In Wheatstone bridge, the resistances in four arms are `10Omega,10Omega,10Omega` and `20Omega`. To make the bridge balance, resistance connected across `20Omega` is

Options

A. `10Omega`
B. `5Omega`
C. `20Omega`
D. `40Omega`

Answer

C

Explanation

`R_(1)=10Omega, R_(2)=10Omega, R_(3)=10Omega, R_(4)=20Omega, x=?`
To balance bridge, `(R_(1))/(R_(2))=(R_(3))/(R_(4))`
In the given problem, `(10)/(10)!=(10)/(20)`
Thus, the bridge is unbalanced. Hence to balance the bridge, equivalent resistance of `R_(4)` and shunt x should be `10 Omega`.
`:.R_(p)=(R_(4).x)/(R_(4)+x)`
`10=(20x)/(20+x)`
`20x-10x=200`
`10x=200`
`x=20 Omega` .