Current Electricity

JEE Main level, single correct, medium

161 of 380

Question 161 of 380

Six equal resistances are connected between points P,Q\mathrm{P}, \mathrm{Q} and R\mathrm{R} as shown in figure. Then net resistance will be maximum between :

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Answer: (B) P and Q

Resistance between P\mathrm{P} and Q\mathrm{Q}
rPQ=r11(r3+r2)=r×56rr+56r=511rr_{\mathrm{PQ}}=r_{11}\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
rQR=r211(r+r3)=r2×43rr2+43r=411r\mathrm{r}_{\mathrm{QR}}=\frac{\mathrm{r}}{2} 11\left(\mathrm{r}+\frac{\mathrm{r}}{3}\right)=\frac{\frac{\mathrm{r}}{2} \times \frac{4}{3} \mathrm{r}}{\frac{\mathrm{r}}{2}+\frac{4}{3} \mathrm{r}}=\frac{4}{11} \mathrm{r}
Resistance between P\mathrm{P} and R\mathrm{R}
rPR=r3l(r2+r)=r3×32rr3+32r=311r\mathrm{r}_{\mathrm{PR}}=\frac{\mathrm{r}}{3} \mathrm{l}\left(\frac{\mathrm{r}}{2}+\mathrm{r}\right)=\frac{\frac{\mathrm{r}}{3} \times \frac{3}{2} \mathrm{r}}{\frac{\mathrm{r}}{3}+\frac{3}{2} \mathrm{r}}=\frac{3}{11} \mathrm{r}
Hence, it is clear that rPQ\mathrm{r}_{\mathrm{PQ}} is maximum

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