Question

Two resistances are joined in parallel whose resultant is $\frac{6}{8}$ohm. One of the resistance wires is broken and the effective resistance becomes$2\Omega$. Then the resistance in ohm of the wire that got broken was

• A. 3/5
• B. 2
• C. 6/5
• D. 3

Answer 1

Answer: (C)

6/5

Answer 2

If resistances are $R_{1}$ and $R_{2}$ then $\frac{R_{1}R_{2}}{R_{1} + R_{2}} = \frac{6}{8}$ …..(i)

Suppose $R_{2}$ is broken then $R_{1} = 2\Omega$ …..(ii)

On solving equations (i) and (ii) we get $R_{2} = 6/5\Omega$