AC Voltage Applied to a Capacitor Video Lecture | Physics Class 12 - NEET

today we are going to talk about
capacitors and these are used in radio
circuits along with resistors and
inductance
now first of all I need your help to
answer this question I have a capacitor
which is connected to a DC source and I
find that no current flows through this
so if I have a little lamp over here the
lamp does not glow which means no
current is flowing through this
capacitor this seems to make sense
because you know that it is an
insulating medium present between the
plates of the capacitor so current can't
flow through it now if I look at this AC
source which is connected to the
capacitor I find that a current is
flowing through it so say I put a little
lamp over here in the circuit this lamp
closed which means current is flowing in
this now how is it that current is
flowing in this case but it is not
flowing in the case where a DC source is
connected
see in case of the DC source let's call
these plates let's say a and V now the
plates a and B acquire a positive charge
and a negative charge respectively now
during that very short while when the
capacitor is getting charged to a
voltage which is given by Q over C
during this very short interval while
the capacitor is getting charged which
means current is flowing in the wires
during that time this lamp will glow but
then very soon it goes off the current
stops flowing now let's see how current
is flowing in this case say we have an
alternating voltage which continuously
charges and then discharges the
capacitor while while charging the
capacitor the voltage across the plates
of the capacitor rises and the charge
also builds up let's say this is the
charge now the charge on the plates of
the capacitor is going to increase when
the voltage across the plates of the
capacitor decreases the charge on the
plates of the capacitor will also
decrease then when the voltage increases
in the opposite direction the capacitor
gets charged in the reverse direction
which means that this will become
positive and this will be negative so
this is when the voltage reverses and
you can see that this is going to be the
charge is going to be in phase with the
voltage it's in step with it now as the
voltage decreases and once again comes
back to zero the charge on the plates of
the capacitor also come back comes back
to zero now this is true because Q is
given by C UB but if you take a look at
the current the current which is flowing
in the circuit is given by DQ over DT
now we need to understand this a little
more in detail so that we understand the
relation between
the voltage and the current when an
alternating voltage is applied to a
capacitor
the alternating voltage which is applied
to the capacitor is a time varying
voltage which can be represented by a
sine function V is equal to VM sine
Omega T where VM is the amplitude of the
voltage and Omega is the angular
frequency now this voltage is applied
across the capacitor so the charge Q on
the capacitor is C times V which is C
times VM sine Omega T and we just saw
that the charge is also in step with the
voltage which means that there is no
phase difference between the voltage and
the charge next we can write I as DQ
over DT since the current is given by
the variation of the charge so
differentiating the charge we get C and
VM are constants and differentiation
gives us Omega times cos Omega T and now
we are going to write this as V M over 1
by Omega C times sine Omega T plus PI by
2
writing the current I equals I M sine
Omega T plus PI by 2 we can see that the
current also varies sinusoidally though
there is a phase difference of Pi by 2
we'll just look at this a little while
later now coming back to this term over
here I M is equal to VM over and we
write this as X C so XC is 1 by Omega C
and this is the capacitive reactance
so over here view using the term
reactance rather than resistance for
instance we have AI is equal to V over R
so R is the resistor
but over here we are using the term
reactance
now the SI unit of capacitive reactance
ohms same as that of resistance
and if you plot a graph between the
frequency of the alternating voltage
which is applied and XC now you can see
XC is 1 by Omega C so this will vary
inversely as the frequency for low
frequencies the reactance is extremely
high and for high frequencies the
reactance decreases this also explains
what we saw in the beginning that when
frequency is 0 then the capacitive
reactance is infinite since this is 1 by
Omega C so a capacitor does not allow DC
current to flow through it because the
capacitive reactance is in finite
will show the phase difference between
the current and the voltage on the graph
and then using a phasor diagram so we
know that V is VM sine Omega T and for
this case I was given as given by I M
sine Omega T plus PI by 2 so first if
you draw the current sorry
first if you draw the voltage and say
with we show the voltage as a sine curve
now you can see that let's say this is
Omega T over here and this is V so this
is 0 this will be PI and this is 2 pi
now we will just write the current as a
cos curve because that makes it easier
for us to plot it now you know that cos
0 is 1 so the current will have its
maximum value which is IM cos pi by 2 is
0 cos pi is negative 1 so the current
will be minus i am at this instant cos 3
pi by 2 is again 0 and cos 2 pi would
again be 1 now joining these points
shows us the variation of the current in
case of the capacitor so you can draw a
continuous curve for this and this is
the variation of current and the black
curve shows the variation of voltage and
you can see that there is a phase
difference of 90 degrees or PI by 2
between them with the current leading
the voltage by 90 degrees which means
that the current passes its maximum of
water cycle ahead of the voltage
next we'll use faizal's to show the
variation of voltage and current we have
the voltage V which is shown over here
and this is the amplitude of the voltage
see if you recall a projection on the
y-axis which would be this quantity over
here
this gives the instantaneous value of
the voltage cv is VM sine Omega T V is
the instantaneous value and VM is the
amplitude so the length of the phasor
over here gives us the amplitude and we
are using the amplitudes to show the
phase difference between the voltage and
the current similarly we can draw a
current phaser and the current phaser
will need the voltage phasor by 90
degrees so let's say this is the current
phaser and the difference over here
between these two is 90 degrees and this
would be Omega T the angular speed with
which these phases rotate is omega which
is the angular frequency of the
alternating quantities so you can just
think of this for instance you can think
that you have an imaginary clock where
you have two needles and these keep
rotating at an angular speed of Omega
and they are at all times at a
difference of 90 degrees from each other
try these questions to test your
understanding the answers are given a
short while later
SI unit of capacitance is farad SI unit
of capacitive reactance is ohms when the
voltage across the capacitor is maximum
the charge on its plates is also maximum
but the current is minimum the capacitor
oppose lower frequencies more not higher
lower frequencies more than it opposes
higher frequencies voltage phasor lags
the current phaser current is ahead of
the voltage for an applied voltage whose
peak value is V not the peak value of
the current I naught will be equal to V
naught over the capacitive reactance X C
so the RMS value of the current is I
naught by root 2 and in this case it
will be V naught over Omega C which is 1
by Omega C times 1 over root 2 if you've
got all the questions right don't forget
to give yourself a pat on your back and
do tell me your name you can write it in
the comment box below if you haven't got
them right don't bother
go back to the lesson go through it once
again and then try the question thank
you