Four resistances of `16 Omega , 12 Omega , 4 Omega and 9 Omega` respectively are connected in cycle order so from a Wheatstone bridge. Calculate the resistance in be connected in parallel with `9 Omega` resistance to balance the bridge.

Answer

` 13.1 Omega `

Explanation

Here `P = 16 Omega Q = 12 Omega R = 9 Omega , s= 4 Omega`
let ` 9 Omega ` be shoutted with resistance `x` to balance the bright .Then effective rsistence of ` 9 Omega` and `x`
ohm in parallel `R = (9s)/(9 + s)`
For nbalanced bridge `(P)/(Q) = (R)/(S)`
`(16)/(12) = (9 s // (9+ s))/(4) or (4)/(3) = (9s)/(4(9+ s))`
or` 144 + 16 s = 27 s or s = 13.1 Omega `