In a meter-bridge if the left and right gaps are connected with `2 Omega` and `3 Omega` resistances, respectively then the bridge is balanced. The resistance to be connected with `3 Omega` resistance to get the balancing point at midpoint of the bridge wire is

Options

A. `3 Omega` in series
B. `3 Omega` in parallel
C. `6 Omega` in series
D. `6 Omega` in parallel

Answer

D

Explanation

In a meter- bridge, if the balancing point is at the midpoint of the bridge wire, then it shows the resistance gaps have equal resistance.
Given, `R_(1) = 2 Omega`
and `R_(2) = 3 Omega`
For meter-bridge,
`(R_(1))/(R_(2)) = (l_(1))/(L-l_(1))`
Where, `l_(1)` is balancing length,
`(R_(1))/(R_(2)) = (l_(1))/(l_(1)) " "(because l_(1) = L - l_(1))`
`rArr R_(1) = R_(2)`
`rArr 2 Omega = 3 Omega||x Omega`
Here, x is the parallel balancing resistance to `3 Omega` resistor.
Hence, `(3x)/(3+x) = 2`
`rArr = x = 6 Omega`
Hence, the correct option is (d).