In the gives circuit, a meter bridge is shows in a balanced state. The bridge wire has a resistance of `1 Omega cm^(-1)`. Find the value of the unknown resistance `X` and the current draws from the battery of negligible internal resistance.

Answer

`4 Omega, 0.66 A`

Explanation

For the balanced bridge, the ratio of the two resistances is equal to the ratio of the lengths of the two parts `AJ` and `JB` of the wire, i.e.,
`(X)/(6 Omega) = (40 cm)/(60 cm)` or `X = 4 Omega`
No current flows through the galvanometer `G`. The resistances of the parts `AJ` and `JB` are `40 Omega` and `60 Omega`, respectively. If `R` is the equational resistance between the point `A` and `B`, then we have
`(1)/(R ) = (1)/((X + 6) Omega) + (1)/((40 + 60) Omega)`
or `R = (100)/(11) Omega`
`i = (V)/(R ) = (6 V)/((100//11) Omega) = 0.66 A`