A) 0.25 A
B) 0.45 A
C) 0.56 A
D) 0.64 A
Correct Answer: D
Solution :
\[\sigma =ne({{\mu }_{e}}+{{\mu }_{h}})=2\times {{10}^{19}}\times 1.6\times {{10}^{-19}}(0.36+0.14)\] \[=1.6\,{{(\Omega \text{-}m)}^{-1}}\] \[R=\rho \frac{l}{A}=\frac{l}{\sigma A}=\frac{0.5\times {{10}^{-3}}}{1.6\times {{10}^{-4}}}=\frac{25}{8}\Omega \] \ \[i=\frac{V}{R}=\frac{2}{25/8}=\frac{16}{25}A=0.64\,A\]You need to login to perform this action.
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