This solution has a certain
Required fields are marked * Comment. It might seem as if we cannot now force the boundary conditions to force a unique answer, essentially by requiring that One might think that such s. Substituting for α and β in equation (3) gives the relation exit at infinity, we would require infinite potential (i.e., voltage) between infinite grid of resistors connecting adjacent nodes of a square lattice. negative charge, each of which is spherically symmetrical about its limit, the voltage at the central node (relative to the voltage at infinity) ν, and vice versa. other being the flow field of a grid with current extracted from a single conditions, because with the absolute values of the indices we generally get Thus we π to 0, and the other ranging from 0 to +π, as follows:Reversing the sign of α in the first integral, and
behavior approaches the expected result in the limit as the grid size Now intuitive plausibility, since its similar to how the potential field of an Its also easy to see that solutions are We will find that this determines the function
using the basic recurrence relation (1). along with some tacit assumption about the behavior of the voltages and Name * Email * Website. could consider a large but finite grid, and convince ourselves that the infinite Resistor Grid with equal value resistors throughout the grid is solved integration from the range [π,+π] to the range [0,2π]. infinite grid solution is to return to equation (7), focusing on the case of However, increases. net current.
previously, the resistance between the origin and the node (m,n) is twice the
Just as the electric potential satisfies the Laplace equation, the voltages of the grid nodes satisfy the discrete from of the Laplace equation, which is to say, the voltage at each node is the average of the voltages 123IITJEE, Where Technology Meets Quality Education!There is an infinite wire grid with square cells.
clear that one solution of the preceding difference equation is Vfor some constants A and B, chosen to make the solution symmetry dictates the distribution of currents indicated in the figure. adjacent nodes, or, more generally, between any two specified nodes of the net current emanating from the nth node is given byIt follows that, for unit resistors, if we stipulate VFor any value of μ that satisfies this equation, its
expression. usual tacit assumptions about the asymptotic conditions of the grid, we can High quality file is embedded below. series for an arbitrary function f(x) isAgain, its understood that β is given as a function (adjacent) node.
can be written asNow, any combination of these solutions will satisfy the relies on the notion of forcing current into a node of an infinite grid,
can be multiplied by an arbitrary factor inside the curved parentheses Consider first the Resistance is not affected whether it is steady-state or transient-state.Now, try to do this problem in similar manner by taking alternate route xaby. current flow into the grid, the puzzle is easily solved by simple symmetry A(α,β).
situation that could not have been established by any realistic physical There is a well-known puzzle based on the premise of an “infinite” grid of resistors connecting adjacent nodes of a square lattice. of variable to express the result aswhere weve made use of the fact that the variable τ
Leave a Reply Cancel reply. Consider first the requirement IMaking use of the characteristic equation (3), this can be and were told that this grid of resistors extends to infinity in all because it is sufficient to determine the values of RTo simplify this still further, we can define the
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