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  • six equal resistances are connected between points P, Q and R as shown in figure. Then net resistance will be maximum between :  <img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEU

six equal resistances are connected between points P, Q and R as shown in figure. Then net resistance will be maximum between : 

Option: 1

P and R 


Option: 2

P and Q


Option: 3

Q and R 


Option: 4

Any two points 


Answers (1)

best_answer

 

 

 

 

 

 

\text {Resistance in arm} \ P Q,\Rightarrow R_{1}=r$ \\ Resistance in arm $Q R$ \\ $ \Rightarrow R_{2}=\frac{r \times r}{r+r}=\frac{r}{2}$ \\ Resistance in arm $P R$\\ $\Rightarrow R_{3}=\frac{1}{\frac{1}{r}+\frac{1}{r}+\frac{1}{r}}=\frac{r}{3}

\text{Let net resistance between} \ \mathrm{P}$ and $\mathrm{Q}$ is $\mathrm{R}_{\mathrm{PQ}}$. \\ $\frac{1}{R_{P Q}}=\frac{1}{r}+\frac{1}{\frac{r}{3}+\frac{r}{2}}$ $=\frac{1}{r}+\frac{6}{5 r}=\frac{11}{5 r}$ $\therefore R_{P Q}=\frac{5 r}{11}

\text{Let net resistance between} \ Q$ and $R, R_{Q R}$ \\ $\frac{1}{R_{\mathrm{QR}}}=\frac{1}{\frac{r}{2}}+\frac{1}{r+\frac{r}{3}}$ $=\frac{2}{r}+\frac{3}{4 r}=\frac{11}{4 r}$ $\therefore \quad R_{\mathrm{QR}}=\frac{4 r}{11}

\text{Let net resistance between} \ \mathrm{P}$ and $\mathrm{R}, \mathrm{R}_{\mathrm{PR}}$ \\ $\frac{1}{R_{P Q}}=\frac{1}{r / 3}+\frac{1}{r+r / 2}$ $=\frac{3}{r}+\frac{2}{3 r}=\frac{11}{3 r}$ $\therefore R_{P R}=\frac{3 r}{11}

\text { Hence } R_{P Q}>R_{Q R}>R_{P R}

 

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