When two unequal resistances are connected in series, the value increases to 9 ohm from 2 ohm when they are connected in parallel. If one of the resistance is $6 Ω, then the$ value of the other resistances is [[1]] $Ω .$

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Detailed Solution

The total electric resistance, when the resistances are connected in parallel, is:

R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}

And, the total electric resistance, when the resistances are connected in series, is:

R_{s} = R_{1} + R_{2}

From the above relations, we can have:

R_{p} = \frac{R_{1} R_{2}}{R_{s}}

Therefore,

R_{1} R_{2} = R_{p} R_{s}

If we consider

R_{1} = 6 Ω

, then the other resistance is:

R_{2} = \frac{R_{p} R_{s}}{R_{1}}

R_{2} = \frac{(2) (9)}{6}

R_{2} = 3 Ω

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