The given infinite grid consists of hexagonal cell of six resistor each of resistance `R`. Then `R_(12)` is equal to

Options

A. `(R )/(3)`
B. `(2R)/(3)`
C. `(4R)/(3)`
D. `(3R)/(4)`

Answer

B

Explanation

(b) Here we apply superposition priciple.
(ii) Let us feed a current `I` to point 1, but do not draw the current from 2 and allow the current to spread to infinity, Then current in branch `'12` will be `I//3`.
(ii) Secondly, feed current `I` to the current at at take it out from 2, now again current through branch '12' will be `I//3`.
Now carry out both the above steps (i) and (ii) simultaneouly. Then current in branch '12' will be sum of above two currents i.e.,`(I)/(3) + (I)/(3) = (2 I)/(2)`
Resistance between 1 and 2 is `R`. Let remaining resistance is `R_(1)`
Then `R_(1) (I)/(3) = R (2 I)/(3)`
`R_(eq) = (R R_(1))/(R + R_(1)) = (R 2 R)/(R + 2 R) = (2 R)/(3)`