Two resistances are joined in parallel whose reusltant is `6//8 ohm`. One of the resistance wire is broken and the effective resistance becomes `2 Omega` . Then the resistance in ohm of the wire that got broken was

Options

A. `3//5`
B. 2
C. `6//5`
D. 3

Answer

C

Explanation

(C ) If resistance are `R_(1)` and `R_(2)` then `(R_(1) R_(2))/(R_(1) + R_(2)) = (6)/(8)` ……..(i)
Suppose `R_(2)` is broken then `R_(1) = 2 Omega ………(ii)`
On solving equations (i) and (ii) we get `R_(2) = 6//5 Omega`