As shown, the circuit is made of $8$ different resistors. It is found that when $R_1 = 4\,\,\Omega,$ the resistance between $A$ and $B$ is $2\,\,\Omega.$ Now replace $R_1$ by a $6\,\,\Omega$ resistor, what is the resistance between $A$ and $B$?
A$1$
B$2$
C$6$
D$2.4$
Medium
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D$2.4$
d $(i)$ $\frac{\mathrm{R}_{1} \times \mathrm{x}}{\mathrm{R}_{1}+\mathrm{x}}=2$
$\Rightarrow x=4$
$(ii)$ $\mathrm{R}_{1}=6$
Now $\mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{R}_{1} \times \mathrm{x}}{\mathrm{R}_{1}+\mathrm{x}}=\frac{6 \times 4}{10}=2.4$
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