question_answer

Six equal resistances are connected between points P, Q and R as shown in the figure. The net resistance will be maximum between

A) P and Q                       

B) Q and R

C) P and R            

D)        Any two points.

Correct Answer: A

Solution :

Let resistance of each resistor be r. Resistance in arm PQ, \[{{R}_{1}}=r\] Resistance in arm QR, \[{{R}_{2}}=\frac{r\times r}{r+r}=\frac{r}{2}\] Resistance in arm PR, \[{{R}_{3}}=\frac{1}{\frac{1}{r}+\frac{1}{r}+\frac{1}{r}}=\frac{r}{3}\] Let net resistance between P and Q is \[{{R}_{PQ}}.\]        \[\frac{1}{{{R}_{PQ}}}=\frac{1}{r}+\frac{1}{\frac{r}{3}+\frac{r}{2}}=\frac{1}{r}+\frac{6}{5r}=\frac{11}{5r}\]             \[\therefore \]      \[{{R}_{PQ}}=\frac{5r}{11}\]             Net resistance between Q and R \[{{R}_{QR}},\]        \[\frac{1}{{{R}_{QR}}}=\frac{1}{\frac{r}{2}}+\frac{1}{r+\frac{r}{3}}=\frac{2}{r}+\frac{3}{4r}=\frac{11}{4r}\]             \[\therefore \]      \[{{R}_{QR}}=\frac{4r}{11}\]             Net resistance between P and R, \[{{R}_{PR}}\]             \[\frac{1}{{{R}_{PQ}}}=\frac{1}{r/3}+\frac{1}{r+r/2}=\frac{3}{r}+\frac{2}{3r}=\frac{11}{3r}\] \[\therefore \]      \[{{R}_{PR}}=\frac{3r}{11}\] Hence \[{{R}_{PQ}}>{{R}_{QR}}>{{R}_{PR}}\]