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

Options

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

Answer

To solve the problem, we need to use the relationship between the change in collector current (\( \Delta I_C \)) and the change in base current (\( \Delta I_B \)) given by the current amplification factor (\( \beta \)).

### Step-by-Step Solution:

1. **Understand the given parameters**:
- Current amplification factor (\( \beta \)) = 0.8
- Change in base current (\( \Delta I_B \)) = 6 mA

2. **Use the formula for current amplification factor**:
\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]
Here, \( \Delta I_C \) is the change in collector current we need to find.

3. **Rearrange the formula to find \( \Delta I_C \)**:
\[ \Delta I_C = \beta \times \Delta I_B \]

4. **Substitute the known values into the equation**:
\[ \Delta I_C = 0.8 \times 6 \text{ mA} \]

5. **Calculate \( \Delta I_C \)**:
\[ \Delta I_C = 4.8 \text{ mA} \]

### Final Answer:
The change in collector current (\( \Delta I_C \)) when the base current changes by 6 mA is **4.8 mA**.

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