In a Wheatstone's network, `P=2Omega, Q=2Omega, R=2Omega and S=3Omega`. The resistance with which S is to be shunted in order that the bridge may be balanced is

Options

A. `4Omega`
B. `1Omega`
C. `6Omega`
D. `2Omega`

Answer

C

Explanation

Give, `P=2Omega, Q = 2Omega, R = 2 Omega` and `S = 3 Omega`
So, by using of x resistance, S is to be shunted So, according to the Wheatstones network, we have
`(P)/(Q)=(R )/(S)=(R )/((S xx x)/(S+x))`
`(2)/(2)=(2)/((3 xx x)/(3+x))=1=((3+x)2)/(3x)`
`3x=6+2x rArr 3x - 2x = 6`
`rArr " " x = 6 Omega`