L20 : Numericals - Electric Circuits, Physics, Class 12 Video Lecture

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will solve some problems based on the
Wheatstone bridge and meet enriched so
the first problem here is the four arms
of a Wheatstone bridge have the
following resistances that is first let
us name it this is a b c and d so the
four arms have these resistances as you
can see in the figure this is your
galvanometer the galvanometer has a
resistance of 15 ohms calculate the
current through the galvanometer when a
potential difference of 10 volt is
maintained across AC so you have to
calculate the value of the current which
flows through the galvanometer right the
first length let us look at the current
distribution let us suppose some current
comes out of the battery this I reaches
here I 1 flows on to the side so I 2
comes this side I one reaches here IG
flows this side so I 1 minus IG will
flow across DC this I 2 will come here
and IG from this side will also come
here so i2 plus IG then flow across this
side they both will join together and
again form I that is i1 plus i2 so it
comes back to the battery so this is my
current distribution now in this case as
it looks like a Wheatstone bridge first
you check whether the bridge is balanced
or not because if the bridge is balanced
then the deflection and through the
galvanometer will be zero so there will
be no current in the galvanometer but in
this case P by Q is equal to hundred by
10 that is 10 and R by s is 60 by 5 that
is 12 so that is P by Q is not equal to
here P by Q is not equal to R by s so
this is an unbalanced bridge so how will
you solve this so here you will apply
your loop log at a scale chops loop law
so first you apply it to this first loop
that is a B D a so in the first loop abd
a let us assume one direction say this
direction so what does the EMF in this
loop there is no EMF so you can write it
as zero what are the resistances
involved that is hundred ohms 15 ohms
that is the resistance of the
galvanometer and 60 ohms so how much
current flows 200 ohms that is i1
current flowing through 15 ohms is IG
and current through 60 ohms is i2 now
put the signs here as per my convention
I 1 is along my path assume direction
IG is also along my usual direction by
i2 is in the opposite direction so I 2
will have a negative sign the rest both
of them will have positive signs so from
this we get an X equation 20 I 1 plus 3i
g -12 I 2 is equal to 0 this is my first
equation now let us apply loop law to BC
DB now I am solving this problem in the
same manner as I solve another previous
problem so here also let me assume this
as my direction my alakazoom direction
so here what happens here also the net
EMF there is no EMF involved in the
circuit so that is that becomes equal to
0 so what are the resistances involved
resistances are 10 ohms 15 ohms and 5
ohms that is 10 ohms 15 ohms and 5 ohms
so what is the current flowing through
this 10 norms it is i 1 minus i g
current through 15 ohms
IG and cutting to 5 ohms as i2 + IG now
let us look at the signs this is my
assumed direction i 1 minus i g is in
along my assumed direction but i 2 + IG
is in the opposite direction and IG is
also in the opposite direction so these
two men have negative signs but this
will have a positive sign so from this I
get the second equation that is 2 I 1
minus 6i g minus i2 is equal to 0 so now
let us apply the loop law to a DC e a
that is a DC e a to this loop now let us
assume my direction as this one so here
what is the net EMF I mean how many
hang-ups are there there is a 10 volt
EMF so then put it as 10 what are the
resistance is involved that is 60 ohms
and five forms how much current flows
through 60 ohms that is i2 how much
current flows through five ohms that is
i2 plus IG right now comparing with my
direction I 2 is in my along my assumed
direction I 2 plus IG is also along my
Asian direction so both of them will
have positive signs so from this I will
get my third equation that is 13 I 2
plus IG is equal to 2 that is equation 3
so now I have three equations and I also
have three variables that is i1 i2 and
IG so how do I solve it I multiply
equation 2 but 10 and that gives me 20 I
1 minus 60 i G minus 10 i2 is equal to 0
and we call this equation as equation 4
so now we subtract equation 4 from
Equation
and that gives 20 I 1 plus 3i g minus
12i 2 is equal to 0 and this is 20 I 1
minus 60 I G minus 10 I 2 is equal to 0
now I reverse the signs so what do we
get we get 63 i g minus 2 I 2 is equal
to 0 we call this as equation number 5
so now from Equation 3 we found that I 2
is equal to 2 minus IG divided by 30 and
from Equation 5 we find that I 2 is
equal to 63 IG divided by 2 therefore
from these two we can see that both
right-hand sides can be equalized
therefore 2 minus IG by 13 is equal to
63 IG / - so calculating this value we
get IG is equal to 4 point 8 milli
amperes so this is the value of the
current that will flow through the
galvanometer right so now you can see
that how much useful kill chops law is
it it helps us so much we solve
complicated circuits to solve circuits
where wheatstonebridge fails to for 2
short circuits where Ohm's law fails so
there comes the Q chops law so whenever
you have closed loops the chops law is
there to help you now let us look at the
next problem it says that in a meter
wrench the balance point is 4
to be at thirty nine point five
centimeter from the left and a if an
unknown resistor X is in the left gap
and unknown resistor Y of resistance
twelve point five is in the right gap
determine the resistance of X so let us
remember how the metre bridge looks like
this is how it looks like when R is my
known resistance and s is my unknown
resistance right and this is my balance
point right so in this problem but what
are the values that are given it says
that the balance point is found to be at
39 point five centimeter from the left
end so this is the left end so balance
point is found to be thirty nine point
five centimeter from here that means L 1
is given as 39 point five centimeters
right if an unknown resistor X is in the
left gap so that means here s is equal
to X and a known resistor by of
resistance 12 point 5 ohms is in the
right gap now just have a look here
unknown resistor is in the left gap so
this is my left gap right so now it says
that this R is unknown in this problem
and this s is known according to this
problem because the right gap resistor
is known here here the resistor which
isn't that right gap is Y and that is
known so this is my right gap so the
right gap resistor is known here that
means s is equal to 12 point 5 ohms and
an is the unknown resistor X so we have
to calculate the value of this X so now
since since we are talking about balance
point what is balance point balance
point means that the
the jockey is at this point the bridge
is balanced right so since we are
talking about balance point therefore
when it is at the birth balance point
the bridge should be balanced since
balance point therefore it refers to a
balanced bridge okay so under this
condition R by s will be equal to L 1
divided by 100 minus L 1 right so what
is I here R is X and s is 12.5 this is
equal to L 1 is thirty nine point five
divided by 100 minus thirty nine point
five therefore your X that is the
unknown resistance will come out to be
thirty nine point five divided by sixty
point five into twelve point front which
comes out to be eight point one six
spoons so this is the value of the
unknown resistor right so if you
remember the construction of the meter
bridge it will be very much easy to
solve problems on leader in so whenever
you get problems on meter bridge roughly
draw the meter bridge and then try to
jot down the points let us look at the
third problem it says that determine the
balance point of the previous bridge if
x and y are interchanged
that means you remember the previous you
remember the same bridge here it says
that x and y are interchange that means
now X comes here and y goes here so what
will happen in that case it will be
somewhat like this that means now this
will be your Y and this will be X so
this is again unknown and this is known
right so it says that if it is not
unknown now now we know both x and y so
it just sees that if X is here and Y is
here where will be the balance point
right so that means here we have to
calculate the value of
now we don't know what is l1 it sees
that whatever value you calculated for X
in the previous question let us suppose
that it with the value of X but now we
see that X is not here instead X is here
and why this here so for balance point
the condition that should be true is
that our by s should be equal to L 1
divided by 100 minus L 1 so R is why X
is the S is X this is L 1 divided by 100
minus L 1 now what is y the value of y
was if you look at the previous question
it was 12.5 and what was X X be
calculated as 8.16 so this is equal to L
1 divided by 100 minus L 1 so from this
we find that L 1 is equal to sixty point
five centimeters therefore the balance
point is sixty point five centimeters
away from left and a so it is this much
distance away from this end a right
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