A) \[0.01mA\]
B) \[0.04mA\]
C) \[0.02mA\]
D) \[0.03mA\]
Correct Answer: B
Solution :
\[V={{V}_{CE}}+{{i}_{C}}{{R}_{L}}\] \[\Rightarrow \] \[15=7+{{i}_{C}}\times 2\times {{10}^{3}}\] \[\Rightarrow \] \[{{i}_{C}}=4mA\] \[\because \] \[\beta =\frac{{{i}_{C}}}{{{i}_{B}}}\Rightarrow \,\,\,{{i}_{B}}=\frac{4}{100}=0.04\,\,mA\]You need to login to perform this action.
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