(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 $$t=\frac{E}{P}=\frac{2 \times 10^{-17}}{6} \approx 10^{-17} \mathrm{~s}$$