The given infinite grid consists of hexagonal cells of six resistors each of resistance R. then `R_(12)`

Options

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

Answer

B

Explanation

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

`R_(1)(I)/(3) = R(2I)/(3) or R_(1) =2R`
`R_(eq) = (R R_(1))/(R +R_(1)) = (R2R)/(R +2R) = (2R)/(3)`