$(a)$ Consider the circuit in the 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 thermal energy. What time scale would one associate with the energy in problem $(a)$? Given: $n = 10^{29} \, m^{-3}$,length of circuit $l = 10 \, cm$,cross-section $A = (1 \, mm)^2$.

(N/A) Current in the circuit $I = \frac{V}{R} = \frac{6}{6} = 1 \, A$.
Number density of electrons $n = 10^{29} \, m^{-3}$.
Length of the wire $l = 10 \, cm = 0.1 \, m$.
Area of cross-section $A = (1 \, mm)^2 = 10^{-6} \, m^2$.
Volume of the wire $V_{vol} = A \times l = 10^{-6} \times 0.1 = 10^{-7} \, m^3$.
Total number of electrons $N = n \times V_{vol} = 10^{29} \times 10^{-7} = 10^{22}$.
Drift velocity $v_d = \frac{I}{nAe} = \frac{1}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}} = \frac{1}{1.6 \times 10^4} = 0.625 \times 10^{-4} \, m/s = 6.25 \times 10^{-5} \, m/s$.
Total kinetic energy $K.E. = N \times \frac{1}{2} m_e v_d^2 = 10^{22} \times 0.5 \times 9.1 \times 10^{-31} \times (6.25 \times 10^{-5})^2$.
$K.E. = 4.55 \times 10^{-9} \times 39.06 \times 10^{-10} \approx 1.78 \times 10^{-17} \, J$.
$(b)$ Power dissipated $P = I^2 R = (1)^2 \times 6 = 6 \, W$.
Since $P = \frac{E}{t}$,the time scale $t = \frac{E}{P} = \frac{1.78 \times 10^{-17}}{6} \approx 2.97 \times 10^{-18} \, s \approx 3 \times 10^{-18} \, s$.