L19 : Examples; Numericals - Current Electricity, Physics, Class 12 Video Lecture

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all the videos to the part 18 before
going ahead with this part the first
problem see is that a potential
difference of 200 volts is applied
across the ends of a conductor of
resistance 5 resistance 50 ohms
calculate the number of electrons
flowing through it in one second let us
suppose we have a conductor let us say
this is the conductor a potential
difference is applied across this
conductor which is say 200 volts and
this conductor has a resistance of 50
ohms so we have to calculate the number
of electrons which will flow through
this conductor in one second so
basically here the potential difference
is given as 200 volts and the resistance
is given as 50 ohms so using these data
we can calculate the amount of current
which will flow through the conductor
using Ohm's law we can say that V is
equal to IR so from this we can
calculate the amount of current that is
V divided by R which is equal to 200
divided by 50 so that is 4 amperes that
means 4 amperes of current will flow
through this conductor we have to
calculate the number of electrons which
will flow through it in one second now
we define current as charge per unit
time and how do we define charge charge
is nothing but number of electrons
flowing into the charge on each electron
so this is equal to n P divided by T so
from this we can calculate the number of
electrons which is flowing through it so
n will be equal to I into T divided by E
I use the amount of current flowing
through the conductor which is four
amperes time is one second because we
want to calculate the number of
electrons in one second so this will be
one divided by the charge on each
electron which is 1.6 into 10 to the
power minus 19 coulombs so this value
comes out to be 2 point 5 into 10 to the
power 19 so these many electrons will
flow through the conductor in one second
now let us look at the next problem it
says that in the Bohr model of hydrogen
atom the electron circulates around the
nucleus in a path of radius 5 point 1
into 10 to the power minus 11 meters at
a frequency of 6 point 8 into 10 to the
power 15 revolutions per second
calculate the equivalent current so it
says that let us suppose this is the
nucleus and this is my electron so the
electron revolves around the nucleus
that is it circulates around the nucleus
in a path of radius the radius of the
path is given as let us say R so this
radius is given as 5 point 1 into 10 to
the power minus 11 meters now the
frequency at which the electron moves is
6 point 8 into 10 to the power 15
revolutions per second so we have to
calculate the value of current so how
much current will be produced due to the
motion of these electrons so we know
that current is defined as charge per
unit time what is the charge in this
case in this case the charge is nothing
but the charge carried by the electron
because here we are considering only one
electron in case of hydrogen atom there
is only one electron so this will be the
charge carried by electron divided by
time taken so what is that we have to
calculate we have to calculate the time
by the electron to complete one
revolution right so this T is nothing
but the time taken by the electron to
complete one revolution right so the
time taken to complete one revolution
will be the distance traveled or the
distance covered by the electron in one
revolution divided by the speed of the
electron right so what is the distance
traveled in one revolution it is nothing
but the circumference of this circle so
that will be 2 PI R divided by the speed
of the electron let us say the speed is
V now we know that V is equal to R into
Omega that is speed is equal 8 H to the
angular velocity so V is equal to R into
Omega where Omega is angular velocity
and what is angular velocity angular
velocity can be written as 2 pi F where
F is the frequency so therefore my time
period t becomes equal to 2 pi r divided
by r into 2 pi F so R 2 5 everything
gets cancelled so this time taken
becomes equal to 1 by F therefore they
say that the current produced will be
equal to e divided by T where T is 1 by
F so this will be I is equal to e into F
so what is V that is 1.6 into 10 to the
power minus 19 and what is frequency
frequency is 6 point 8 into 10 to the
power 15 so this comes out to be one
point zero eight 8 into 10 to the power
minus three amperes so this would be the
equivalent current in this
while we considered only one electron
here because in case of hydrogen there
is only one electron which revolves
around the nucleus right so let us look
at the next problem this problem says
that in a discharge tube the number of
hydrogen ions drifting across a
cross-section per second is 1 into 10 to
the power 18 in a discharge tube the
number of hydrogen ions drifting across
a cross-section per second is 1 into 10
to the power 18 by the number of
electrons drifting in the opposite
direction
across another cross-section is 2 point
7 into 10 to the power 18 meter per
second if the supply voltage is 230
volts what is the effective resistance
of the so that means it is somewhat like
this let us suppose you have a discharge
tube I am just considering it as some
area okay so here the number of hydrogen
ions which are drifting across a
cross-section per second is 10 to the
power 1 into 10 to the power 18 so let
us suppose these are the hydrogen ions
which are drifting this way okay while
the number of electrons drifting in the
opposite direction across another
cross-section that means let us suppose
the electron drifting along this
direction in another cross-section is 2
point 7 into 10 to the power 18 per
second
the supply voltage is 230 volts what is
the effective resistance of the tube now
what will be the effective resistance
across this discharge tube it will be
the voltage applied across that you've
divided by the current through the
cross-section so this would be the
current through the cross-section right
so according to Ohm's law effective
resistance will be V by I now what will
be the total current through the
cross-section it will be the current
through this cross-section plus the
current through this cross-section that
means current is charged per unit time
and charges equal to number of electrons
into charge on each electron for this
cross-section plus the number of
hydrogen ions plus the charge on each
ion right so if these many ions are
moving that means that many electrons
are also moving in the opposite
direction so we can say that this is n E
and this is n P and E is number of
electrons and here n P is number of
protons let so as when if four protons
are moving here that means four
electrons will also move in the opposite
direction so we can say that this will
be equal to n e+ n P into charge on an
electron divided by T so this will be
equal to what is n V number of electrons
is 2 point 7 into 10 to the power 18 so
this is 2 point 7 into 10 to the power
18 what is n P that is the hydrogen ions
which is 1 into 10 to the power 18 this
into e that is charge on an electron
which is 1.6 into 10 to the power minus
19 divided by time taken so what is the
time taken it is one second because both
these terms are defined in terms of one
second so this comes out to be zero
point five nine two ampere so this is
the amount of current which flows
through this frode's
flows through that discharge tube
therefore the effective resistance of
the tube will be equal to V upon I where
V is equal to 230 volts divided by I is
zero point five nine two which comes out
to be three point nine into 10 to the
power 2 ohms so this is the effective
resistance of the discharge tube so in
this case what we did we did nothing the
simple we use the same logic here also
in this case the total current was equal
to the total current through both the
cross-section in this cross section the
electrons were moving so n V that is the
number of electrons into charge and each
electron similarly in this row
section the hydrogen ions removing the
same number of electrons will also move
in the opposite direction so we said n P
that is the number of hydrogen ions into
the charge on each electron right okay
so now let us go ahead and look at the
next problem problem 4 says a wire of
resistance 5 ohms is drawn out so that
its length is increased by twice its
original length calculate its new
resistance so initially let us suppose
we have a wire of resistance 5 ohms the
length of the wire is L and the area of
cross section of the wire is a what
happens later later let us suppose that
we do not know the resistance so
resistance we have to calculate the
length of the wire is increased by twice
its original length that means the new
length becomes original length plus
twice its original length that means the
increase is twice that means if
initially it was L now it is L plus two
n it is increased by twice please
understand this statement clearly
increased by twice means the increases
twice the original length so that Delta
is two L therefore the new length will
be the original length plus twice the
original length okay and what is the new
cross sectional area so that new cross
sectional area we will calculate now we
know that the volume of the wire will
still remain constant so volume of the
wire will remain constant that means
area into length will be equal to the
area and length later so eh L will be
equal to a - L - so from this a - will
be equal to a L divided by n - and what
is L - L - is nothing but 3 L so we can
write it as 3 L therefore area becomes
equal to a by 3 so a - we can put it as
a by 3 right so now we have to
chill eight the new resistance that is
our - so our - will be equal to from the
expression for resistance we know that
our - will be rho l - divided by a - so
this will be equal to Rho into L dash
can be written as 3 L a - can be written
as e by 3 so this will be equal to Rho L
by a into 9 so this is 9 into Rho L by a
what is Rho L by a Rho L by a is nothing
but this resistance R so this will be
equal to 9 r so 9 into r is 5 ohms so
this is 45 ohms so that new resistance
becomes 45 ohms the catch in this
problem was the length increased by
twice its original length many students
make this mistake that they take the new
length L dash is equal to 2m but please
understand it is increased by twice its
original length that means the increases
twice the original length the increases
- L so the new length will be L plus 2 n
right
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