In the arrangement shown, the magnitude of each resistance is 1Ω. The equivalent resistance between O and A is given by |
$\frac{14}{13} \Omega$ $\frac{3}{4} \Omega$ $\frac{2}{3} \Omega$ $\frac{5}{6} \Omega$ |
$\frac{3}{4} \Omega$ |
Both B and D are symmetrically located with respect to points O. Hence, the figure can be folded as shown in figure. The given circuit forms Wheatstone bridge. ∴ $R_{S}=\frac{1+1}{2}=\frac{3}{2} \Omega$ Resistance, $R_{\text {net }}=\frac{\frac{3}{2} × \frac{3}{2}}{\frac{3}{2}+\frac{3}{2}}=\frac{3}{4} \Omega$ |