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(D) The circuit can be simplified into two parallel branches connected between points $A$ and $B$. One branch contains the resistor $R_1$,and the othe

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$2.4\,\,\Omega$

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(D) The circuit can be simplified into two parallel branches connected between points $A$ and $B$. One branch contains the resistor $R_1$,and the other branch represents the equivalent resistance $x$ of the remaining $7$ resistors.$(i)$ When $R_1 = 4\,\,\Omega$,the equivalent resistance $R_{AB} = 2\,\,\Omega$. Since $R_1$ and $x$ are in parallel:$\frac{R_1 	imes x}{R_1 + x} = R_{AB}$$\frac{4 	imes x}{4 + x} = 2$$4x = 8 + 2x$$2x = 8 \Rightarrow x = 4\,\,\Omega$$(ii)$ Now,replace $R_1$ with $6\,\,\Omega$. The new equivalent resistance $R'_{AB}$ is:$R'_{AB} = \frac{R_1 	imes x}{R_1 + x} = \frac{6 	imes 4}{6 + 4} = \frac{24}{10} = 2.4\,\,\Omega$.

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 $B$?

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