NUmericals: Cell parallel; series (Part - 10) - Electric Circuits, Physics, Class 12 Video Lecture

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circuits part 10 is brought to you by
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please make sure that you have watched
all the videos till part 9 before going
ahead with part 10 just look at the
third problem it says that a
galvanometer together with an unknown
resistance in series is connected across
two identical batteries each of 1.5
volts when the batteries are connected
in series the galvanometer records a
current of 1 ampere and when the
batteries are connected in parallel the
current is 0.6 ampere what is the
internal resistance of the battery so
that means in this case we have two
scenarios
the first scenario tells that the two
cells are connected in series and we
have a galvanometer with an unknown
resistance in series let us call this
resistances R and this is my
galvanometer such that they are 1.5 volt
1.5 volts right so in this case it says
that the galvanometer reads a current of
1 ampere the galvanometer reads 1 ampere
there is another scenario that these two
are connected in parallel selectors let
me draw the second scenario here this is
my second scenario where these two cells
are connected in parallel and we have
this galvanometer and the unknown
resistance so this is 1.5 volts this is
also 1.5 volts so in this case it reads
current as 0.6 ampere so we have to
calculate the internal resistance of the
battery
since they are identical batteries so
their internal resistances will also be
the same so we have to calculate the
value of AD right so in this first
scenario what will be the equivalent EMF
of the circuit that will be nothing but
the sum of the EMF that is e1 plus e2
which is 1.5 plus 1.5 latter's 3 volts
and what is the total resistance of the
circuit total resistance will be
external resistance plus internal
resistances plus the resistance of the
galvanometer because nowhere it said
that the galvanometer does not have a
resistance because the galvanometer also
has a resistance of its own so this will
be R plus 2 R plus R G right so what
would be the current through this
circuit the current flowing through this
circuit would be the equivalent EMF
divided by the total resistance that is
3 divided by R + 2 R + RG now according
to a question this current I is given as
1 ampere so that means this is equal to
1 so we can see that 3 is equal to R + 2
r + RG so this is one expression that we
get right so let me call this as
expression 1 okay so we can also write
it as R + RG is equal to 3 minus 2i fine
so now let us forget about the series
formulation and let us concentrate on
the parallel combination what happens in
the parallel combination in parallel
combination the current flowing through
the circuit is again e equivalent
divided by the total resistance so what
will be the equivalent resistance in
parallel combination that will be 1.5
volts because in parallel
the nation the resistance is safe across
each each cell I mean the EMA the same
across each cell so e-equivalent will be
1.5 and what will be the total
resistance that will be R + RG + R
equivalent this R equivalent is nothing
but the equivalent resistance of these
two so what is this R equivalent
our equivalent is nothing but r1 into r2
divided by r1 + r2 here R 1 and R 2 both
are equal to R so this will be R square
by 2 R which is equal to R by 2 so you
can say I is equal to 1 point 5 divided
by R + RG + R by 2 now here also in the
problem it is given that this I is 0.6
ampere so we can write it as 0.6 so now
if we solve this expression we get zero
point six is equal to 1 point 5 divided
by R + RG + R by 2 or we can say zero
point six R plus zero point six RG plus
zero point three R is equal to one point
five or zero point 6 R plus zero point 6
RG is equal to one point five minus zero
point 3 and or odd + RG is equal to one
point five minus zero point 3 R divided
by 0.6 so this is equal to two point
five minus zero point five ad so this is
my expression T now if you compare
equation 1 and equation two you see that
left hand side of
both the expressions are saying that is
our plus LG here also too as our plus IG
so we can equate that I can say that
means we can say that 3 minus 2 I is
equal to 2.5 minus 0.5 I so from this we
can write 3 minus 2.5 is equal to 2 R
minus 0.55 or we can say 0.5 is equal to
1.5 R or R is equal to 0.5 by 1.5 which
is 1 by 3 that is 0.33 3 ohms so this is
the value of the internal resistance so
if you look at this problem carefully
you can see that the main funder while
looking at problems which has multiple
source of EMF s that is multiple number
of cells the main point that you need to
remember is when they are in series
combination the equivalent EMF will get
added up when they are in parallel
combination the equivalent EMF across
each of them will be the same and also
you have to calculate the net or the
total resistance of the circuit and with
the help of this equivalent EMF and
total resistance you can calculate the
total current that will flow through the
circuit and once you are able to
calculate the total current that will
flow through the circuit I think you are
almost done with any kind of problem
because the problem will always ask you
something which you can calculate by
calculating the current flowing through
the circuit okay
so now let us look at yet another
problem it says that three sets of EMF 2
volts 1.8 volts and 1.5 volts are
connected in series their internal
resistances at 0.5 0 ohms 0.7 ohms and
one ohms respectively ok so let us draw
it simultaneously I have three cells
which are connected in series okay let
me call them
as e1 e2 and e3 the battery is connected
to an external resistor via a very low
emitter what would be the reading of the
ammeter I wish to draw the emitter later
suppose this is a meter
so there is an ammeter and then you have
this external resistance I so you have
to calculate the reading of the amitabh
what does an ammeter do and a meter
reads current it measures current so you
have to calculate the value of current
and what are the values that are given t
1 is given as 2 volts e 2 is given as 1
point 8 volts e 3 is given as 1 point 5
volts and also the internal resistances
are given as zero point five zero ohms
zero point seven ohms and one ohm and
the external resistor is given as four
ohms so these are the values that are
given so we have to calculate the total
current through the circuit so if you
calculate the equivalent EMF through the
circuit what will be it it will be the
sum of the EMF because it is connected
in series that will be e1 plus e2 plus
e3 that is equal to two plus one point
eight plus one point 5 that is equal to
five point three volts and God will be
the total resistance of the circuit that
will be the external resistor I plus the
equivalent internal resistance so the
external resistor R plus equivalent
internal resistance which is R 1 plus R
2 plus R 3 so this will be equal to four
plus zero point zero five zero plus R 2
is zero point seven plus R 3 is 1 so
this comes out to be five point seven
five ohms
therefore what is the current that will
be flowing through the circuit it will
be the equivalent EMF
divided by the total resistance in the
circuit so that is 5.3 volts divided by
five point seven five ohms so this comes
out to be 0.92 amperes so this is the
amount of current that will flow through
them circuit right well I made a mistake
here here I have taken R one and zero
point two zero five zero but in the
problem it was zero point five zero so
you can either consider it it has zero
point zero five zero or you can
calculate it with this value okay so
this is how we solve this problem so I
hope that after solving some three four
problems it is now clear to you how do
we approach problems related to grouping
of cells in series and parallel
combination so now it is time that we
will talk so till now we were talking
only about sense because cell was
something which we had not discussed
earlier so now it is time that we will
start solving electric circuits to solve
electric circuits there are certain
rules which we generally follow however
I have told you about many small tricks
as well as rules to calculate equivalent
resistances which often helps a lot in
solving complicated circuits but now we
will talk about circuits in terms of
battery as well as resistances when you
have back when you have multiple
batteries and multiple resistances in a
circuit how do you solve such
complicated circuits so for that we will
we will study about certain laws like we
even have kids chops long we will also
study about wheatstonebridge we will
study about the metre bridge or which is
also known as the slidewire bridge so
when I talk about these two three
methods and we will solve some circuits
we will try to solve some four to five
circuits and then we will conclude the
chapter with some miscellaneous problems
so let us go ahead and look at fit chops
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