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Question

(A) The circuit is symmetric about the horizontal axis $PQ$. Due to symmetry,the potential at the central junction points $A$ and $B$ is the same,mean

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The circuit is symmetric about the horizontal axis $PQ$. Due to symmetry,the potential at the central junction points $A$ and $B$ is the same,meaning no current flows through the vertical resistors $2R$ connected between $A$ and $B$. Alternatively,we can simplify the circuit by identifying series combinations. The top branch consists of two $2R$ resistors in series,giving $4R$. The middle branch consists of two $r$ resistors in series,giving $2r$. The bottom branch consists of two $2R$ resistors in series,giving $4R$. These three branches are in parallel. Thus,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,$R_{eq} = \frac{2Rr}{R + r}$.

What is the equivalent resistance between $P$ and $Q$ in the given circuit?

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