Find the current in the resistor shown in figure.

Explanation

In AHGBA
` 2+(i-i_1)-2 = 0 `
` rArr i-i_1 = 0 `
In circuit CFEDC,
` - (i_1 -i_2)+2 +i_2 -2 = 0 `
` rArr i_2 - i_1 + i_1 = 0 `
` 2i_2 - i_1 = 0 `...........(i)
In circuit BGFCB,
` rArr -(i- (i_2)) + 2 - i_2-2 = 0 `
` rArr i_2 -i_1 -i_1 = 0 `
` rArr 2i_1 - i_2 = 0 ` .........(ii)
from (i) and (ii), we get
` i_2 = 0 `
` i_1 -i_2 = 0 `
Current in the three registers is 0 A.