MCQ
In a common base circuit of a transistor, the amplification factor is $0.95.$ The base current when the emitter current is $2\,mA,$ is ....$mA$
- ✓$0.1$
- B$0.2 $
- C$0.19 $
- D$1.9$
✓
Answer
Correct option: A.
$0.1$
a
Given $\alpha=\frac{I_{e}}{I_{E}}=0.95$
Given $\alpha=\frac{I_{e}}{I_{E}}=0.95$
Thus $ \mathrm{I}_{\mathrm{c}}=0.95 \times \mathrm{I}_{\mathrm{E}} $
$=0.95 \times 2 \mathrm{\,mA} $
$=1.9 \mathrm{\,mA}$
The base current is, therefore,
$ \mathrm{I}_{\mathrm{B}} =\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{C}} $
$=2-1.9 \mathrm{\,mA} $
$=0.1 \mathrm{\,mA} $
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