MCQ
For a transistor, the current amplification factor is$\alpha 0.8$. The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by $6\, mA$ is.....$mA$
- A$6$
- B$4.8$
- ✓$24$
- D$8$
✓
Answer
Correct option: C.
$24$
c
(c)$\alpha= 0.8 \Rightarrow \beta = \frac{{0.8}}{{(1 - 0.8)}}=4$
Also $\beta = \frac{{\Delta {i_c}}}{{\Delta {i_b}}}$$ \Rightarrow \Delta {i_c} = \beta \times \Delta {i_b}$
(c)$\alpha= 0.8 \Rightarrow \beta = \frac{{0.8}}{{(1 - 0.8)}}=4$
Also $\beta = \frac{{\Delta {i_c}}}{{\Delta {i_b}}}$$ \Rightarrow \Delta {i_c} = \beta \times \Delta {i_b}$
$= 4 × 6 = 24\,mA.$
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