The given Wheatstone Bridge shows no deflection in the galvanometer joined between the points B and D. Calculate the value of ‘X’.

Explanation

Because given Wheatstone Bridge is in a balanced position,
`therefore (P)/(Q)= (R )/(S)`
`(100)/(Q)= (200)/(40)` ..(i)
But in arm BC, `100Omega and X Omega` resistances are in parallel
Let their net resistance be Q
So `(1)/(Q) =(1)/(100) + (1)/(X)`
`rArr (1)/(Q) = (100+X)/(100X)`
`rArr Q= (100X)/(100+ X)`
Putting value of Q in eqn (1)
`(100 (100 +X))/(100X)= (200)/(40)`
`rArr (100+ X.)/(100X) = (1)/(20)`
`rArr 2000 + 20X= 100X`
`80X= 2000`
`rArr X= (2000)/(80) = 25Omega`