A) 6 mA
B) 4.8 mA
C) 24 mA
D) 8 mA
Correct Answer: C
Solution :
\[\alpha =0.8\] \[\beta =\frac{\alpha }{1-\alpha }\] \[=\frac{0.8}{1-0.8}=4\] \[\beta =\frac{\Delta \,{{I}_{C}}}{\Delta \,{{I}_{B}}}\] or \[\Delta \,{{I}_{C}}=\beta \,\Delta \,{{I}_{B}}\] \[=4\times 6\] \[=24\,mA\]You need to login to perform this action.
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