question_answer

(i) Consider 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?

(ii) Electrons give up energy at the rate of \[R{{I}^{2}}\]per second to the thermal energy. Estimate the time scale associated with energy in problem (i) ? Given, n = number of electron per volume \[={{10}^{29}}\] per \[{{m}^{3}}.\] Length of circuit = 10 cm cross-section \[=A={{(1mm)}^{2}}.\]

   

Answer:

(i) 9600 J           (ii) 1600 s (i) Given,           \[n={{10}^{29}}{{m}^{-3}}\] \[l=10\,cm=0.01\,m\] \[A=1\,m{{m}^{2}}={{10}^{-6}}{{m}^{2}}\]             Energy absorbed by the electrons = VIt                         \[I=\frac{V}{R}=\frac{6}{6}=1A\] Time taken by the free electrons to cross the length of the conductor,\[t=l/{{\text{v}}_{d}}\] Where, \[{{\text{v}}_{d}}\] is the drift velocity?             \[I=neA{{\text{v}}_{d}}\] \[\therefore \]      \[{{\text{v}}_{d}}=I/neA\]             \[=\frac{1}{{{10}^{29}}\times 16\times {{10}^{-19}}\times {{10}^{-6}}}\]             \[=0.625\times {{10}^{-4}}=6.25\times {{10}^{-5}}m{{s}^{-1}}\] \[\therefore \]      \[t=\frac{l}{{{\text{v}}_{d}}}=\frac{0.01}{6.25\times {{10}^{-5}}}\]             \[=0.16\times {{10}^{3}}=160\,\,s\] \[\therefore \]      Energy = VIt             \[=6\times 1\times 1600=9600\,\text{J}\] (ii) \[t=1600\,\,s\] (as obtained before)