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Question

(A) The circuit exhibits symmetry about the horizontal axis passing through the central branch. The vertical resistors of resistance $2R$ are connecte

Vedclass · Accepted Answer

$2Rr/(R + r)$

Vedclass · Answer

(A) The circuit exhibits symmetry about the horizontal axis passing through the central branch. The vertical resistors of resistance $2R$ are connected between nodes that are at the same potential due to this symmetry. Therefore,no current flows through these vertical resistors,and they can be removed.After removing the vertical resistors,the circuit simplifies to three parallel branches:$1$. The top branch has two $2R$ resistors in series,giving a total resistance of $2R + 2R = 4R$.$2$. The middle branch has two $r$ resistors in series,giving a total resistance of $r + r = 2r$.$3$. The bottom branch has two $2R$ resistors in series,giving a total resistance of $2R + 2R = 4R$.Now,we have three resistors $4R$,$2r$,and $4R$ in parallel. The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{4R} + \frac{1}{2r} + \frac{1}{4R} = \frac{2}{4R} + \frac{1}{2r} = \frac{1}{2R} + \frac{1}{2r} = \frac{r + R}{2Rr}$Therefore,$R_{eq} = \frac{2Rr}{R + r}$.

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

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