Class 12 Exam > Class 12 Videos > L24 : Numericals - Capacitor & Capacitance, Physics, Class 12
L24 : Numericals - Capacitor & Capacitance, Physics, Class 12 Video Lecture
FAQs on L24 : Numericals - Capacitor & Capacitance, Physics, Class 12 Video Lecture
| 1. What is a capacitor and what is its role in circuits? |  |
Ans. A capacitor is an electronic component that stores and releases electrical energy in a circuit. It consists of two conductive plates separated by an insulating material, known as a dielectric. When a voltage is applied across the capacitor, it charges up by storing electric charge on its plates. In circuits, capacitors are used for various purposes such as energy storage, filtering, smoothing, timing, and coupling.
| 2. How is the capacitance of a capacitor determined? | |
Ans. The capacitance of a capacitor is determined by its physical characteristics, such as the area of the plates, the distance between the plates, and the type of dielectric material used. The capacitance (C) is directly proportional to the area (A) of the plates and the permittivity (ε) of the dielectric material, and inversely proportional to the distance (d) between the plates. Mathematically, the capacitance can be expressed as C = εA/d, where C is measured in Farads (F), A in square meters (m²), d in meters (m), and ε is the permittivity of the dielectric material.
| 3. What factors affect the capacitance of a capacitor? | |
Ans. The capacitance of a capacitor is influenced by several factors. The main factors include the area of the plates, the distance between the plates, the type of dielectric material used, and the permittivity of the dielectric material. Increasing the area of the plates or decreasing the distance between them will result in a higher capacitance. The type of dielectric material used also affects the capacitance, as different materials have different permittivity values. Additionally, the thickness and composition of the dielectric material can also influence the capacitance.
| 4. How does a capacitor store and release electrical energy? | |
Ans. A capacitor stores and releases electrical energy through the process of charging and discharging. When a voltage is applied across the capacitor, electrons accumulate on one plate, creating a negative charge, while the other plate becomes positively charged. This separation of charges creates an electric field between the plates. The capacitor stores energy in this electric field. When the capacitor discharges, the stored energy is released as the charges on the plates redistribute and the electric field collapses.
| 5. How can the capacitance of a capacitor be increased? | |
Ans. The capacitance of a capacitor can be increased by increasing the area of the plates, decreasing the distance between the plates, or using a dielectric material with a higher permittivity. By increasing the plate area, there is more surface area available for charge storage, leading to a higher capacitance. Decreasing the distance between the plates reduces the separation between the charges and increases the capacitance. Additionally, using a dielectric material with a higher permittivity increases the ability of the capacitor to store charge, resulting in a higher capacitance.
Text Transcript from Video
hello friends this video on capacitors
and capacitance part 24 is brought to
you by exam vo dot-com
no more fear from exam please make sure
that you have watched all the videos to
buy 23 be for going ahead with part 24
so let us look at the first problem it
says find the potential difference
between the left and the right plate of
each capacitor in the circuit so in this
you have to find out the potential
difference between the plates that means
the potential difference for c1 and the
potential difference for c2 so in this
circuit what all do you have you have
two capacitors and two voltage sources
you are and with EMF C 1 and V 2 so now
it is up to our assumption how do we
consider the flow of panic right now
whenever in a circuit you have more than
one source of voltage you assume that
one voltage source is greater than the
other so that you can follow one
particular convention so let us suppose
here I assume so I assume it is my
assumption that p1 is greater than p2
let me as human like that so if I assume
that P 1 is greater than P P whichever
is greater that will dominate so the
motion will start the motion of the
charge will start from that point so I
say even is greater than P so let me
start from 1 I assume this is positive
this is negative this is negative this
will make this negative so this will
make this positive so as a result now
let us assume that even is greater than
t2 so v's assume that the motion starts
from this battery because this dominates
so I assume the direction of charge flow
in this direction so this is positive so
positive charge comes on this take
therefore this has negative charge like
for this battery this is positive this
is negative similarly this is negative
so this will make this negative this
positive click and I assume this as the
direction of charge group now let me
write down the equation
you chop snow according to Kirchhoff's
law let me start from p1 so what would
be the sign of V 1 it is departing from
positively right so it will be plus even
now I go to the first capacitance so
what is the voltage across the first
capacitance I don't know the voltage
across first capacitance actually that
is what I have thank you later
so voltage across this capacitance
little suppose that is v1 so this V 1
will be equal to Q divided by c1
similarly let us assume that the voltage
across C 2 is V 2 so which will be equal
to Q which is the charge on this
capacitor divided by C 2 right so why
the charges will be same because if the
charges are coming from the same battery
this is sending plus Q to this and -
whew - this right so this is plus V 1
and then I go to C 1 so at C 1 what is
the potential it will be equal to Q by C
1 but what will be the sign it is
departing from the negative link so this
will be minus Q by C 1 then I go to keep
you what will be the second sign for
YouTube it is departing from negative
clip so this will be minus T 2 and then
I go to C 2 or C 2 it is what is the
potential for C 2 it is Q by C 2 what is
the sign it is departing from the
negative plate so this will be negative
this is equal to 0 so from this what do
we get we can find out the value of Q
from here because if we get the value of
Q we can calculate the values of V 1 and
people right so our aim is to calculate
Q from this so we get T 1 minus T 2 is
equal to Q divided by C 1 plus 1 by C 2
therefore we say that Q is equal to C 1
C 2 divided by C 1 plus C 2 into T 1
minus T 2 so this is the value of Q so
now once I have calculated the value of
Q I can find out V 1
is a potential difference for first
capacitor capacitor so v1 will be Q
divided by C 1 that is equal to C 2
divided by C 1 plus C into e1 minus e2
similarly I can calculate the potential
difference across the second capacitor
that is C 2 so that will come out to be
Q divided by C 2 so that comes out to be
C 1 divided by C 1 plus C 2 into e1
minus e2 so these are the values of the
potential differences between the left
and the right plates of food the
capacitors right now friends let us see
I told you in the previous slide right
that we can assume any direction now let
us suppose you must be thinking how can
I give the same result if I assume the
current flow of charge and if I assume
the charge flow in the opposite
direction okay now also here I have
assumed this e 1 is greater than if that
was also a assumption if I assume that T
2 is greater than T 1 then also you will
get the same result if you want approve
let us solve it like that as well let us
assume so let us now write it in as
moving the opposite direction so now let
us assume that e 2 is greater than T 1
so if I assume that e 2 is greater than
1 then if you will dominate right so for
e 2 this is positive this is negative so
according to that keeping will make this
positive and since this is negative so e
2 will make this negative so please
remember whatever I am marking in blue
therefore this assumption when e 2 is
greater than 5 so if this becomes
positive this becomes negative similarly
if this becomes negative this becomes
positive and for this it is still this
is positive and this is negative because
you cannot change the positive
indicative of the battery right so this
is your circuit now so now since we are
assuming that t2 is greater than t1 so
we assume that the current flows from it
starts flowing from people I mean I am
talking about charge flow so let us
suppose now that the charge flows in
this direction so write down the
equation for Q chops law in this case so
we will start with ET so if you look at
e 2 e 2 will be positive because it
departs here the charge departs from the
positive terminal then we go to C 1 or C
1 it will be negative because it divides
from the negative terminal so it will be
minus Q by C 1 then we reject even for
even it is departing from the negative
terminal so this will be minus C 1 then
we reach C 2 for C 2 again it is
departing from the negative terminal so
this will be minus Q by C 2 is equal to
0 so therefore from this also if you see
e 2 minus T 1 is equal to Q into 1 by C
1 plus 1 by C 2 so isn't it not the same
as this equation if you look at this
equation this equation and this equation
are the same so that means even if we
assume the other way round even if we
assume the opposite direction of current
we still get the same thing as long as
we put the signs correctly so you start
solving such problems first you assume
if you have multiple battery sources
first you as you that this battery the
value of this battery is the highest so
that battery will dominate so in this
case is whichever I assume I am I assume
that that will dominate right because
let us suppose if I assume that this
battery is higher so if this is positive
so this will make this positive and if
this is negative this will make this
negative the melanie if this battery has
higher value and this will make if it is
positive it will make me reven positive
and this will make the second so
whichever will be higher that battery
will dominate so whenever you have two
batteries in a single circuit you assume
that one is higher and also for
contention you assume that the charge
flow starts from that battery so
starting from that battery you assume
the direction of the charge and then
from the equation following the
convention that the sign which you put
plus or minus depends on the plate from
which it is departing from which the
charge is departing so I hope I am able
to explain this to you however I will
solve some more problems now where we
will use the same concept to make things
even easier so let us look at the next
problem thank you please visit exam
viacom to watch free educational videos
try free online tests get the best
quality study materials study from the
best tutors and mentors and much moon
thank you once again
and capacitance part 24 is brought to
you by exam vo dot-com
no more fear from exam please make sure
that you have watched all the videos to
buy 23 be for going ahead with part 24
so let us look at the first problem it
says find the potential difference
between the left and the right plate of
each capacitor in the circuit so in this
you have to find out the potential
difference between the plates that means
the potential difference for c1 and the
potential difference for c2 so in this
circuit what all do you have you have
two capacitors and two voltage sources
you are and with EMF C 1 and V 2 so now
it is up to our assumption how do we
consider the flow of panic right now
whenever in a circuit you have more than
one source of voltage you assume that
one voltage source is greater than the
other so that you can follow one
particular convention so let us suppose
here I assume so I assume it is my
assumption that p1 is greater than p2
let me as human like that so if I assume
that P 1 is greater than P P whichever
is greater that will dominate so the
motion will start the motion of the
charge will start from that point so I
say even is greater than P so let me
start from 1 I assume this is positive
this is negative this is negative this
will make this negative so this will
make this positive so as a result now
let us assume that even is greater than
t2 so v's assume that the motion starts
from this battery because this dominates
so I assume the direction of charge flow
in this direction so this is positive so
positive charge comes on this take
therefore this has negative charge like
for this battery this is positive this
is negative similarly this is negative
so this will make this negative this
positive click and I assume this as the
direction of charge group now let me
write down the equation
you chop snow according to Kirchhoff's
law let me start from p1 so what would
be the sign of V 1 it is departing from
positively right so it will be plus even
now I go to the first capacitance so
what is the voltage across the first
capacitance I don't know the voltage
across first capacitance actually that
is what I have thank you later
so voltage across this capacitance
little suppose that is v1 so this V 1
will be equal to Q divided by c1
similarly let us assume that the voltage
across C 2 is V 2 so which will be equal
to Q which is the charge on this
capacitor divided by C 2 right so why
the charges will be same because if the
charges are coming from the same battery
this is sending plus Q to this and -
whew - this right so this is plus V 1
and then I go to C 1 so at C 1 what is
the potential it will be equal to Q by C
1 but what will be the sign it is
departing from the negative link so this
will be minus Q by C 1 then I go to keep
you what will be the second sign for
YouTube it is departing from negative
clip so this will be minus T 2 and then
I go to C 2 or C 2 it is what is the
potential for C 2 it is Q by C 2 what is
the sign it is departing from the
negative plate so this will be negative
this is equal to 0 so from this what do
we get we can find out the value of Q
from here because if we get the value of
Q we can calculate the values of V 1 and
people right so our aim is to calculate
Q from this so we get T 1 minus T 2 is
equal to Q divided by C 1 plus 1 by C 2
therefore we say that Q is equal to C 1
C 2 divided by C 1 plus C 2 into T 1
minus T 2 so this is the value of Q so
now once I have calculated the value of
Q I can find out V 1
is a potential difference for first
capacitor capacitor so v1 will be Q
divided by C 1 that is equal to C 2
divided by C 1 plus C into e1 minus e2
similarly I can calculate the potential
difference across the second capacitor
that is C 2 so that will come out to be
Q divided by C 2 so that comes out to be
C 1 divided by C 1 plus C 2 into e1
minus e2 so these are the values of the
potential differences between the left
and the right plates of food the
capacitors right now friends let us see
I told you in the previous slide right
that we can assume any direction now let
us suppose you must be thinking how can
I give the same result if I assume the
current flow of charge and if I assume
the charge flow in the opposite
direction okay now also here I have
assumed this e 1 is greater than if that
was also a assumption if I assume that T
2 is greater than T 1 then also you will
get the same result if you want approve
let us solve it like that as well let us
assume so let us now write it in as
moving the opposite direction so now let
us assume that e 2 is greater than T 1
so if I assume that e 2 is greater than
1 then if you will dominate right so for
e 2 this is positive this is negative so
according to that keeping will make this
positive and since this is negative so e
2 will make this negative so please
remember whatever I am marking in blue
therefore this assumption when e 2 is
greater than 5 so if this becomes
positive this becomes negative similarly
if this becomes negative this becomes
positive and for this it is still this
is positive and this is negative because
you cannot change the positive
indicative of the battery right so this
is your circuit now so now since we are
assuming that t2 is greater than t1 so
we assume that the current flows from it
starts flowing from people I mean I am
talking about charge flow so let us
suppose now that the charge flows in
this direction so write down the
equation for Q chops law in this case so
we will start with ET so if you look at
e 2 e 2 will be positive because it
departs here the charge departs from the
positive terminal then we go to C 1 or C
1 it will be negative because it divides
from the negative terminal so it will be
minus Q by C 1 then we reject even for
even it is departing from the negative
terminal so this will be minus C 1 then
we reach C 2 for C 2 again it is
departing from the negative terminal so
this will be minus Q by C 2 is equal to
0 so therefore from this also if you see
e 2 minus T 1 is equal to Q into 1 by C
1 plus 1 by C 2 so isn't it not the same
as this equation if you look at this
equation this equation and this equation
are the same so that means even if we
assume the other way round even if we
assume the opposite direction of current
we still get the same thing as long as
we put the signs correctly so you start
solving such problems first you assume
if you have multiple battery sources
first you as you that this battery the
value of this battery is the highest so
that battery will dominate so in this
case is whichever I assume I am I assume
that that will dominate right because
let us suppose if I assume that this
battery is higher so if this is positive
so this will make this positive and if
this is negative this will make this
negative the melanie if this battery has
higher value and this will make if it is
positive it will make me reven positive
and this will make the second so
whichever will be higher that battery
will dominate so whenever you have two
batteries in a single circuit you assume
that one is higher and also for
contention you assume that the charge
flow starts from that battery so
starting from that battery you assume
the direction of the charge and then
from the equation following the
convention that the sign which you put
plus or minus depends on the plate from
which it is departing from which the
charge is departing so I hope I am able
to explain this to you however I will
solve some more problems now where we
will use the same concept to make things
even easier so let us look at the next
problem thank you please visit exam
viacom to watch free educational videos
try free online tests get the best
quality study materials study from the
best tutors and mentors and much moon
thank you once again
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Dec 02, 2024 Last updated |
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