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(B) Let each resistance be $r$. The circuit consists of three branches between the nodes $P, Q,$ and $R$.Branch $PQ$ has one resistance $r$.Branch $QR

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$P$ and $Q$

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(B) Let each resistance be $r$. The circuit consists of three branches between the nodes $P, Q,$ and $R$.Branch $PQ$ has one resistance $r$.Branch $QR$ has two resistances in parallel,so its equivalent resistance is $r_{QR} = r/2$.Branch $PR$ has two resistances in parallel,so its equivalent resistance is $r_{PR} = r/2$.$1$. Resistance between $P$ and $Q$ $(R_{PQ})$:$R_{PQ}$ is the parallel combination of the branch $PQ$ (resistance $r$) and the series combination of branches $PR$ and $QR$ (resistance $r/2 + r/2 = r$).$R_{PQ} = \frac{r 	imes r}{r + r} = \frac{r}{2} = 0.5r$.$2$. Resistance between $Q$ and $R$ $(R_{QR})$:$R_{QR}$ is the parallel combination of the branch $QR$ (resistance $r/2$) and the series combination of branches $PQ$ and $PR$ (resistance $r + r/2 = 1.5r$).$R_{QR} = \frac{(r/2) 	imes (1.5r)}{(r/2) + (1.5r)} = \frac{0.75r^2}{2r} = 0.375r$.$3$. Resistance between $P$ and $R$ $(R_{PR})$:$R_{PR}$ is the parallel combination of the branch $PR$ (resistance $r/2$) and the series combination of branches $PQ$ and $QR$ (resistance $r + r/2 = 1.5r$).$R_{PR} = \frac{(r/2) 	imes (1.5r)}{(r/2) + (1.5r)} = 0.375r$.Comparing the values: $R_{PQ} = 0.5r$,$R_{QR} = 0.375r$,and $R_{PR} = 0.375r$.Thus,the net resistance is maximum between $P$ and $Q$.

Six equal resistances 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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