A) the base current will be 1 mA
B) the base current will be 10 mA
C) the emitter current will be 1.1 mA
D) the emitter current will be 9 mA
Correct Answer: A
Solution :
Key Idea: Emitter current is sum of base current and collector current. We know that amplification facto\[\alpha =\frac{\Delta {{I}_{C}}}{\Delta {{I}_{E}}}\] \[\therefore \] \[\Delta {{I}_{E}}=\frac{\Delta {{I}_{C}}}{\alpha }=\frac{10}{0.9}=11mA\] Also, \[\Delta {{I}_{E}}=\Delta {{I}_{B}}+\Delta {{I}_{C}}\] \[\Rightarrow \]\[\Delta {{I}_{B}}=\Delta {{I}_{E}}-\Delta {{I}_{C}}\] \[=11-10\] \[=1\,\text{mA}\] Alternative: The collector current \[{{I}_{C}}=10\,mA\] As given, 90% of emitter current = collector current or \[\frac{90}{100}{{I}_{E}}={{I}_{C}}\] \[{{I}_{E}}=\frac{100}{90}{{I}_{C}}\] \[=\frac{100}{90}\times 10\] \[\simeq 11\,\text{mA}\] Thus, \[{{I}_{B}}={{I}_{E}}-{{I}_{C}}\] \[=11-10=1\,\text{mA}\]
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