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

\frac{2 R r}{(R + r)}

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\frac{8 R (R + r)}{(3 R + r)}

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2 r + 4 R

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\frac{5 R}{2 + 2 r}

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Solution

The correct option is A $\frac{2 R r}{(R + r)}$
Due to mirror symmetry intermediate resistances can be removed as shown

Now, using the formula for parallel combination of resistors for the $3$ branches, we get
$\frac{1}{R_{e q}} = \frac{1}{4 R} + \frac{1}{4 R} + \frac{1}{2 r}$
$R_{e q} = \frac{2 r R}{r + R}$

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