A) 4.9mA, 0.1 mA
B) 4.9 mA, 0.2 mA
C) 5.9mA, 0.3mA
D) 5.9 mA, 0.8 mA
Correct Answer: A
Solution :
[a] : Given \[\alpha =0.98\]and \[\Delta {{I}_{E}}=5.0mA\] From the definition of,\[\alpha =\frac{\Delta {{I}_{C}}}{\Delta {{I}_{E}}}\] Change in collector current \[\Delta {{I}_{C}}=(\alpha )(\Delta {{I}_{E}})=0.98\times 5.0=4.9mA\] Change in base current, \[\Delta {{I}_{B}}=\Delta {{I}_{E}}-\Delta {{I}_{C}}=\left( 5.0-4.9 \right)mA=0.1mA\]
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