The figure can be redrawn as follows :

The charged particles whose flow in a definite direction constitutes the electric current are called current carriers. In metals the valence electrons of the atoms do not remain attached to individual atoms but are free to move throughout the volume of the conductor. Their velocity under normal conditions is of the order of a fraction of $$mm/s.$$

The specific resistance (resistivity) of a metallic conductor nearly increases with increasing temperature as shown in figure. This is because, with the increase in temperature the ions of the conductors vibrate with greater amplitude and the collisions between electrons and ions become more frequent, over a small temperature range (upto $${100^ \circ }C$$  ). The resistivity of a metal can be represented approximately by the equation
$${\rho _t} = {\rho _0}\left( {1 + \alpha t} \right)$$

The factor $$\alpha $$ is called the temperature coefficient of resistivity.

Let voltage at $$C = x$$  volt
From kirchhoff’s current law,
$$KCL : {i_1} + {i_2} = i$$
$$\eqalign{ & \frac{{20 - x}}{2} + \frac{{10 - x}}{4} = \frac{{x - 0}}{2} \Rightarrow x = 10 \cr & \therefore i = \frac{V}{R} = \frac{x}{R} = \frac{{10}}{2} = 5A \cr} $$

$$\eqalign{ & {\rho _B} = 2{\rho _A} \cr & {d_B} = 2{d_A} \cr & {R_B} = {R_A} \Rightarrow \frac{{{\rho _B}{\ell _B}}}{{{A_B}}} = \frac{{{\rho _A}{\ell _A}}}{{{A_A}}} \cr & \therefore \frac{{{\ell _B}}}{{{\ell _A}}} = \frac{{{\rho _A}}}{{{\rho _B}}} \times \frac{{d_B^2}}{{d_A^2}} = \frac{{{\rho _A}}}{{2{\rho _A}}} \times \frac{{4d_A^2}}{{d_A^2}} = 2 \cr} $$

Consider a potentiometer wire of length $$L$$ and a resistance $$r$$ are connected in series with a battery of emf $${{E_0}}$$ and a resistance $${{r_1}}$$ as shown in figure. Current in wire $$AB = \frac{{{E_0}}}{{{r_1} + r}}$$

Potential gradient,
$$x = \frac{{Ir}}{L} = \left[ {\frac{{{E_0}}}{{{r_1} + r}}} \right]\frac{r}{L}$$
emf produced across $$E$$ will be given by
$$E = x \cdot l = \left[ {\frac{{{E_0}r}}{{{r_1} + r}}} \right]\frac{l}{L}$$

Balancing condition of Wheatstone bridge is used to calculate the value of unknown resistance.
The situation can be depicted as shown in figure.

As resistances $$S$$ and $$6\,\Omega $$ are in parallel their effective resistance is
$$\frac{{6S}}{{6 + S}}\Omega $$
As the bridge is balanced, hence it is balanced Wheatstone bridge.
For balancing condition,
$$\eqalign{ & \frac{P}{Q} = \frac{R}{{\left( {\frac{{6S}}{{6 + S}}} \right)}} \cr & {\text{or}}\,\,\frac{2}{2} = \frac{{2\left( {6 + S} \right)}}{{6S}} \cr & {\text{or}}\,\,3S = 6 + S \cr & {\text{or}}\,\,S = 3\,\Omega \cr} $$

Problem Solving Strategy
Find the equivalent resistance of the circuit and then calculate the power dissipated by the circuit.
Resistances $$1\,\Omega $$  and $$3\,\Omega $$  are connected in series, so, effective resistance
$$R' = 1 + 3 = 4\,\Omega $$
Now, $${R'}$$ and $$8\,\Omega $$  are in parallel. We know that potential difference across resistances in parallel is same.

So, from Kirchhoff’s law, $${V_1} = {V_2}$$
$$\eqalign{ & R' \times {i_1} = 8{i_2}\,\,{\text{or}}\,\,4 \times {i_1} = 8{i_2} \cr & {\text{or}}\,\,{i_1} = \frac{8}{4}{i_2} = 2{i_2} \cr & {\text{or}}\,\,{i_1} = 2{i_2}\,......\left( {\text{i}} \right) \cr} $$
Power dissipated across $$8\,\Omega $$  resistance is
$$\eqalign{ & i_2^2\left( 8 \right)t = 2\;W\,\,\left[ {\because P = iRt} \right] \cr & {\text{or}}\,\,i_2^2t = \frac{2}{8} = 0.25\,W\,.......\left( {{\text{ii}}} \right) \cr} $$
Power dissipated across $$3\,\Omega $$  resistance is
$$\eqalign{ & H = i_1^2\left( 3 \right)t \cr & = {\left( {2{i_2}} \right)^2}\left( 3 \right)t = 12\,i_2^2t \cr} $$
but $$i_2^2t = 0.25\,W$$
$$\therefore H = 12 \times 0.25 = 3\,W$$

The equivalent circuits are shown in the figure.

The circuit represents balanced Wheatstone Bridge. Hence $$6R\,\Omega $$  resistance is ineffective

$$\frac{1}{{{R_{eq}}}} = \frac{1}{{3R}} + \frac{1}{{6R}},\,\,\,{R_{eq}} = \frac{{\left( {3R} \right)\left( {6R} \right)}}{{\left( {3R} \right) + \left( {6R} \right)}} = 2R$$
For Max. Power
External Resistance = Internal Resistance
$$2R = 4\Omega \,\,\therefore R = 2\Omega $$