Six equal resistances are connected between points $P$, $Q$ and $R$ as shown in the figure. Then the net resistance will be maximum between
IIT 2004, Diffcult
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(a) Resistance between $P$ and $Q$
${R_{PQ}} = R||\left( {\frac{R}{3} + \frac{R}{2}} \right)$$ = \frac{{R \times \frac{5}{6}R}}{{R + \frac{5}{6}R}}$$ = \frac{5}{{11}}R$
Resistance between $Q$ and $R$
${R_{QR}} = \frac{R}{2}||\left( {R + \frac{R}{3}} \right)$$ = \frac{{\frac{R}{2} \times \frac{{4R}}{3}}}{{\frac{R}{2} + \frac{{4R}}{3}}}$$ = \frac{4}{{11}}R$
Resistance between $P$ and $R$
${R_{PR}} = \frac{R}{3}||\left( {\frac{R}{2} + R} \right)$$ = \frac{{\frac{R}{3} \times \frac{{3R}}{2}}}{{\frac{R}{3} + \frac{{3R}}{2}}}$$ = \frac{3}{{11}}R$
Hence it is clear that ${P_{PQ}}$ is maximum.
${R_{PQ}} = R||\left( {\frac{R}{3} + \frac{R}{2}} \right)$$ = \frac{{R \times \frac{5}{6}R}}{{R + \frac{5}{6}R}}$$ = \frac{5}{{11}}R$
Resistance between $Q$ and $R$
${R_{QR}} = \frac{R}{2}||\left( {R + \frac{R}{3}} \right)$$ = \frac{{\frac{R}{2} \times \frac{{4R}}{3}}}{{\frac{R}{2} + \frac{{4R}}{3}}}$$ = \frac{4}{{11}}R$
Resistance between $P$ and $R$
${R_{PR}} = \frac{R}{3}||\left( {\frac{R}{2} + R} \right)$$ = \frac{{\frac{R}{3} \times \frac{{3R}}{2}}}{{\frac{R}{3} + \frac{{3R}}{2}}}$$ = \frac{3}{{11}}R$
Hence it is clear that ${P_{PQ}}$ is maximum.
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