The effective resistance between points P and | vedclass

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Question

(A) The circuit is symmetrical about the axis $POQ$. Due to symmetry,the potential at nodes $A$ and $B$ is the same as the potential at node $O$. Thus

Vedclass · Accepted Answer

$\frac{2Rr}{R + r}$

Vedclass · Answer

(A) The circuit is symmetrical about the axis $POQ$. Due to symmetry,the potential at nodes $A$ and $B$ is the same as the potential at node $O$. Thus,no current flows through the vertical resistors of resistance $2R$ connected to $O$. The circuit simplifies to three parallel branches between $P$ and $Q$: $1$. Top branch: Two $2R$ resistors in series,total $= 4R$. $2$. Middle branch: Two $r$ resistors in series,total $= 2r$. $3$. Bottom branch: Two $2R$ resistors in series,total $= 4R$. The equivalent resistance $R_{PQ}$ is given by: $\frac{1}{R_{PQ}} = \frac{1}{4R} + \frac{1}{2r} + \frac{1}{4R} = \frac{2}{4R} + \frac{1}{2r} = \frac{1}{2R} + \frac{1}{2r} = \frac{r + R}{2Rr}$. Therefore,$R_{PQ} = \frac{2Rr}{R + r}$.

The effective resistance between points $P$ and $Q$ of the electrical circuit shown in the figure is

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