Four resistances arranged to form a Wheatstone's network are `8Omega, 12Omega, 6 Omega,` and `27Omega`. The resistance that should be connected across `27Omega` resistance to balance the bridge is

Options

A. `13.5Omega`
B. `15.5Omega`
C. `27Omega`
D. `12Omega`

Answer

A

Explanation

`R_(1)=8 Omega, R_(2)=12 Omega, R_(3)=6 Omega, R_(4)=27Omega, x=?`
To balance bridge
`(R_(1))/(R_(2))=(R_(3))/(R_(4))` i.e. `(8)/(12)!=(6)/(27)`
Thus, the bridge is unbalanced. Hence to balance bridge the equivalent resistance of `R_(4)` and shunt x should be `9 Omega`.
`:.R_(p)=(R_(4)xx x)/(R_(4)+ x)`
`9=(27x)/(27+x)`
`27x-9x=9xx27`
`x= (9xx27)/(18)=13.5 Omega`