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(C) Let the central node be $O$. The circuit consists of four resistors connected between $x$ and $O$,and $y$ and $O$. Specifically,the resistors $8\,

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$(4 / 3)\, \Omega$

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(C) Let the central node be $O$. The circuit consists of four resistors connected between $x$ and $O$,and $y$ and $O$. Specifically,the resistors $8\, \Omega$ and $2\, \Omega$ are connected in parallel between $x$ and $O$. Their equivalent resistance $R_1$ is given by $1/R_1 = 1/8 + 1/2 = 5/8$,so $R_1 = 1.6\, \Omega$. The resistors $4\, \Omega$ and $6\, \Omega$ are connected in parallel between $y$ and $O$. Their equivalent resistance $R_2$ is given by $1/R_2 = 1/4 + 1/6 = 5/12$,so $R_2 = 2.4\, \Omega$. Now,the resistors $4\, \Omega$ (top) and $4\, \Omega$ (bottom) are connected in parallel between $x$ and $y$. Their equivalent resistance $R_3$ is $4/2 = 2\, \Omega$. However,looking at the circuit,the path $x-O$ and $O-y$ are in series. $R_{xOy} = R_1 + R_2 = 1.6 + 2.4 = 4\, \Omega$. Finally,this $R_{xOy}$ is in parallel with the $4\, \Omega$ resistor connected directly between $x$ and $y$. $1/R_{xy} = 1/4 + 1/4 = 2/4 = 1/2$. Therefore,$R_{xy} = 2\, \Omega$. Wait,re-evaluating the circuit: The top $4\, \Omega$ and bottom $4\, \Omega$ are in parallel with the series combination of $(8||2)$ and $(4||6)$. $R_{xy} = (4 || 4) || ( (8||2) + (4||6) ) = 2 || (1.6 + 2.4) = 2 || 4 = (2 	imes 4) / (2 + 4) = 8 / 6 = 4/3\, \Omega$.

The total resistance between $x$ and $y$ is ...... $\Omega$.

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