Effective resistance of $${R_2}$$ and $${R_4}$$ in series,
$$R' = 10 + 10 = 20\,\Omega $$
Effective resistance of $${R_3}$$ and $${R_5}$$ in series,
$$R'' = 10 + 10 = 20\,\Omega $$
Net total resistance of $${R'}$$ and $${R''}$$ in parallel is
$${R_p} = \frac{{20 \times 20}}{{20 + 20}} = 10\,\Omega $$
∴ Total resistance between $$A$$ and $$D$$
$$\eqalign{ & = 10 + 10 + 10 \cr & = 30\,\Omega \cr} $$

Net resistance of a metal wire having resistivity $$\rho ,$$ we have

$$\eqalign{ & {R_1} = {\rho _1}\frac{L}{A} \cr & {\text{Similarly,}}\,{R_2} = {\rho _2}\frac{L}{A} \cr & {\text{Then, net effective resistance of two metal wires,}} \cr & {R_{{\text{eq}}}} = {R_1} + {R_2} \Rightarrow \rho \frac{{2L}}{A} = {\rho _1}\frac{L}{A} + {\rho _2}\frac{L}{A} \cr & \Rightarrow 2\rho = {\rho _1} + {\rho _2} \cr & {\text{As, conductivity}}\,\sigma = \frac{1}{\rho },\,{\text{we have}} \cr & \frac{2}{\sigma } = \frac{1}{{{\sigma _1}}} + \frac{1}{{{\sigma _2}}} \Rightarrow \frac{2}{\sigma } = \frac{{{\sigma _1} + {\sigma _2}}}{{{\sigma _1}{\sigma _2}}} \cr & \Rightarrow {\text{Net effective conductivity of combined wires,}} \cr & \sigma = \frac{{2{\sigma _1}{\sigma _2}}}{{{\sigma _1} + {\sigma _2}}} \cr} $$

KEY CONCEPT : The heat produced is given by
$$\eqalign{ & H = \frac{{{V^2}}}{R}\,{\text{and}}\,R = \frac{{\rho \ell }}{{\pi {r^2}}}\,\,\therefore H = {V^2}\left( {\frac{{\pi {r^2}}}{{\rho \ell }}} \right) \cr & {\text{or,}}\,H = \left( {\frac{{\pi {V^2}}}{\rho }} \right)\frac{{{r^2}}}{\ell } \cr} $$
Thus heat ($$H$$) is doubled if both length ($$\ell $$) and radius ($$r$$) are doubled.

When the temperature increases, resistance increases. As the $$e.m.f.$$   applied is the same, the current density decreases the drift velocity decreases. But the $$rms$$  velocity of the electron due to thermal motion is proportional to $$\sqrt T .$$  Therefore, the thermal velocity increases.

$$\eqalign{ & V = IR = \left( {neA{v_d}} \right)\rho \frac{\ell }{A} \cr & \therefore \rho = \frac{V}{{{V_d}{\text{lne}}}} \cr} $$
Here $$V$$ = potential difference
$$l$$ = length of wire
$$n$$ = no. of electrons per unit volume of conductor.
$$e$$ = no. of electrons
Placing the value of above parameters we get resistivity
$$\eqalign{ & \rho = \frac{5}{{8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 4}} \times 0.1}} \cr & = 1.6 \times {10^{ - 5}}\Omega m \cr} $$

Amount of heat produced in a conductor is equal to work done in carrying a charge $$q$$ from one end of conductor to other end of conductor having potential difference $$V.$$
$$\eqalign{ & \therefore H = W = Vq = Vit = {i^2}Rt\,\,\left[ {{\text{as,}}\,V = IR} \right] \cr & \therefore H = {i^2}Rt\,J \cr & \Rightarrow R = \frac{H}{{\left( {{i^2}t} \right)}} \cr & {\text{Given,}}\,\,H = 80\,J,i = 2\,A,t = 10\,s \cr & {\text{So,}}\,\,R = \frac{{80}}{{{{\left( 2 \right)}^2} \times 10}} \cr & = 2\,\Omega \cr} $$

KEY CONCEPT : At any instant of time $$t$$ during charging process, the transient current in the circuit is given by
$$I = \frac{{{V_0}}}{R}{e^{ - \frac{t}{{RC}}}}$$
$$\therefore $$ Potential difference across resistor $$R$$ is
$$\eqalign{ & {V_R} = \left[ {\frac{{{V_0}}}{R}{e^{ - \frac{t}{{RC}}}}} \right] \times R \cr & = {V_0}{e^{ - \frac{t}{{RC}}}}\,......\left( {\text{i}} \right) \cr} $$
$$\therefore $$ Potential diff. across $$C$$
$$\eqalign{ & {V_c} = {V_0} - {V_0}{e^{ - \frac{t}{{RC}}}} = {V_0}\left( {1 - {e^{ - \frac{t}{{RC}}}}} \right)\,......\left( {{\text{ii}}} \right) \cr & \because \quad {V_c} = 3{V_R}\,\,\,\,\,\,\,\left( {{\text{given}}} \right) \cr & \because \quad {V_0}\left( {1 - {e^{ - \frac{t}{{RC}}}}} \right) = 3{V_0}.{e^{ - \frac{t}{{RC}}}} \cr & \Rightarrow 1 - {e^{ - \frac{t}{{RC}}}} = 3{e^{ - \frac{t}{{RC}}}} \Rightarrow 1 = 4{e^{ - \frac{t}{{RC}}}} \cr} $$
Taking log on both sides
$$\eqalign{ & {\log _e}1 = 2{\log _e}2 + \left( { - \frac{t}{{RC}}} \right) \cr & \Rightarrow 0 = 2 \times 2.303{\log _{10}}2 - \frac{t}{{RC}} \cr & \Rightarrow t = \left[ {2 \times 2.303{{\log }_{10}}2} \right] \times 2.5 \times {10^6} \times 4 \times {10^{ - 6}} = 13.86\,{\text{sec}} \cr} $$

No. of electron in the wire $$ = nA\ell \left( {n = {e^ - }{\text{density}}} \right)$$     and momentum $$\left( {nA\ell } \right){m_e}{v_d}$$
$$\eqalign{ & = neA{v_d}\frac{{\ell {m_e}}}{e} = \frac{{I\ell {m_e}}}{e} \cr & = \frac{{70 \times 1000}}{{1.6 \times {{10}^{ - 19}}}} \times 9.1 \times {10^{ - 31}} \cr & = 40 \times {10^{ - 8}}N - \sec \cr} $$

The figure can be redrawn as follows:

$${R_{AB}} = R + \frac{{3R}}{4} + R = \frac{{11R}}{4}$$

$$\eqalign{ & I = \frac{{dQ}}{{dt}} = 10t + 3 \cr & {\text{At}}\,t = 5s,I = 10 \times 5 + 3 = 53\,A \cr} $$