The current enters at A and comes out at D. Some of the resistance are shown. What should be resistance of wire CB so that it draws double of the current that enters the wire BF.

Options

A. `((9)/(10))R`
B. `((8)/(9))R`
C. `((7)/(9))R`
D. `((3)/(20))R`

Answer

D

Explanation

Let current in AB be `I_(1)` in BF be `I_(2)` in BC be `2I_(2)`, in FC be `I_(3)`
in FE be `I_(3)` and in CD be `(I_(3) + 2I_(2))`
At point `B, I_(1) =3I_(2)`
At point `F, I_(2) = 2I_(3)`
Along ABFCD,
`V_(A) -V_(D) =V =I_(1) ((2R)/(5)) +I_(2)((R )/(5)) + I_(3) ((R )/(5)) + (I_(3)+2I_(2))(R )/(5) =2I_(2)R`
Along ABCD,
`V_(A) -V_(D) =V =2I_(2)R`
` =I_(1)((2R)/(5)) +2I_(2)R' +(I_(3) +2I_(2))(R )/(%)`
or `2I_(2)R' = (3)/(10)I_(2)R or R' =(3)/(20)R`