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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....
00:00so hello friends last from last couple of weeks we were into continuous lecture session and now
00:11again current electricity and CRT and this is salt example and already in that direction we
00:19have covered one video in which we had current electricity 3.1 salt example now we'll have 3.5
00:29and that previous video a link I'll be sharing in my video so that you could go through that also so
00:38here comes 3.5 what is given in this problem we would be needing Kirchhoff's law and for that
00:46already in theory we have covered Kirchhoff's law how to tackle Kirchhoff's junction rule and Kirchhoff's
00:54voltage law both the laws how do we this you can go through from my current electricity 05
01:02now in this question this is 10 volt battery this is source of current in this complex network having
01:15the size of Q that means this complex network is given in the form of Q which has 4 4 8 and 4 12
01:26all 12 edges has equal resistances say it would be R and here it is given 1 ohm and all the 12 edges
01:37that means all 12 edges are 12 resistors of reserve 1 ohm each now we have to find the equivalent
01:48resistance of the network such that equivalent resistance between diagonally opposite ends so
01:55here in the diagram as question depicts equivalent resistance between A and C dash these are two
02:05diagonally opposite points firstly here the point first and foremost point we need to understand is
02:13we are to take that means consider assume 3i of current battery is sending 3i at the junction A and such
02:26that iii that means it is splitting in equal proportions in all the three directions that is i i and i why what's the
02:39reason behind this kind of assumption because the circuit is the given circuit is symmetrical and from all the
02:47route whichever route current decides to proceed it will have to cover shortest three edges
02:56say this this this and this second route this this and this then third route this this and this then this this this this that means any
03:10route we follow we'll have to cover three edges that means potential is equally distributed for all the three routes that's why here in the given
03:22situation equal division so hence current i and further i will be splitting into i by two and this i also i by two i by two and this i also i by two and i by two then in next next stage all i by two i by two will be combining in to see this i by two and this i by two will be combined to form
03:52i then this i by two and this i by two and this i by two and this i by two and this i by two and this i by two again will be combining to form i so hence all these three i i i i and i will be combining to reform three i which is going back to bat hence this is rational
04:22assumption of currents of currents across all the edges and roots now how do we apply kirchhoff's law so this was from symmetry after considering of all the currents in each of the edges across all the resistors let us consider one closed loop here in the circuit
04:49let us first decide a b c c c dash e a this closed loop i'm applying one different color across this so this closed loop already i have pointed out i won't discuss i'm not going to discuss how to apply kirchhoff's law i have chosen the closed loop now directly i i am to apply because already i have explained how to apply kirchhoff's laws
05:17in detail in my theory lecture notes so this closed loop a b c c c dash e a and in brief let me remind for closed loop we apply the technique sigma i r is equal to sigma e so all i are sum i r plus i by two r plus i r is equal to ten and let us
05:47recall again recall again what we call again what we are to find we are to find the equivalent resistance between a and c dash to find r equivalent that is r a c dash this we are to find and this has become
06:05it implies that one i r two i r two i r two plus one and a half that means five by two five by two r is equal to ten let us consider e only this is equation number one and also across these two points a and c dash e is also equal to ir also across the points a c dash e is equal to
06:35the main potential is equal to main current main current main current that means three i already we are considering main current as three i so three i here also three i multiplied by required equivalent resistance r a c dash and this has become equation number two from equation one and two five by two i
07:05r is equal to five by one and two now what do we cancel it implies that r a c dash is equal to five by six r so we can generalize also now as far as the answer is concerned here in the given question capital r was given one ohm you can see here the each of resistance one so capital r is given therefore
07:34r r a c dash is given one that means five by six so hence it is done now as we can see here in the question determine the equivalent resistance of the network and second part the current along each edge of the cube edge of the cube that means if we consider
08:03now already from equation number two now already from equation number two or from equation number one also we can find if we use five by two i this r is given one and this e is given ten so i could be found and here across some of the edges i by two current is also going that means whatever value of i will be obtaining for these edges ad
08:31and a a a dash and ad and this was uh and ad across these three edges i otherwise uh this this this this this and and and this all these are i by two and again lastly this this and these three edges will have a current of i so let's first find the i now from equation number one
08:57second part from equation number one yeah five by two ir is equal to e five by two ir is equal to e five by two ir is equal to e five by two i two one is equal to ten it implies that i is equal to four ampere hence i'll be naming i'll be listing all the edges hence following are the list of edges
09:27along along with their currents along with their currents now i'll be naming edges a a dash a d a b then lastly these were the three and d dash c dash b dash c dash and c c dash all these are having i that means four ampere
09:53three and then четы and four and three are seven seven two four five five five six now remaining edges now again i'll be
10:01listing concern edges which has got i by two a dash d dash a dash b dash d dash d d dash d c then b b dash all these six edges has a current of i by two that means two ampere so determine the current
10:21in each branch of the network shown again this diagram is given in the question sometimes in
10:31the given diagram current is not shown in that case we need to assume from our side what are
10:42the currents but here it is more feasible to use now directly we don't have to judge anything
10:52or anticipate directly we we are to start writing the equations of decided closed loops
11:00and wherever we need let us see this is i1 this is i2 and this is i1 minus i2 that means
11:12junction rule from junction rule incoming current is equal to outgoing current so here it is
11:21not appearing proper to use junction rule because directly currents are shown here this these
11:29this and this i2 plus i3 again directly junction rule is applied while assuming currents so no
11:38need to apply again that means directly we need to decide the closed loops a d c a this closed
11:50loop we are to decide first and here only applying the jobs second rule on the closed loop a d c
12:03a d c a a d c a 4 into i1 minus i2 traversing circuit in opposite direction with respect to current so
12:15negative minus 2 into i2 plus i3 minus i1 now again along the direction of current so 1 into i1 plus 1 to
12:32i1 is equal to i1 is equal to 10 volt for i1 minus 4 i2 this is equation number one considering close loop
12:47a b c a now this this time this close loop how many unknown variables are there i1 i2 and i3 and what is to
13:00be found determine the current in each branch of the network and for this purpose for this purpose we
13:10need to find i1 i2 and i3 first then only we could find all the branches currents because all these currents
13:18are expressed in the form of i1 i2 i3 only this time a b c a 4 i2 then plus 2 i2 plus i2
13:30i3 plus 1 i1 is equal to 10 it implies that i1 plus 6 i2 plus 2 i3 is equal to 10 this has become
13:46equation number two equation number one two now third equation this one b c t e b like this for three
14:00unknown variables at least we would be need needing three linear equations simple law of algebra now for
14:08this closed loop again i'll be writing closed loop b c d e b keep checking from the diagram two into
14:21i2 plus i3 all these are coming positive because we are traversing our circuit along the direction of
14:33current otherwise we would be having negative this is not a rule this is uh our convention it is up to you
14:43which convention you follow but i suggest you simply apply this only plus two into i2 plus i3 minus i1
14:57that is equal to 5 just now we saw uh equation number one and two then after that close loop b c d e d i'll show you
15:13the initially equation number one was a d c a this is the equation number one which we opened this now a b c a
15:25a b c a from here we get equation number two this one now from the loop b c d e b this loop now let's see
15:41what do we get 2 i2 plus i3 plus 2 2 i2 plus i3 minus i1 that is equal to 5 okay it implies that let's see
15:55minus 2 i1 2 i2 2 i2 plus 4 i2 2 i3 and 2 i3 again plus 4 i3 is equal to 5 now third equation this has become
16:17third equation clearly visible from third equation clearly visible from equation 1 and 2 if we add up
16:23these two equations now adding equation 1 plus 2 we get we get simply 7 plus 1 8 i1 equals
16:39plus 20 4 plus 4 i1 is equal to 5 by 2 equals 2.5 ampere one value now we have got now remaining i2 for i2
16:58and i3 what do we let's do one thing multiplying by 2 if we uh multiply equation number 2 by 2 then here we
17:12obtain 4i3 and here 4i2 then we we can subtract and in the process of subtracting this i3 will cancel out
17:25and only an i1 value and only an i1 value already we have got that is 2.5 and this way i2 could be found
17:36now my equation number 2 multiplied by 2 we have 2i1 2i1 plus 12i2 plus 4i3 is equal to
17:37and then equation number 2 and then equation number 3 minus 2i1 plus 4i2 plus 4i3 is equal to and then
17:51equation number 3 minus 2i1 plus 4i2 plus 4i3 now subtracting these two cancel then
18:034i1 plus 4i1 plus 4i1 plus 4i1 plus 4i1 plus 4i2 is equal to 15 but i1 value already we have
18:14that is 2.5 this is 2.5 this is 2.5 into 4 that is 10 8i2 is 15 minus 10 is equal to 5 it implies that
18:27i2 is 5 by 8 let's check yeah it's correct i2 is also done and hence from any of the values simply
18:41these two value could be substituted in equation number one or any equation whichever you like then
18:50from equation number one so let's pick equation number three equation number three minus 2i1 plus 4i2
19:01plus 4i3 4i2 plus 4i3 is equal to 5 now let's let's put the values for i1 we'll put 2.5 for i1
19:16we'll put 2.5 for i2 5 by 8 let's apply that minus 5 plus 5 by 2 plus 4i3 is equal to 5 15 by 8 this way
19:33after obtaining these three values i1 i2 and i3 we have got all the three values for all the branches
19:45current it was a determine the current in each branch of the network shown and figure i1 was 2.5 ampere and
19:56i2 was 5 by 8 hence all the branches ab that is i2 5 by 8 ampere ac i1 5 by 2 ampere ad i1 minus i2
20:145 by 2 5 by 2 5 by 8 5 by 4 3 minus dd is i2 plus i3 5 by 8 plus 15 by 1 plus 3 4 4 5 1 20 by 8 4 5 1 24
20:34bd bd pc i2 plus i3 5 by 8 5 by 2 ampere then cd i2 plus i3 minus i1 5 by 2 minus oh it should be 0
20:54d ed last branch it is i3 i3 plus 15 by 8 after getting these three values i1 i2 i3 i3 i have arranged
21:09all the branches currents hence this question is done here comes example 3.7 ncrt salt current electricity
21:19this is wheatstone branch unbalanced it is not balanced wheatstone bridge so we need to apply
21:27it jobs law we are supposed to find the current ig in this branch provided 10 volt of battery is is
21:40connected across these two points a and c that means 10 volt is maintained across ac so let's consider
21:53a b d a closed loop then thereafter b c d p these two closed loops we'll be considering for kirchov's law
22:06purpose b a d b is equal to zero equation number one then considering closed loop b c d b then in this
22:16loop 10 i1 minus ig and i1 minus ig minus 5 i2 plus ig minus this has become equation number two now this time
22:29closed loop a d c e a t e a and this is our equation number three so let us first try to eliminate
22:45i2 from equation number three so that we could find ig from equation one and two from equation one and two
22:56let's try to eliminate i1 for this we need to multiply equation number two by 10 so that in equation number two also
23:07we get 20 i1 and already 20 i1 we have in equation number one also so then subtracting both the equation
23:18then multiplying equation number two by 10 so 20 i1 minus 10 i2 minus minus 12 i2 plus 3 ig is equal to zero
23:35now we need to subtract by subtracting we obtain it implies that i2 is equal to and this is
23:45equation number four is equation number four then equation number from equation number three and
23:534 13 i2 plus ig is equal to 2 now we are to 409.5 plus 1 and therefore ig is 2 upon
24:10410.5 it implies that ig is 4.87 milliampere hence done so we have got the answer which was given in
24:27ncrt salt example and this way we have finished all the salt examples ncrt current electricity 12 cbs so
24:41thanks for enjoying all of you my physics loving friends and keep waiting for another videos until then bye
