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Question: For a transistor, the current amplification factor is 0.8. The transistor is connected in common emi...

For a transistor, the current amplification factor is 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.

A

6 mA

B

4.8 mA

C

24 mA

D

8 mA

Answer

24 mA

Explanation

Solution

α = 0.8 ⇒β=0.8(1−0.8)\Rightarrow \beta = \frac{0.8}{(1 - 0.8)}=4

Also β=ΔicΔib⇒Δic=β×Δib\beta = \frac{\Delta i_{c}}{\Delta i_{b}} \Rightarrow \Delta i_{c} = \beta \times \Delta i_{b}= 4 × 6 = 24mA.