Hello friends in this video we are going
to see whenever AC supply is given to a
pure capacitor how the nature of current
will be look like and also we will see
what is the power consumed by a
capacitor so say up down over here a
diagram where AC voltage is given to
pure capacitor C so whenever voltage is
supplied the capacitor will charge so
let's take the applied voltage is VM
sine Omega T so charge develop through
this plate is Q and given by C into V so
the charge develop is directly
proportional to the voltage applied and
proportionality constant is a
capacitance so Q equal to cv v we have
given VM sine Omega T so I am getting Q
equal to C into VM sine Omega T here I
have assumed voltage as a reference
phasor having a phase angle zero now we
know current is rate of flow of charge
so I equal to DQ by DT so that will be
derivative of Q with respect to time and
Q is given as C VM sine Omega T so
ultimately I will get I as derivative of
C VM sine Omega T C VM is a constant I
take it outside the derivative so I am
getting cv m derivative of sine Omega T
which is nothing but cos Omega T into
Omega so I will write this I in terms of
VM divided by 1 by Omega C into sine
Omega T plus Pi by 2 so what I have
done over here cos Omega T I have
written in terms of sine as sine Omega T
plus PI by 2 because we know cos theta
can be represented as sine theta plus PI
by 2 what is the reason behind doing
this so that I can compare this equation
of I with a standard equation I of T
equal to I am sine Omega T plus Phi now
if I compare I will get maximum value of
current I am as VM divided by 1 over
Omega C and Phi as PI by 2 radians or 90
degree now 1 over Omega C I written as
XE that I will explain in next slide so
ultimately what is the conclusion that I
can draw over here if I consider V as a
reference signal starting from 0 I will
get I as I am sine Omega T plus 90
degree so I can say in purely capacitive
circuit the current leads the voltage
applied by PI by 2 Radian or 90 degree
that is the most important conclusion
that we have drawn so suppose I want to
draw waveforms it will be like this
as I mentioned voltage is a reference
phasor or a reference signal starting
from a zero because we have considered
phase angle of voltage is  zero current
flowing through the capacitor will already
start before zero degree and it will
reach maximum at zero so I can say it is
started at minus PI by 2 so I can say
current leads voltage by angle PI by 2
Radian or 90 degree if I have draw a
phasor diagram it will be like this
voltage is the reference along the
positive x-axis direction current leads
voltage by 90 degree so in
anti-clockwise sense it will be upward
along positive J axis now we have
denoted 1 upon Omega C as XC and that we
call as capacitive reactance so X is
given as 1 over Omega C Omega if I
replace by 2 pi F I will get X C as 1
upon 2 pi FC which is capacitive
reactance so let define what is the
capacitive
reactances so in a previous slide we
have considered 1 upon Omega C as XC and
that is called as the capacitive
reactance so if I substitute 1 upon
Omega C as XC I will get I am as V M
upon XC and XC will be 1 upon Omega C if
I substitute Omega as 2 pi F and getting
XC as 1 over 2 pi FC which we call as
capacitive reactance is defined as the
opposition of word by the capacitance of
a circuit to the flow of an alternating
sinusoidal current so just like a
inductive reactance it is the opposition
offered by a capacitance and it is like
a resistance
hence it is measured in ohms now let 
see what is the nature of power consumed
by a capacitor so instantaneous power is
given by P equal to instantaneous
voltage multiplied by instantaneous
current instantaneous voltage we have
taken as VM sine Omega T and instant in
his current that we obtain is I am sine
Omega T plus PI by 2 which is nothing
but cos Omega T so ultimately I am
getting instantaneous power as VM I M
sine Omega T into cos Omega T if I
adjust the two term I ultimately I will
get instantaneous power as vm im by 2
into sine 2 Omega T so ultimately
instantaneous power is a sinusoidal
quantity so I have drawn over here so
being a sinusoidal quantity the average
if I took over a one complete cycle it
will be zero so average power consumed
by a capacitor is nothing but P average
over a one complete cycle integral 0 to
2pi vm im by 2 into sine 2 Omega T D
Omega T that is nothing but zero so what
is the conclusion pure capacitance never
consumes a
power so whenever a pure element is
given either a capacitor or the inductor
it never consumes a power in AC circuit
the only passive element which is
responsible for consumption of a power
is resistance only capacitor and
inductor never consumes a power
