There are 8 equal resistances R. Two are conn | vedclass

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(B) $1$. Each group consists of $2$ resistors of resistance $R$ connected in parallel.$2$. The equivalent resistance of one such parallel group is giv

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$2 R$

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(B) $1$. Each group consists of $2$ resistors of resistance $R$ connected in parallel.$2$. The equivalent resistance of one such parallel group is given by $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$,which implies $R_p = \frac{R}{2}$.$3$. There are $4$ such groups connected in series.$4$. The total equivalent resistance $R_{eq}$ is the sum of the resistances of the $4$ groups: $R_{eq} = R_p + R_p + R_p + R_p = 4 	imes R_p$.$5$. Substituting the value of $R_p$,we get $R_{eq} = 4 	imes \frac{R}{2} = 2R$.

There are $8$ equal resistances $R$. Two are connected in parallel,and such four groups are connected in series. The total resistance of the system will be:

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