Six equal resistances R are connected between | vedclass

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(A) Based on the figure,the connections are as follows:Between $P$ and $Q$: One resistor $R$ is in parallel with a combination of two resistors in ser

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$P$ and $Q$

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(A) Based on the figure,the connections are as follows:Between $P$ and $Q$: One resistor $R$ is in parallel with a combination of two resistors in series (which are in parallel with another resistor). Wait,let's re-examine the diagram: - Branch $PQ$ has $1$ resistor.- Branch $QR$ has $2$ resistors in parallel,equivalent to $R/2$.- Branch $PR$ has $3$ resistors in parallel,equivalent to $R/3$.Now,calculating equivalent resistance between points:$1$. Between $P$ and $Q$: The path $PQ$ $(R)$ is in parallel with the path $PRQ$ $(R/3 + R/2 = 5R/6)$.$R_{PQ} = \frac{R 	imes (5R/6)}{R + 5R/6} = \frac{5R^2/6}{11R/6} = \frac{5}{11}R \approx 0.454R$.$2$. Between $Q$ and $R$: The path $QR$ $(R/2)$ is in parallel with the path $QPR$ $(R + R/3 = 4R/3)$.$R_{QR} = \frac{(R/2) 	imes (4R/3)}{R/2 + 4R/3} = \frac{2R^2/3}{11R/6} = \frac{2}{3} 	imes \frac{6}{11}R = \frac{4}{11}R \approx 0.363R$.$3$. Between $P$ and $R$: The path $PR$ $(R/3)$ is in parallel with the path $PQR$ $(R + R/2 = 3R/2)$.$R_{PR} = \frac{(R/3) 	imes (3R/2)}{R/3 + 3R/2} = \frac{R^2/2}{11R/6} = \frac{1}{2} 	imes \frac{6}{11}R = \frac{3}{11}R \approx 0.272R$.Comparing the values,$R_{PQ} = \frac{5}{11}R$ is the maximum.

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

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