A) \[{{R}_{B}}=15\,k\Omega \]
B) \[{{R}_{B}}=200\,k\Omega \]
C) \[{{R}_{B}}=1\,k\Omega \]
D) \[{{R}_{B}}=185\,k\Omega \]
Correct Answer: D
Solution :
[d] By the definition \[\therefore \,\,\,\,\,\,\,\,\,\,{{i}_{B}}=\frac{{{i}_{C}}}{{{\beta }_{dc}}}=\frac{4mA}{100}=0.04mA\] From figure \[{{i}_{B}}{{R}_{B}}+{{V}_{BE}}=\,\,\,\,{{V}_{CC}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{R}_{B}}=\frac{{{V}_{CC}}-{{V}_{BE}}}{{{i}_{B}}}=\frac{(8-0.6)}{0.04\times {{10}^{-3}}}=185\,k\Omega \] Also \[{{R}_{L}}{{i}_{C}}+{{V}_{CE}}={{V}_{CC}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,{{R}_{L}}=\frac{{{V}_{CC}}-{{V}_{CE}}}{{{i}_{C}}}=\frac{8-4}{8\times {{10}^{-3}}}=1\,k\Omega \]You need to login to perform this action.
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