Numericals : Equivalent Capacitance (Part - 17) - Capacitor & Capacitance, Physics, Class 12 Video Lecture

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and capacitance part 17 is brought to
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going ahead with Pat's MD now let us
solve a bit different type of a problem
here also however the question asks to
calculate the equivalent capacitance
between points T and V so what did you
notice what is the difference in this
problem that the previous problems in
previous problems your endpoints were
open and that made it very easy for you
to understand which is in series in
which is inside of them here you have a
closed circuit and in that the question
asks you to calculate the equivalent
capacitance between two points so now
you have to keep your reference at these
two points and then watch which is in
series and which is in parallel so it is
always advisable to draw the circuit
diagram as per your comfort first you
draw the two points between which you
have to conclude so what do you see
there is one capacitor in between these
points so let us draw that capacitor
first and that capacitance c1 what about
the other capacitors the other
capacitors they start from a and C 2
then C 3 and C 4 then ends at B so it is
somewhat like this it starts from a C 2
then C 3 and C 4 and then ends in D so
this is C 2 C 3 C 4 now if you look at
this diagram it becomes very simple you
have C 2 C 3 C 4 in series and then that
in parallel with C 1 right so what do
you see here here C 2 C 3 and C 4 makes
a series combination
let us call its capacitances c dash and
then C dash and C 1 make a parallel
combination right so then equivalent
capacitance of these three capacitor C
dash let us first calculate C dash so 1
by C dash will be equal to 1 by C 2
one by CP plus 1 by C 4 so that is equal
to 1 by 9 plus 1 by 9 plus 1 by 9
because you to secrecy is given as 9
microfarads so this is equal to 3 by 9
therefore C Dash comes out to be cream
and profile right now C Dash and C 1 are
in parallel therefore the equivalent
capacitance will be going to C dash plus
C 1 C Dash is 3 and C 1 is also 3 so
this is 6 micro farad therefore the
equivalent capacitance between points C
and B will be equal to 6 micro fans
now let me one important thing now let
us suppose instead of point a and B if
the question would have asked to
calculate the equivalent capacitance
between points C and B then how would
you calculate in that is the entire
arrangement will change in that case
this will be your C this will be your
dealing and this will be C 3 and the
rest 3 will go like this C 2 C 1 and
then C 4 right so this will be C 3 and
this will be C 2 C 1 and C 4 so the
value of your equivalent capacitance
will change in that case because the
values of C 1 right so understand the
question carefully first of all focus on
where you have to find the equivalent
capacitance what are the two points
between which you have to calculate the
equivalent capacitance that is very much
important ok let us look at next problem
it again asks us to calculate the
equivalent capacitance between points P
and Q ok so let us look at it again here
also we will draw this diagram as per
our convenience we have to calculate
between P and Q let us first draw now
you see from P there is one capacitor C
1 and what is the other capacitor
connected to P if you look at it the
other capacitor which is connected to B
is C 3 so from P you have C 1 and you
also
have see three red letters fall this is
c1 and this is C key now focus on qu
what happens to Q choose one point is a
negative C 3 that means the C 3 is
connected with you also Q is again
connected to the other end of C 1 as
well as to C 2 right so that means it is
connected to this end of C 1 and also
one end of C 2 and C 2 is this end is
also connected to P so what do we get we
get that C 1 C 2 C 3 have their starting
and ending points and the scene that
means C 1 C 2 and C key make a parallel
combination so basically looking at the
figure first of all you have to identify
the points between which you have to
calculate equivalent capacitance then
you draw a separate diagram as per your
convenience right then you see what all
capacitors are connected to which point
and then you draw your own diagram and
you can see it for yourself that things
become so simple so here you are the
equivalent capacitance of the circuit
will be given as C 1 plus C 2 plus C key
let us look at a bit different kind of a
problem however this is also related to
combination of capacitors it says that
an electrical technician requires the
capacitance of 2 micro farad in a
circuit across a potential difference of
1 kilo volt so this is this requirement
he wants a capacitance of 2 micro farad
across a potential difference of 1 kilo
volt a large number of 1 micro farad
capacitors are available to him each of
which can withstand a potential
difference of not more than 400 moles
suggest a possible arrangement that
requires the minimum number of
capacitors that means he is maybe
designing some circuit for that he needs
a
Center of ambassadors - microfinance
across the potential difference of one
fill hole okay
he has several capacitors of one
microfarad so he has two now you will
say okay to join to my to one microfarad
capacitor in parallel combination so the
equivalent capacitance will be one plus
one - but it also has a pending that it
should be able to withstand the
potential difference of one kilo volt
whereas in case of these 1 micro farad
capacitors each of these capacitor
cannot withstand anything more than 400
volts so that means you have to arrange
the capacitors in such a way that the
equivalent capacitance is 2 micro farad
at the same time the capacitor should be
able to withstand the potential
difference of one EU voltage so let us
see how do we approach such kind of
problems so how do we approach this
problem what do we do first of all we
assume we assume that we arrange end
capacitors in series
okay we are assuming this that we have
arranged n capacitors in series and
there are M such rows because I do we
don't know how it will be arranged but
there are only two possible still it
will be arranged in series or it will be
arranged in parallel so I am assuming
something like this that it is suppose I
have arranged some n capacitors and that
n can be anything
surrender passengers in series and there
are M such rows that means these kind of
arrangement is repeated some M number of
things that means basically what I am
saying hansung that there are n
capacitors in series and each of these n
capacitors are in parallel with in such
arrangements right so this arrangement
is repeating some n number of so your
arrangement is somewhat like this
together so you have to find out what is
the value of this thing that is how many
capacitors you are arranging in series
and how many rows of such series
capacitors you have such that your
equivalent capacitance
out to be to microfarads and the
potential difference of one kilowatt now
what does the potential difference that
I want along this circuit my desired
potential difference along this circuit
is 1 kilo volt that is 1,000 volts so
this is the potential difference between
point B and right now the potential
difference across this is the potential
difference across each row right so
across each of the row this is the
potential difference now what will be
the potential difference across each
capacitor so if I talk of each capacitor
what would be the potential difference
if the potential difference along this
row is thousand and there are n
capacitors in one row so the potential
difference across each capacitor will be
thousand by n words right but at the
same time we know that each of this
capacitor has a capacitance of 1 micro
farad and it can withstand a potential
difference of maximum 400 volts that is
accordingly the question that the
maximum potential difference that it can
withstand is equal to 400 foods and as
per our assumption the potential across
each capacitor should be thousand by n
so we can equate these two as for the
problem thousand by n that is the
potential across each capacitor can be
maximum 400 it cannot be greater than 48
therefore the value of n comes out to be
thousand divided by 400 which comes out
to be 2.5 now two point five where were
two n here n is the number of capacitors
a number of cannot capacitor cannot be
in decimals right so that means the
minimum number of capacitors that we can
have has to be key because it has to be
more than right that is quite obvious
so more than 2 has to be 3 so that means
we see that there are three capacitors
in one group so how many such
there for that we have to calculate the
value of the thing we have to calculate
M na now let us look at let us try to
find out the equivalent capacitance so
what would be the equivalent capacitance
of the three capacitors so now we know
that there are three capacitors arranged
in series right
so the series 1 by C is of III right now
I am talking about equivalent
capacitance of one row that should be
equal to 1 by C 1 plus 1 by C 2 plus 1
by C 3 we know that C 1 C 2 C 3 are all
1 micro farad so this will be equal to 3
so that means C s is equal to 1 by 3
micro farad this is what is this this is
that equivalent capacitance across each
row right so this for this row take
valid capacitances 1 by 3 micro farad's
so we have to calculate how many such
rules are there so what we can do is
instead of drawing so many capacitors
now we can simplify this diagram like
this so we have capacitors like this we
have to calculate the number of Blues
this is point B and this is point a now
each of this capacitor is CS because
those three capacitors I have removed
with CS right so we have to calculate
the value of M and we also know that the
equivalent capacitance of the circuit
the net equivalent capacitance of the
circuit should be equal to two
microphones so how many such capacitors
should be there in parallel such that
the net equalent inheritance comes out
to be 2 micro farad
so then equivalent capacitance of this
circuit which I have drawn VD equal to c
s+ c s+ c s m times length so that is
equal to M into C is now what I want
that the equivalent capacitance
of this circuit should be equal to do
microfarads so therefore M becomes equal
to two micro farad's / c s for two CS
CS is 1 by 3 micro farad's so therefore
the value of m comes out to be 2 divided
by 1 by 3 that is equal to 6 so we found
the value of n as 3 and M as 6 that
means how the possible arrangement
should be the arrangement should be
somewhat like this there should be 3
capacitors in series and there should be
six such rules that means 1 3 4 5 6
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