Combination of Resistors

Resistance of each piece: $r = R/8$. When 4 pieces are connected in parallel, the equivalent resistance of one set is $R_p = \frac{r}{4} = \frac{R/8}{4} = \frac{R}{32}$. When two such identical parallel sets are connected in series, the net resistance is $R_{net} = R_p + R_p = \frac{R}{32} + \frac{R}{32} = \frac{2R}{32} = \frac{R}{16}$.

In parallel: $R_{eq} = \frac{R}{4} = 0.25\ \Omega \Rightarrow R = 1\ \Omega$ (each wire) In series: $R_{series} = 4R = 4 \times 1 = 4\ \Omega$

B and C are in parallel, so $V_B = V_C$.
Equivalent resistance of B and C: $\frac{1.5R\times 3R}{1.5R + 3R} = R$
This combination is in series with A (resistance R). In series $V \propto R$, so the voltage across A equals that across the combination.
$V_A = V_B = V_C$

The two arcs have resistances $R_1$ and $R_2$ (proportional to $l_1$ and $l_2$) and are in parallel between A and B.
$R_1 + R_2 = 12$ ...(i)
$\frac{R_1R_2}{R_1 + R_2} = \frac{8}{3} \Rightarrow R_1R_2 = 32$ ...(ii)
Solving: $R_1 = 4\ \Omega$ and $R_2 = 8\ \Omega$
$\frac{l_1}{l_2} = \frac{R_1}{R_2} = \frac{1}{2}$

In the circuit the 9 $\Omega$ and 6 $\Omega$ resistors are in parallel, and this combination is in series with the 2 $\Omega$ resistor.
Power in the 9 $\Omega$ resistor: $P = I_1^2R$, so $36 = I_1^2\times9$, giving $I_1 = 2$ A
Potential difference across the 9 $\Omega$ resistor = $I_1R = 2\times9 = 18$ V
The same 18 V is across the 6 $\Omega$ resistor, so $I_2 = \frac{18}{6} = 3$ A
Total current through the 2 $\Omega$ resistor = $I_1 + I_2 = 2 + 3 = 5$ A
Potential difference across the 2 $\Omega$ resistor = $5\times2 = 10$ V

Length of the wire = circumference = $2\pi r = 2\pi\times0.1 = 0.2\pi$ m
Total resistance of the wire = $12\times0.2\pi = 2.4\pi\ \Omega$
Diametrically opposite points divide the circle into two equal semicircles, each of resistance $\frac{2.4\pi}{2} = 1.2\pi\ \Omega$
Between A and B these two halves are in parallel: $R_{AB} = \frac{1.2\pi\times1.2\pi}{1.2\pi + 1.2\pi} = \frac{1.2\pi}{2}$
$R_{AB} = 0.6\pi\ \Omega$

Current through the $4\,\Omega$ resistor: $i_1 = 1$ A.
Potential difference across the $4\,\Omega$ resistor: $V_{PM} = 1 \times 4 = 4$ V. This is the potential difference between P and M.
The $3\,\Omega$ resistor is in parallel with $4\,\Omega$, so the current through it is $i_2 = \dfrac{4}{3}$ A.
Lower branch: the two $0.5\,\Omega$ resistors in parallel give $\dfrac{0.5 \times 0.5}{0.5 + 0.5} = 0.25\,\Omega = \dfrac{1}{4}\,\Omega$ between P and N.
This is in series with $1\,\Omega$ between N and M, so the resistance of the lower branch $= \dfrac{1}{4} + 1 = \dfrac{5}{4}\,\Omega$.
The lower branch is also connected between P and M, so the current through it is $i_4 = \dfrac{V_{PM}}{5/4} = \dfrac{4 \times 4}{5} = \dfrac{16}{5} = 3.2$ A.
Potential difference across N and M: $V_{NM} = i_4 \times 1\,\Omega = 3.2 \times 1 = 3.2$ volt
Second approach: $V_{PN} + V_{NM} = 4$ V and $\dfrac{V_{PN}}{V_{NM}} = \dfrac{1/4}{1} = \dfrac{1}{4}$, so $\dfrac{1}{4}V_{NM} + V_{NM} = 4 \Rightarrow V_{NM} = \dfrac{16}{5} = 3.2$ V.

When the wire is bent into a circle, two diametrically opposite points A and B divide it into two equal semicircles.
Each semicircle has half the length of the wire, so its resistance is $\dfrac{R}{2}$.
Between A and B the two semicircles are in parallel.
$R_{AB} = \dfrac{(R/2)(R/2)}{(R/2) + (R/2)}$
$R_{AB} = \dfrac{R^2/4}{R} = \dfrac{R}{4}$

For n equal resistances r in parallel: $\dfrac{1}{R} = \dfrac{1}{r} + \dfrac{1}{r} + \dots$ (n times) $= \dfrac{n}{r}$
So according to the question $R = \dfrac{r}{n} \Rightarrow r = nR$
For the same n resistances in series: $R_{series} = r + r + \dots$ (n times) $= nr$
Substituting $r = nR$: $R_{series} = n \times nR = n^2R$
So the series combination has a resistance of $n^2R$ ohm.