A) \[\frac{2}{9}\]
B) \[\frac{1}{2}\]
C) \[\frac{9}{2}\]
D) \[\frac{1}{6}\]
Correct Answer: C
Solution :
\[k\Omega \] \[{{n}_{e}}=3.13\times {{10}^{15}}\] \[{{n}_{p}}=3.12\times {{10}^{15}}\] Since, the two lines intersect. So, putting above values in second line, we get \[i={{n}_{e}}{{q}_{e}}+{{n}_{p}}{{q}_{p}}\] Taking 1 st and 3rd terms, we get 2r ? 2 = 4r + 1 \[=3.13\times {{10}^{15}}\times 1.6\times {{10}^{-17}}\] r = - 3/2 Also, taking 2nd and 3rd terms, we get 3r ? 1 ? k = 8r + 2 \[+3.12\times {{10}^{15}}\times 1.6\times {{10}^{-19}}\] \[=1\times {{10}^{-3}}=1mA\]
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