(a) (i) In absence of any electric field, the free electrons in metals move haphazardly in all possible directions and hence, develop no net flow of current. When an electric field is applied, a force acts on the electrons and the electrons now tend to move in the direction of the force. When electron collides with lattice, its velocity becomes instantaneously zero and then again it starts moving in a specific direction due to the applied electric field. If the average time between two collisions (relaxation time) is $\tau$, then $l = \frac{1}{2} a \tau^2$ Where $l =$ average drift distance $a = \text{ acceleration } = \frac{E e}{m}$ $E = \text{ electric field intensity }$ $e = \text{ charge of electron }$ $m = \text{ mass of electron }$ $\therefore l = \frac{1}{2} \frac{E e}{m} \tau^2$ $\text{ Drift velocity } = \frac{l}{\tau} = v_d$ $= \frac{1}{2} \frac{E e}{m} \tau$ $\frac{e \tau}{2 m} = K$, a constant, which depends on the nature of the material and the temperature. $\therefore v_d = K \times E$ Thus, free electrons in a metal at constant temperature under the action of an electric field attain a constant average velocity. (b) Since, the wires are joined in series current flowing through then will be same. Let the current in both $A$ and $B$ be I. Diameter being same, there areas of cross section are also same. Let it be $A$. So, in wire $A$ $I = n_A e A v_{dA}$ In wire $B$, $I = n_B e A v_{dB}$ Taking the ratio, $1 = \frac{n_A}{n_B} \times \frac{v_{dA}}{v_{dB}}$ $\text{Or, } 1 = 1.5 \times \frac{v_{dA}}{v_{dB}}$ $\therefore \frac{v_{dA}}{v_{dB}} = \frac{1}{1.5} = 2:3$ OR (b) (i) Terminal voltage vs. load resistance graph: $V = E - i r$ And $i = \frac{E}{R+r}$ ∴ $V = E - \frac{E r}{R+r}$ Or, $V = E \frac{1-r}{r+R}$ ∴ $V = \frac{E R}{r+R}$ Terminal voltage vs. current graph: (ii) Three cells combination diagram is given below, Current through cells $(i) = \frac{E_{\text{eq}}}{r_{\text{eq}}} = \frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}$ .....(i) And $\frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$ Or, $\frac{1}{r_{\text{eq}}} = \frac{1}{2r} + \frac{1}{3r} + \frac{1}{6r}$ $\therefore$ Equivalent resistance of internal resistance $r_{\text{eq}} = r$ Putting in equation (i), $i = \frac{E_{\text{eq}}}{r} = \frac{E}{2r} + \frac{E}{3r} + \frac{E}{6r}$ $i = \frac{6E}{6r} = \frac{E}{r}$ $\therefore E_{\text{eq}} = E$ So, the equivalent circuit is, (i) Current flowing through the circuit $= i = \frac{E}{R+r}$(ii) Terminal potential difference, $V = i R$$\text{Or, } V = \left( \frac{E}{R+r} \right) R$$\therefore V = \frac{E R}{r+R}$