IIT JEE , AIEEE Forum - Physics , Electricity - the infinite grid again..i dont want the solution as whole..i want to be taught something

biki

Question:-There is an infinite wire grid with square cells.the resistance of each wire between neighbouring joint connections is equal to r.find the resistance of the whole grid b/w points A and B

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My Confusion (read it carefully please...... the whole one)

we apply potential +V and -V on the two terminals.

for +V, say 4i comes out. Then it gets distributed in 4 branches as i,i,i and i.
for -V, say 4i comes in. Then in the part AB,we get i+i = 2i.

Now connceting the whole cell of emf = V - (-V) = 2V

why do we write .. V - (-V) = 4iR = potential drop across equivalent resistance
and V - (-V) = 2ir = potential drop across arm AB of resistance r
or 4R =2r
i,e R = r/2

is the current not 5i ?????
i,i,i through remaining three branches and 2i through AB

Again let us see it like this..........

The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine currents within the modified network for each power source separately. Then, once currents have been determined for each power source working separately, the values are all "superimposed" on top of each other (added algebraically) to find the actual currents with all sources active.

if current in a circuit is i, we can take that the +ve terminal gives current i/2 and -ve terminal absorbs current i/2. Now both being in same direction give total current = i for the circuit with both +ve and -ve terminal connected.

Here we have cosidered that the cell is of emf 2V

So +ve terminal contributes +V and -ve terminal contributes -V.

for +V, we take current = 4i(going out)

for -V , we take current = 4i(coming in)

So total current for the cell should be 8i.

once 8i .......again 5i.........again 4i....???????

please tell me whether there is any conception problem.....

if no, please tell why such different values come......??

if yes, correct my mistake and make me understand...please

PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE

biki

please experts.......help me out......

biki

so many pleases ........ still no answer........

i know that goIITian experts are very helping.......
please keep up my belief........

help me

PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE

bvsatyaram

Hello Biki,

I understand ur confusion.

First to get started, let me make the principle of super position more clear.

In principle of super position, u take 2 things happening independently. (1 thing is independent of other. Any of these can exist without the existance of the other.).
U just take only first thing, observe the effect. Now take only second thing observe.
Principle of superposition says, the observation u make due to existence of both will be sum of the observations u make due existence of each of the things alone.

Now, lets get started.....
take the grid, trow in a current 4i into the grid at any grid point. (remember we are not goin to collect this current from the adjacent grid point.).Since the grid is symmetric, current divides equally that is., i, i, i, i.
Now, take a different case, take out current 4i from any grid point. (once agoin, we are not inputting current any where). So, what ever current u draw, has to be collected symmetrically. So, i, i, i, i.
Now club these 2 cases, so I am sending current 4i through A and collect 4i through B.

Forget about V and -V u were talking about. The current passing through wire B is 2i. So, potential difference accross A and B is 2i *r

Now, let the equivalent resistence be R.

So, replace the whole grid by a resistance R between A and B.
I am sending current 4i at one end and collecting the same current from the other end. (now there is no more grid, so, dont think about i,i,i,i and i. Just think abt outside 4i). So the potential difference accross A and B is R*4i
Therefore, 4i*R=2i*r
which implies R = r/2

kishan12

Toooooo Gooooooooooooooooooooooddddddddd Sir,,,,,,,,,,,,,,,,,,,,,,,,,,,

biki

80% CLEAR NOW

still sir ..

please tell me what happens in the case we superimpose the observations and look at the grid after superposing.....
we get i,i,i thru three branches and 2i thru AB which gives a total current of 5i.

please clear my this doubt sir ....please

i beg you for another reply bvsatyaram sir

bvsatyaram

Well, it's valid doubt. U say when u superimpose, the currents are i,i,i and 2i.
U say the last current is 2i, by adding currents in both cases. But what abt the first thre currents? U r just considering the first case. there will be a significant contribution by the second case, even in the other three branches, of course, we can't find their exact magnitudes.
So, the net currents will be i+i1, i+i2, i+i3 and i+i, where i1, i2 and i3 are currents in 3 branches emerging from A other than AB due to the second case
so, the net current will be 5i+i1+i2+i3. Here u don't know the value of i1, i2 and i3. So, we cant say the answer.. Infact, i1+i2+i3 will turn out to be (-i), making the net current to 4i.

Hope every thing is clear now.

biki

@bvsatyaram sir

is there any proof of this..." i1+i2+i3 will turn out to be (-i) "

if there is one, please write it sir.........and i will start worshipping you ........ !!!!!

how can we be so sure about it that i1+i2+i3 = -i

biki

@bvsatyaram sir
is there any proof of this..." i1+i2+i3 will turn out to be (-i) "

if there is one, please write it sir.........and i will start worshipping you ........ !!!!!

how can we be so sure about it that i1+i2+i3 = -i

bvsatyaram

The equation i1+i2+i3 = -i can be proved usind advanced maths, which u haven't yet seen. So, for the time being, the equation i1+i2+i3 = -i cannot be proved.

But y do u want to prove it?
You get the total current leaving A is 5i+i1+i2+i3. But u don't know a way by which u can find i1+i2+i3. So, go for an alternate method. look from out the circuit. What ever current enters A has to leave A. (as no charge can be stored a a point). So, we conclude that current leaving A is 4i. So, we are getting an answer passing trough a tunnel hiding from 3 monsters i1,i2 and i3.

If u go directly towards the monsters, u have to either give up and loose, or fight against them to win. To fight against them, u need weapons (advanced maths), which are not yet trained to use. So, try escaping it...... Its not a joke... I dont write this to entertain u.... This will be the case with any tough problem u encounter. Follow the same rule.

biki

thank you sir........

pardesi

actually the best thing about this is the solution lie in Random walks a completely(seemingly atleast) different field and the boundary conditions + laplacians

bvsatyaram

I don't get where random walk and laplacian are getting in here?
What is random here? Every thing is well defined and symmetric. Nothing is random.
The same is the case with laplacian....

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