Question
The effective resistance between points $$P$$ and $$Q$$ of the electrical circuit shown in the figure is
The effective resistance between points $$P$$ and $$Q$$ of the electrical circuit shown in the figure is
A.
$$\frac{{2Rr}}{{R + r}}$$
B.
$$\frac{{2R\left( {R + r} \right)}}{{3R + r}}$$
C.
$$2r + 4R$$
D.
$$\frac{{5R}}{2} + 2r$$
Answer :
$$\frac{{2Rr}}{{R + r}}$$
Solution :
The circuit is symmetrical about the axis $$POQ.$$
The circuit above the axis $$POQ$$ represents balanced wheatstone bride. Hence the central resistance $$2R$$ is ineffective. Similarly in the lower part (below the axis $$POQ$$ ) the central resistance $$2R$$ is ineffective.
Therefore the equivalent circuit is drawn.
$$\eqalign{ & \therefore \frac{1}{{{R_{PQ}}}} = \frac{1}{{4R}} + \frac{1}{{4R}} + \frac{1}{{2r}} = \frac{{r + r + 2r}}{{4Rr}} \cr & {R_{PQ}} = \frac{{2Rr}}{{R + r}} \cr} $$
The circuit is symmetrical about the axis $$POQ.$$
The circuit above the axis $$POQ$$ represents balanced wheatstone bride. Hence the central resistance $$2R$$ is ineffective. Similarly in the lower part (below the axis $$POQ$$ ) the central resistance $$2R$$ is ineffective.
Therefore the equivalent circuit is drawn.
$$\eqalign{ & \therefore \frac{1}{{{R_{PQ}}}} = \frac{1}{{4R}} + \frac{1}{{4R}} + \frac{1}{{2r}} = \frac{{r + r + 2r}}{{4Rr}} \cr & {R_{PQ}} = \frac{{2Rr}}{{R + r}} \cr} $$
