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Hi, myself Sufal Kumar,
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Hi, myself Sufal Kumar,
Physics Faculty.
Dear friends, viewers & students, my channel is about Physics Education. I am to concentrate mainly on JEE, NEET, CBSE & ISC in near future.
#sufalphysicsforum
#jee
#neet
#physics
#cbse
#iitjee
#igcse
#cbse12thexam
Pls like, share & subscribe, if you like my educational video....
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LearningTranscript
00:00Current electricity 5th part in which I am starting with heating effect of current.
00:07When a current is passed across any resistor then some heat is dissipated because of the
00:16obstruction posed by resistor and this phenomenon is called joule heating.
00:22What's the cause of this heating effect?
00:30Let me try to find out some key words.
00:34So whenever potential difference is applied across the ends of a conductor its free electrons
00:41get accelerated in the opposite directions of the applied field but the speed of electron
00:46does not increase beyond a constant drift speed.
00:52This is because during the course of their motion the electrons collide frequently with
00:58the positive metal ions.
01:00The kinetic energy gained by the electrons during the intervals of free acceleration
01:06between collisions is transferred to the metal ions at the time of collision.
01:11The metal ions begin to vibrate about their mean positions more and more violently.
01:19The average kinetic energy of the ions increases.
01:22This increase, this increases the temperature of the conductor.
01:28Because of this increase in average kinetic energy temperature increases.
01:33Thus the conductor gets heated due to the flow of current.
01:38Eventually the electrical energy supplied by the source of emf is converted into heat.
01:45So these are the key words and phrases we need to remember while we are understanding
01:53heating effect of current.
01:55From this pink box power P is H by T I square R P by T it implies that P is I square R is
02:09equal to V square by R is equal to V I. At the same time this also we need to remember
02:19because power is rate of doing work or rate of heat dissipation.
02:26Obviously we are referring this power as electric power.
02:31Here from the above pink box we can see it is mentioned calorie also.
02:37Obviously Joule can be converted into 1 calorie is equal to 4.18 Joule.
02:47Kinetic energy the total work done electrical energy consumed in the circuit power in series.
02:57Series may 1 upon P equivalent is 1 upon P1 plus 1 upon P2 1 upon P3 so on so on.
03:11That means if n number of electrical devices are connected in series such that P1 P2 P3
03:19are individual powers then total power consumed that is equivalent power should be equal to
03:28reciprocal sum like equivalent resistance and parallel similar but here this is series
03:37power in parallel P equivalent is P1 plus P2 plus P3 so on P1 P2 P3 are individual power
03:49consumption such that they are connected in parallel of a source of EMF.
03:56So this is the output upon input output is of course V I and E I is input power then
04:05after cancellation of I V upon E and this way finally we obtain always efficiency is
04:15output upon input that means total external resistance is equal to total internal resistance
04:23of the circuit.
04:24It shows laws there are two laws first law junction rule second law AVL voltage rule
04:34AVL it shows voltage law total current across any junction is zero some are approaching
04:43some are approaching but some are going away that means incoming if we consider positive
04:51then I1 plus I2 it implies that it implies that total incoming currents across any junction
05:04this is basically first law total current across any junction is zero or total incoming
05:12current these are two forms in which this first law could be expressed.
05:18If we see any illustration for the given circuit find the value of I let's identify
05:26which all are incoming and which all are outgoing these are incoming incoming incoming which
05:37are outgoing this is outgoing this is outgoing and even this is also outgoing so if we try
05:47to solve here only at this interface incoming is equal to outgoing it imply that incoming
05:56means 0.2 plus 0.6 plus 0.7 now outgoing green one 0.4 0.5 plus I and we were to find I only
06:141.3 that means 1.5 is equal to 0.9 plus I it imply that I is 1.5 minus 0.9 0.6 ampere answer
06:31this way so this was simple illustration just to understand how junction rule is applied
06:39but now voltage law second total potential across any closed loop
06:58how to apply that in general we'll have always closed loop only total potential across any
07:05closed loop is zero or in other words sigma ir is equal to sigma e for any closed loop for any
07:17closed loop total emf sum should be equal to all the potential drops across the resistors in that
07:27very closed loop the following figure shows a complex network of conductors which can be
07:35divided into two closed loops like e a c e that means e a c e this way we are to this is we are
07:48traversing our circuit in this particular circuit one more clarification is not here because here
07:57we are to traverse the circuit so our direction is also there along the direction of current
08:04or opposite to the direction of current because black arrow is suggesting that we are they have
08:12traversed the circuit along the direction of current I'll specify arrow so again low
08:19loop e a to sorry potential drop the current direction will traverse current to positive
08:26i1 r1
08:29here what current we would be taking
08:43now another loop a b c a that means
08:47lower
08:54when we start from a b c a a b this is b c a this i2 r2 is coming negative because
09:08our traverse direction is opposite to direction of current so this time let us start with i4
09:16r4 then come to again plus i5 r5 and this one is negative of i2 r2 is equal to zero
09:30because in this closed loop there is no battery or cell connected so hence it is equal to zero
09:39hence so through these two equations we could get the desired value whatever thing is unknown
09:46but this illustration has been given only to explain the formation of equations it is not
09:55proper numerical so we'll be further solving it could be written as i4 r4 plus i5 r5 is equal to
10:06i2 r and all these are numericals if we if we will have any problem now let's try to frame the
10:15now let us consider two equations again one thing more i i would tell you here we have how many
10:26equations chances probability of close how many closed loops do we have one one a d f e a and
10:37another is e f c b e and one more is there that is a loop so we can always choose this also but
10:51why to take a longer route why to take longer equation to involve so that's why we
11:00rarely move on to this option otherwise we have three options and moreover we would be
11:06needing only two equations and along with these two equations junction rule is also there
11:16so why to choose fourth one but we are given this option we may apply now let's follow
11:24let's write the equations firstly i am willing to opt for a d f e a and what is asked for calculate
11:34the current that flows in the one oh that means let's try to indicate
11:47any direction we choose
11:50that doesn't mean
11:58what i am trying to say
12:04this is i2 at this junction e
12:21no i2 is equal to i1 plus i3 from junction rule i2 is equal to i1 plus i3 it implies that
12:34because they have asked for current across i1 ohm it implies that i3 is equal to i2 minus
12:44i1 let's see whether we would be able to apply this equation for our answer or not so loop a
12:55d f e a
13:18also current is going this way and we are also traversing along the direction of current only
13:24only three resistors that is equal to nine volt in both the loops battery were positive only
13:32so this way if we add up these two equations plus i3 and minus i3 are getting cancelled
13:44from one
13:45one p i1 minus minus i3 is equal to six now just to eliminate i1 what do we do this value
13:59will be substituting over here therefore five i3 now only one variable is left let's see
14:09whether we could tally our answer or not i think we only required i
14:14ah yes that's why i have eliminated i1 i3 is equal to minus 3 by 23 ampere
14:24we had to go two three steps of calculation extra so hello friends for a few days it was
14:34kept aside just to complete another few videos now i am coming back to last edition of current
14:43electricity and here it is wheelstone bridge and few more topics like meter bridge and i guess
14:52for potentiometer since it is not in cbsc boards now anymore so for potentiometer i am to cover
15:02a separate video what is wheatstone bridge let us first have proper know-how in terms of
15:10definition then thereafter with the help of diagram i'm going to explain i must tell you
15:17in which format and this is galvanometer this particular type of circuit this particular
15:24shape circuit which has four resistances in the given pattern first pe q r s and this
15:35middle branch consisting galvanometer and if galvanometer shows no deflection that is ig
15:45is equal to zero or third option i am indicating four points namely what are those four
15:54points a again i'll repeat first let me indicate all the branches currents this is i1 minus
16:04ig and after combining this i2 and ig here i2 plus ig and then thereafter i1 plus i2 is equal to
16:17i and the in the given circuit if galvanometer shows no different or ig is equal to zero
16:28if galvanometer shows no deflection that means in this middle branch no current
16:35that means ig is equal to zero that's why i wrote ig is equal to zero or potential difference between
16:43b and d is equal to zero vb minus vd is equal to zero then p upon q is equal to r upon s and this
16:55is the condition for that means the given circuit will be considered balanced if this condition
17:05fulfills otherwise the given circuit won't be considered balanced and then the then what's
17:12the remedial measure for this kind of circuit if it is not balanced that means p upon q
17:21is not equal to r upon s then the only remedial measure to solve the circuit is to apply kirchhoff's
17:30law for finding out any particular current or any particular resistance in the given format of
17:39circuit now let's derive the this condition if we apply kirchhoff's voltage law in the circuit
17:49abda i1p plus igg igg and then here since current is opposite with respect to our
18:03traversing direction of the circuit so i2r minus is equal to zero why because this closed loop has
18:13no cell or battery since for balance condition ig is equal to zero therefore i1p plus zero
18:24minus i2r is equal to zero it implies that i1p is equal to i2r equation number one b c d b b c
18:39d b i1 minus ig q minus because again we are traversing this resistor in opposite direction
18:51so i2 minus negative of i2 plus ig and then again ig into g negative is equal to zero for the same
19:04reason again right hand side will be zero because there is no cell or battery in this
19:10closed loop then for balance condition ig is equal to zero i1q minus i2s minus zero
19:28is equal to zero it implies that i1q is equal to i2s this is equation number two one divided by
19:39two i1p upon i1q i2r upon i2s i1 i1 cancel i2 i2 cancel therefore p upon q is equal to
19:53r upon s proved then here comes meter bridge meter bridge or slide wire bridge
20:02first most important point is this meter bridge practical form of and another point of interest is
20:11why it is also named as slide wire bridge for this to understand we need to consider
20:19primary circuit diagram for meter bridge i'll be pasting one diagram so in context to the
20:30diagram pasted here firstly i'll explain here battery current i coming here here and at this
20:40junction it is splitting some part went here through resistance box and going this way this
20:48way this way and this s is unknown resistance of platinum platinum wire of unknown resistance
20:58and again it is proceeding and here and again remaining part of current went into this part
21:09constituting and from point d there is one branch at this point it is untouched which consists of
21:18one resistance box just to make the circuit balancing it is included mainly it has one
21:28galvanometer also and again if you proceed then this point is jockey which is sliding thing and
21:38if you keep sliding in the current mode there is one point in the process of sliding
21:47jockey will acquire at one point where the null point would be achieved and at null point
21:55galvanometer shows no deflection that means v b minus v d is zero and at null point from a to b
22:10is balancing length l and this part of resistance would be considered as p and from b to c it is q
22:22and corresponding length will be 100 minus l why how because this a a c part is one meter
22:34uniform resistance wire it is fixed that's why this device is known as meter bridge and
22:43and because of this sliding jockey thing it is known as slide wire and for what this device is
22:53used for just to determine the unknown resistance s with the help of this resistance box let me
23:02clarify resistance box also resistance box is a cuboid thing which has plenty of keys suppose if
23:12you pick out 2 ohm keys then that resistance box will start behaving as 2 ohm resistance further
23:22if you pick out 3 ohm keys then it will start behaving as 2 plus 3 5 ohms and so on so on so
23:30for our sake of convenience and feasibility we we pick out keys just to make ourselves
23:40feasible according to our requirements we keep on adding or subtracting resistances then let's
23:49come over to last part of this meter bridge that is how to find out this unknown resistance s
23:58and again i'll remind this is a c is uniform resistance wire of 100 centimeter or one meter
24:07length i hope i have explained all the points related to meter bridge now few lines from the
24:17e upon q is equal to r upon s firstly i'll indicate left part and q is right part which has to be
24:28p is lr q is 100 minus l into r r upon s where r is the resistance per unit length
24:43which is usually one ohm per centimeter as far as lab experiment is concerned
24:50r r cancel imply that s and hence this unknown resistance is is found by applying this
24:59formula so now i'm ending this session hoping potentiometer has to be covered in
25:08another video separately and in that video potentiometer applications including few
25:15numerical examples all will be covered in one single video and for numericals related to these
25:21topics i am to cover in ncrt exercises solved and even hc bomba thing so that's why here i have left
25:33those numericals so thanks for viewing and enjoying my video and just keep enjoying keep studying
25:43and bye see you take care thank you