Question

For a transistor, the current amplification factor \( \alpha \) is \( 0.8 \). The transistor is then connected in common emitter configuration. The change in the collector current when the base current changes by \( 6 \text{ mA} \) is

• A. $6 \text{ mA}$
• B. $4.8 \text{ mA}$
• C. $24 \text{ mA}$
• D. $8 \text{ mA}$

Hint:

- $\beta = \frac{\alpha}{1 - \alpha}$ - In CE mode: $\Delta I_C = \beta \Delta I_B$

Answer 1

The Correct Option is C

Concept: Relation between current gains: \[ \beta = \frac{\alpha}{1 - \alpha} \] Also, \[ \beta = \frac{\Delta I_C}{\Delta I_B} \]

Step 1: Calculate $\beta$.
\[ \beta = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4 \]

Step 2: Use current relation.
\[ \Delta I_C = \beta \cdot \Delta I_B \]

Step 3: Substitute values.
\[ \Delta I_C = 4 \times 6 = 24\ \text{mA} \] Answer: $24 \text{ mA}$