Two resistor are joined in parallel whose resultant is $\frac{6}{5} Ω$ . One of the resistance wire is broken and the effective resistance becomes $2 o h m s$ . Then the resistance (in ohms) of the wire that got broken is

\frac{6}{5}

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2

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\frac{3}{5}

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3

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Solution

The correct option is C $\frac{6}{5}$
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 $R_{1} = 2 Ω . . . . . . (i i)$

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

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Two resistor are joined in parallel whose resultant is $\frac{6}{5} Ω$ . One of the resistance wire is broken and the effective resistance becomes $2 o h m s$ . Then the resistance (in ohms) of the wire that got broken is

The correct option is C $\frac{6}{5}$
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 $R_{1} = 2 Ω . . . . . . (i i)$

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