In an experiment with a potentiometer, $V_B=10 \mathrm{~V} . R$ is adjusted to be $50 \Omega$ (figure). A student wanting to measure voltage $E_1$ of a battery (approx. 8V) finds no null point possible. He then diminishes R to $10 \Omega$ and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.
Let $R^{\prime}$ be the resistance of the potentiometer wire.
Effective resistance of potentiometer and variable resistor $(R=50 \Omega)$ is given by $=50 \Omega+R^{\prime}$
Effective voltage applied across potentiometer $=10 \mathrm{~V}$.
The current through the main circuit,
$$I=\frac{V}{50 \Omega+R}=\frac{10}{50 \Omega+R}$$
Potential difference across wire of potentiometer,
$$I R^{\prime}=\frac{10 R^{\prime}}{50 \Omega+R}$$
Since with $50 \Omega$ resistor, null point is not obtained it's possible only when
$$\begin{aligned} \frac{10 \times R^{\prime}}{50+R} & <8 \\ 10 R^{\prime} & <400+8 R^{\prime} \\ \Rightarrow\quad 2 R^{\prime}<400 \text { or } R^{\prime} & <200 \Omega . \end{aligned}$$
Similarly with $10 \Omega$ resistor, null point is obtained its possible only when
$$\frac{10 \times R^{\prime}}{10+R^{\prime}}>8$$
$$\begin{array}{llr} \Rightarrow & 2 R^{\prime}> 80 \\ \Rightarrow & R^{\prime}> 40 \\ & \frac{10 \times \frac{3}{4} R^{\prime}}{10+R^{\prime}} <8 \\ \Rightarrow & 7.5 R^{\prime} < 80+8 R^{\prime} \\ \Rightarrow & R^{\prime} > 160 \\ & 160 < R^{\prime}<200 . \end{array}$$
Any $R^{\prime}$ between $160 \Omega$ and $200 \Omega$ wlll achieve.
Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop across 400 cm of wire $>8 \mathrm{~V}$.
This imply that potential gradient
$$\begin{aligned} k \times 400 \mathrm{~cm} & >8 \mathrm{~V} \\ \text{or}\quad k \times 4 \mathrm{~m} & >8 \mathrm{~V} \\ k & >2 \mathrm{~V} / \mathrm{m} \end{aligned}$$
Similarly, potential drop across 300 cm wire $<8 \mathrm{~V}$.
$$\begin{aligned} k \times 300 \mathrm{~cm} & <8 \mathrm{~V} \\ \text{or}\quad \mathrm{k} \times 3 \mathrm{~m}<8 \mathrm{~V}, k & <2 \frac{2}{3} \mathrm{~V} / \mathrm{m} \\ \text{Thus,}\quad 2 \frac{2}{3} \mathrm{~V} / \mathrm{m} & >k>2 \mathrm{~V} / \mathrm{m} \end{aligned}$$
(a) Consider circuit in figure. How much energy is absorbed by electrons from the initial state of no current (Ignore thermal motion) to the state of drift velocity ?
(b) Electrons give up energy at the rate of $R I^2$ per second to the thermal energy. What time scale would number associate with energy in problem (a)? $n=$ number of electron/volume $=10^{29} / \mathrm{m}^3$. Length of circuit $=10 \mathrm{~cm}$, cross-section $=A=(1 \mathrm{~mm})^2$.
(a) By Ohm's law, current $I$ is given by
$$\begin{aligned} & I=6 \mathrm{~V} / 6 \Omega=1 \mathrm{~A} \\ \text{But,}\quad & I=n e t ~ A v_d \text { or } v_d=\frac{i}{n e A} \end{aligned}$$
On substituting the values
For, $$\quad n=\text { number of electron/volume }=10^{29} / \mathrm{m}^3$$
$$\begin{aligned} \text { length of circuit } & =10 \mathrm{~cm}, \text { cross-section }=A=(1 \mathrm{~mm})^2 \\ v_d & =\frac{1}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}} \\ & =\frac{1}{1.6} \times 10^{-4} \mathrm{~m} / \mathrm{s} \end{aligned}$$
Therefore, the energy absorbed in the form of KE is given by
$$\begin{aligned} \mathrm{KE} & =\frac{1}{2} m_e v_d^2 \times n A I \\ & =\frac{1}{2} \times 9.1 \times 10^{31} \times \frac{1}{2.56} \times 10^{20} \times 10^8 \times 10^6 \times 10^1 \\ & =2 \times 10^{-17} \mathrm{~J} \end{aligned}$$
(b) Power loss is given by $P=I^2 R=6 \times 1^2=6 \mathrm{~W}=6 \mathrm{~J} / \mathrm{s}$
Since, $$P=\frac{E}{t}$$
Therefore, $$E=P \times t$$
or $$t=\frac{E}{P}=\frac{2 \times 10^{-17}}{6} \approx 10^{-17} \mathrm{~s}$$