For the network shown in the figure, the value of the current `i` is

Options

A. `(9V)/(35)`
B. `(8V)/(18)`
C. `(5V)/(9)`
D. `(18V)/(5)`

Answer

B

Explanation

(b) The circuit given resimbles the balanced Wheatstone bridge as `(4)/(6) = (2)/(3)`
Thus, middle arm containing `4 Omega` resistance will be indffective and no current flows through it.
The equivalent circuit is shown as below :
Net resistance of `AB` and `BC`
`R' = 4 + 2 = 6 Omega`
Net resistance of `AD and DC`
`R" = 6 + 3 = 9 Omega`
Thus, parallel combination of `R'` and `R"` gives
`R = (R' xx R")/(R' + R") = (6 xx 9)/(6 + 9) = (54)/(15) = (18)/(5) Omega`
Hence, current `i= (V)/(R ) = (V)/(18//5) = (5 V)/(18)`