The equivalent resistance of the resistances (two) joined in parallel is 6/5 Ω. When one of the resistance wire is broken, the effective resistance becomes 2Ω. The resistance of the wire that got broken was :

Concept:

For the resistance connected in series, the equivalent resistance is:

⇒ Req = R1 + R2  ------ (1)

For the resistance connected in parallel combination, the equivalent resistance,

\(⇒ \frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_{2}}}}\)     ----(2)

Calculation:

Given:

Case-I:

Case-II:

Let's assume R1 got broken, therefore R2 = 2Ω 

For parallel combination, Req = 6/5 Ω, R2 = 2Ω as given above;

\(⇒ \frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_{2}}}} \)00

\(⇒ \frac{5}{{{{6}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{{2}}}} \)

\( \frac{1}{{{R_1}}} = \frac{1}{{{{3}}}}\)

∴ R1 = 3Ω