(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)
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