When two resistance are connected in parallel then the equivalent resistance is $\frac {6}{5}\Omega $ . When one of the resistance is removed then the effective resistance is $2\, \Omega $. The resistance of the wire removed will be ................ $\Omega$
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$\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{6}{5}\, \Omega$
Now, one resistance is removed then $\mathrm{R}_{2}=0$
$\therefore \mathrm{R}_{1}=2\, \Omega$
$\therefore \frac{2 \mathrm{R}_{2}}{2+\mathrm{R}_{2}}=\frac{6}{5}$
$10 \mathrm{R}_{2}=12+6 \mathrm{R}_{2}$
$4 \mathrm{R}_{2}=12$
$\mathrm{R}_{2}=3\, \Omega$
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