capacitive AC Circuit Video Lecture | Physics for JEE Main & Advanced

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go ahead and look at power associated
with a capacitor so here also we will
follow the same technique as we did for
inductor as well as resistor so what was
our applied voltage it was B is equal to
VM sine Omega T and what was the current
associated with the capacitor it was I M
sine Omega T plus Phi by 2 therefore we
can see that the instantaneous power
denoted by a small P will be equal to
the product of the instantaneous voltage
and instantaneous current so this will
be equal to VM I M sine Omega T into
sine Omega T plus PI by 2 so this is my
instantaneous power so what would be the
average power over one complete cycle
average power capital T would be equal
to 1 by T integration of T DT from 0 to
time period capital T so this will be
equal to 1 by T integration of VM I M
sine Omega T into sine Omega T plus PI
by 2 DT so this can be written as V M I
M divided by T integration of sine Omega
t + sine Omega T plus Phi by 2 we can
write sine Omega T plus Phi by 2 as cos
Omega T so this would be sine Omega T
cos Omega T DT now we can write this as
integration of sine theta cos theta
would be sine 2 theta divided by 2 that
can sign
to Omega T divided by 2 DT from 0 to
capital T right so we can write this as
equal to V M I M divided by 2 T 2 minus
cos 2 Omega T divided by 2 Omega from 0
to T so this will be equal to VM I M
divided by 4 T Omega into minus cos 2
Omega T plus cos 0 is 1 so this is 1 now
this Omega T can be written as 2 pi
Omega T can be replaced with 2 pi
because we know that Omega is equal to 2
pi by T so this we can write it as VM I
M divided by 4 T Omega into minus cos 4
PI plus 1 now what is the value of cos 4
pi cos 4 pi is equal to 1 so this will
be equal to VM I M divided by 4 T Omega
into minus 1 plus 1 which comes out to
be 0 and now this is the average power
associated with the capacitor so we can
see that the average power associated to
a capacitor over one cycle is 0 that is
average power supplied to a capacitor
over one complete cycle is zero so in
this respect the capacitor behaves in a
similar way to the inductor because in
case of inductor also we saw that the
average power over one complete cycle
was zero right so only in case of
resistor the average power associated
over one cycle was nonzero and was equal
to I square R where I represented the
RMS value of alternating current right
so once we have now studied the power
associated with capacitor it is time
that we
look at the phasor diagram representing
the current voltage relationship for a
capacitive circuit so if you look at
this diagram you can see that first we
have plotted the phasor for voltage as
we do for all the circuits now the
current phaser is ahead of the voltage
phasor by PI by 2 when we were plotting
the inductive circuit that and the
current was behind voltage that's why
the current was plotted like this
because it was behind voltage but now
the current is ahead of voltage
therefore the current is plotted PI by 2
ahead of voltage so I hope now you are
understanding that how do we represent
current voltage relationship with the
help of phasor diagrams just trying to
plot these phasor diagrams yourself so
that you can understand things better so
now that we have reached towards the end
of discussing the basic AC circuits that
is we discussed a resistive circuit we
discussed a capacitive circuit minused
first and inductive circuit and we saw
that how an applied alternative voltage
affect each of them so now it is time
that we discuss a little more
complicated AC circuits where we will
have combination of resistor inductor
and capacitor thank you please visit
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