Each component in the infinite network shwon in fig has a resistance `R =4 Omega`. A battery of emf1 V and neglibible interanal resistance is connected between any two neighboring points, say X and Y.
i. Find current shown by the ammeter.
ii. if the resistance R between X and Y is removed , then find current shown by ammeter.

Explanation

understanding of symmetry and superposiiton could
greatly help in solving the given cirucit. A battery is connected
between any tow neighboring points x and Y . Consider first
the distribution of curent at X as shown in fig. (a). A
current I enters the cirucit at X and distributes in the infinite network. by symmetry, the current will be divided equally
in the four resistance conneced to X. this is beacues of the mutual equivalent of sthe four possible direciton in which the current distributes at X.


Now consider the neighboring point Y, which is connected
to the negative terminal of the battery. By symmetry and
independence of our earlier discussion, if a current I has to
leave the circuit at Y, current each `I//4` will be directed toward
B through the four resistors connected to Y, as shown in fig
Let us now superpose the two cases so that a current I enters
at X and leaves at Y. Superposition of figs. is shown
fin in the resistance R between X and Y, the two currents in figs (b) being in the same direction, add
up and a total current `I//2` ,passes through it. Thus, current I divides at X such that `I//2` goes through this R and the remaining
`I//2` through the rest of circuit, and at Y, these currents are
combined resulting in current I leaving the network.
It implies that we can cosider the network to consist of
two resistance connected in parallel between X and Y. One
of these is the resistane R between X and Y and the other is
the equivalent resistance of the rest of cirucit. This shown in fig.
,
From fig (a) we have `V =(1)/(2)R`
and from (b) we have `V = IR_(eq)`
`IR_(eq) =(I)/(2)R or R_(eq) = (R )/(2)`
Hence, the equivalent resistance resistance fo the network between X and
Y or any two neighboring points si `R//2`
`V = IR_(eq) , I = (V)/(R_(eq)`
`R_(eq) = (R )/(2), I = (2V)/(R )`
Given `V =1 V and R =4 Omega`
`I = (2)/(4) = 0.5A`