VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    A potential difference of \[2V\] is applied between the opposite faces of a Ge crystal plate of area \[1c{{m}^{2}}\] and thickness 0.5 mm. If the concentration of electrons in Ge is \[2\times {{10}^{19}}/{{m}^{2}}\] and mobilitys of electrons and holes are \[\text{0}\text{.36 }{{\text{m}}^{\text{2}}}{{\text{V}}^{\text{-1}}}{{\text{s}}^{\text{-1}}}\]and \[\text{0}\text{.14 }{{\text{m}}^{\text{2}}}{{\text{V}}^{\text{-1}}}{{\text{s}}^{\text{-1}}}\]respectively, then the current flowing through the plate will be

    A)  0.25 A

    B)  0.45 A

    C)  0.56 A

    D)  0.64 A

    Correct Answer: D

    Solution :

    Conductivity,  \[\sigma =ne\,({{\mu }_{e}}+{{\mu }_{h}})\] \[=2\times {{10}^{19}}\times 1.6\times {{10}^{-19}}\] \[(0.36+0.14)=1.6{{(\Omega m)}^{-1}}\] \[R=\rho \frac{l}{A}=\frac{l}{\sigma A}\] \[=\frac{0.5\times {{10}^{-3}}}{1.6\times {{10}^{-4}}}=\frac{25}{8}\Omega \] \[i=\frac{V}{R}=\frac{2}{25/8}\] \[=\frac{16}{25}A=0.64A\]  


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