Four resistors, each of resistance $\displaystyle 2 \cdot 0 \Omega$, are joined end to end to form a square ABCD as shown. Using appropriate formula, determine the equivalent resistance of the combination between its two ends A and B.

Between A and B, $\displaystyle 3$ resistors in series = AD + DC + CB in one branch Equivalent resistance, \(\displaystyle \mathrm{R}_{\mathrm{S}}=2 \Omega+2 \Omega+2 \Omega=6 \Omega\) Now one resistor in arm AD is in parallel combination with the other three. Two branches: \[\mathrm{R}_{1}=2 \Omega \quad \mathrm{Rs}=6 \Omega \] \[\begin{aligned} \therefore & \frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{6}+\frac{1}{2}=\frac{1+3}{6}=\frac{4}{6} \\ & \mathrm{R}_{\mathrm{p}}=\frac{6}{4}=1 \cdot 5 \Omega \end{aligned} \]