For a transistor the current amplification factor is `0.8`. The transistor is connected in common emitter configuration. The change in the collector curren when the base current chages by a `6mA` is

Options

A. `6 mA`
B. `4.8 mA`
C. `24 mA`
D. `8 mA`

Explanation

To solve the problem, we need to find the change in collector current (ΔIc) when the base current (ΔIb) changes by 6 mA, given that the current amplification factor (α) is 0.8.

### Step-by-Step Solution:

1. **Understand the relationship between α and β**:
The current amplification factor (α) is related to the current gain (β) by the formula:
\[ \beta = \frac{\alpha}{1 - \alpha} \]
Given that α = 0.8, we can calculate β.

2. **Calculate β**:
Substitute α into the formula:
\[ \beta = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4 \]

3. **Use the relationship between ΔIc and ΔIb**:
The relationship between the change in collector current (ΔIc) and the change in base current (ΔIb) is given by:
\[ \Delta I_c = \beta \cdot \Delta I_b \]
We know that ΔIb = 6 mA.

4. **Calculate ΔIc**:
Substitute the values of β and ΔIb into the equation:
\[ \Delta I_c = 4 \cdot 6 \text{ mA} = 24 \text{ mA} \]

### Final Answer:
The change in the collector current when the base current changes by 6 mA is **24 mA**.

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