The effective resistance between points $P$ and $Q$ of the electrical circuit shown in the figure is
AIIMS 2013, Medium
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The circuit is symmetrical about axis $POQ$. Therefore the equivalent circuit is drawn
$\therefore $ $\frac{1}{\mathrm{R}_{\mathrm{PQ}}}=\frac{1}{4 \mathrm{R}}+\frac{1}{4 \mathrm{R}}+\frac{1}{2 \mathrm{r}}$ $=\frac{1}{2 \mathrm{R}}+\frac{1}{2 \mathrm{r}}$ $=\frac{\mathrm{R}+\mathrm{r}}{2 \mathrm{Rr}}$
$\Rightarrow \mathrm{R}_{\mathrm{PQ}}=\frac{2 \mathrm{Rr}}{\mathrm{R}+\mathrm{r}}$
If a potential difference is applied across $P$ and $Q,$ there will be no currents in arms $AO$ and $OB$. So these resistance will be ineffective.
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