A) 4mA
B) 2 mA
C) 3.6mA
D) 1.8mA
Correct Answer: D
Solution :
\[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}\] \[\therefore \] \[30=\frac{\Delta {{i}_{C}}}{90-30}\] \[\Delta {{i}_{C}}=30\times 60\] \[=1800\mu A=1.8\,mA\]You need to login to perform this action.
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