# 19. Advance Illustration | Current Electricity | Equivalent Resistance of an Infinite Grid | (GA)

in this illustration we’ll demonstrate you
how to calculate the equivalent resistance of an infinite grid. here the figure shows
an infinite wire grid with square cells, and the resistance of each wire between neighboring
junction is r. and here we are required to find the equivalent resistance of the grid
across these 2 points, ay, and b. and resistance of each wire here is, r. now in this situation,
we solve this problem by using principal of superposition where, you can see. if we connect
a battery, between terminal ay and, it is ground like this. say this is a battery of
voltage v and it is connected to ground. and say if we consider b terminal to be open then,
due to symmetry. every branch connected to all these 4 branches the current will be equally
divided because here we are assuming these 2 be kept open. so if a current i is, getting
into terminal ay. it’ll be divided as, i by 4, i by 4, and so on. in all 4 branches.
so we can say here if. i current. enters at, ay. then. current flowing through the branch
ay b will be, i by 4, this by symmetry if b is kept open. and, if in another case. we
keep this ay open. and we connect this battery b. with the battery in opposite polarity with
low potential terminal at b and high potential is grounded. and we keep, ay open. then here
you can see again the same amount of current i which was flowing, and getting into ay.
will be withdrawn form point b, through the battery because battery is same and the grid
is also symmetric. and again we can see, if. i current. leaves at b. then by symmetry we
can see from point b in all 4 branches i by 4 i by 4 will be contributed. and this current
i is, constituted so here again i ay b is equal to i by 4. so here, if, both of these.
situations. are superposed. then we can say if a current i is getting at point ay and
i is leaving at b, then simultaneously if a battery of voltage v is connected between
the 2. again a current i flows but, the current in the wire ay b will be i by 2. say if both
the situations are superposed, and, a battery of, e m f. v is connected. across, ay and
b. the current through. wire ay b is. in this situation the current through wire ay b will
be i by 4 plus, 1 by 4 that is equal to 1 by 2. so if, the current i is, distributed
in such a way that in wire ay b a current, i by 2 is flowing. that means through rest
of the grid also. the current flowing is i by 2. so here in this situation we can write.
the resistance. of whole grid. across terminals, r, ay and b can be written as resistance of
terminals, ay and b. in parallel combination with the same resistance r because current
between ay b and remaining grid is equally divided. so rest of the grid will also have
resistance equal to that of the wire ay b. then only we can say that current is half,
supplied through wire ay b and remaining half through, the rest of the grid so it’ll be
equals to, r by 2. that will be the final result of this problem.

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