• # question_answer The effective resistance across the points P and Q is A) $\frac{r}{4}$ B)  $\frac{r}{2}$ C)  $\frac{r}{8}$ D)  $\frac{r}{16}$

Since, the potential at${{A}_{1}}F$ is same. The potential at A, O and E is same. The potential at C and D is same. $\therefore$The circuit can be reduced to The equivalent resistance of part (i) is $\frac{1}{{{R}_{1}}}=\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\frac{1}{r}$ or ${{R}_{1}}=\frac{r}{4}$ Similarly, ${{R}_{2}}=\frac{r}{4}$ ${{R}_{PQ}}={{R}_{1}}+{{R}_{2}}=\frac{r}{2}$