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)`
`rArr (100)/(Q) = (75)/(15)` …(1)
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) = (100 +X)/(100X)`
`rArr Q= (100X)/(100+X)`
Putting value of Q in eqn (1)
`(100(100+X))/(100 X)= (75)/(15)`
`rArr (100 + X)/(100X) = (1)/(20)`
`rArr 100X= 2000+ 20X`
`rArr 100X-20 X= 2000`
`rArr 80 X= 2000`
`rArr X= (2000)/(80) rArr X= 25Omega`