For the circuit shown in fig. the equivalent resistance between A and C is

Options

A. `(12)/(11)r`
B. `(13)/(11)r`
C. `(14)/(11)r`
D. `(15)/(11)r`

Answer

D

Explanation

At junction E or F:
`I_(1) +I_(3) +I_(4) = I_(2) - I_(3) or I_(1) - I_(2) +2I_(3) +I_(4) = 0`
Loop AFBA or loop ECDE: `rI_(1) = rI_(2) +rI_(3) or I_(1) =I_(2) +I_(3)`

Loop BEFB or loop EDFE:
`r(I_(2) -I_(3)) +rI_(4) = rI_(3) orI_(2) -2I_(3) +I_(4) = 0`
Loop AFDCMA or ABECMA:
`V = rI_(1) +r(I_(2) - I_(3)) +rI_(2) =rI_(1) + 2rI_(2) l- rI_(3)`
Solve to get `I_(1) =2V//5r, I_(2) =V//3r`
`R_(eq) = (V)/(I) = (V)/(I_(1)+I_(2)) = (V)/(2V //5r + V//3r) = (15r)/(11)`