$(a)$ Given $n$ resistors each of resistance $R,$ how will you combine them to get the $(i)$ maximum $(ii)$ minimum effective resistance? What is the ratio of the maximum to minimum resistance?
$(b)$ Given the resistances of $1\; \Omega, 2\; \Omega, 3\; \Omega,$ how will you combine them to get an equivalent resistance of $(i) \;(11 / 3)\; \Omega,$ $(ii)\;(11 / 5)\; \Omega,$ $(iii)\; 6\;\Omega,$ $(iv)\;(6 / 11)\; \Omega ?$
$(c)$ Determine the equivalent resistance of the networks shown in the figure.

(A) Total number of resistors $= n.$
Resistance of each resistor $= R.$
$(i)$ When $n$ resistors are connected in series,the effective resistance $R_{1}$ is the maximum,given by $R_{1} = nR.$
$(ii)$ When $n$ resistors are connected in parallel,the effective resistance $R_{2}$ is the minimum,given by $R_{2} = R/n.$
$(iii)$ The ratio of the maximum to the minimum resistance is $R_{1}/R_{2} = (nR) / (R/n) = n^{2}.$
$(b)$ Given resistances are $R_{1} = 1\; \Omega, R_{2} = 2\; \Omega, R_{3} = 3\; \Omega.$
$(i)$ To get $(11/3)\; \Omega,$ connect $1\; \Omega$ and $2\; \Omega$ in parallel,then connect the combination in series with $3\; \Omega.$ $R_{eq} = (1 \times 2)/(1+2) + 3 = 2/3 + 3 = 11/3\; \Omega.$
$(ii)$ To get $(11/5)\; \Omega,$ connect $2\; \Omega$ and $3\; \Omega$ in parallel,then connect the combination in series with $1\; \Omega.$ $R_{eq} = (2 \times 3)/(2+3) + 1 = 6/5 + 1 = 11/5\; \Omega.$
$(iii)$ To get $6\; \Omega,$ connect all three in series. $R_{eq} = 1 + 2 + 3 = 6\; \Omega.$
$(iv)$ To get $(6/11)\; \Omega,$ connect all three in parallel. $1/R_{eq} = 1/1 + 1/2 + 1/3 = (6+3+2)/6 = 11/6 \implies R_{eq} = 6/11\; \Omega.$
$(c)$ $(a)$ Each loop has two branches in parallel: one with two $1\; \Omega$ resistors in series $(2\; \Omega)$ and one with two $2\; \Omega$ resistors in series $(4\; \Omega)$. Equivalent resistance of one loop $= (2 \times 4)/(2+4) = 8/6 = 4/3\; \Omega.$ There are four such loops in series. Total $R_{eq} = 4 \times (4/3) = 16/3\; \Omega.$
$(b)$ Five resistors of resistance $R$ are connected in series. Total $R_{eq} = R+R+R+R+R = 5R.$