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(A) Let the top-left node be $C$. The circuit consists of three resistors of resistance $R$ connected between nodes $a$,$b$,and $C$. Specifically,ther

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$\frac{5}{8} R$

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(A) Let the top-left node be $C$. The circuit consists of three resistors of resistance $R$ connected between nodes $a$,$b$,and $C$. Specifically,there is a resistor between $a$ and $C$,a resistor between $C$ and $b$,and a resistor between $a$ and $b$. Additionally,there is a resistor between $C$ and $b$ (the diagonal one) and another resistor between $a$ and $b$ (the bottom one). Wait,looking at the circuit diagram: $1$. There is a resistor $R$ between $a$ and $C$. $2$. There is a resistor $R$ between $C$ and the top-right node. $3$. There is a resistor $R$ between the top-right node and $b$. $4$. There is a resistor $R$ between $C$ and $b$. $5$. There is a resistor $R$ between $a$ and $b$. Actually,the circuit simplifies as follows: The two resistors in series on the top branch (between $C$ and $b$) have a total resistance of $R + R = 2R$. This $2R$ is in parallel with the diagonal resistor $R$. The equivalent resistance of this parallel combination is $R_p = \frac{2R \cdot R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3} R$. Now,this $R_p = \frac{2}{3} R$ is in series with the resistor $R$ connected between $a$ and $C$. So,the total resistance of this upper branch is $R + \frac{2}{3} R = \frac{5}{3} R$. Finally,this $\frac{5}{3} R$ is in parallel with the bottom resistor $R$ connected directly between $a$ and $b$. The equivalent resistance $R_{ab}$ is given by: $R_{ab} = \frac{(\frac{5}{3} R) \cdot R}{(\frac{5}{3} R + R)} = \frac{\frac{5}{3} R^2}{\frac{8}{3} R} = \frac{5}{8} R$.

Find the equivalent resistance between the points $a$ and $b$.

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