For a constant collector-emitter voltage of 8 V, the collector current of a transistor reached to the value of 6 mA from 4 mA, whereas base current changed from $$20 \ \mu A$$ to $$25 \ \mu A$$ value. If transistor is in active state, small signal current gain (current amplification factor) will be

We are given:

Collector-emitter voltage: $$V_{CE} = 8$$ V (constant)

Collector current changes from $$I_{C1} = 4$$ mA to $$I_{C2} = 6$$ mA

Base current changes from $$I_{B1} = 20\ \mu A$$ to $$I_{B2} = 25\ \mu A$$

Recall the formula for small signal current gain. The small signal current gain (AC current amplification factor) is defined as:

$$\beta_{ac} = \frac{\Delta I_C}{\Delta I_B}\bigg|_{V_{CE} = \text{constant}}$$

Calculate the changes in currents. $$\Delta I_C = I_{C2} - I_{C1} = 6 - 4 = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$$ and $$\Delta I_B = I_{B2} - I_{B1} = 25 - 20 = 5\ \mu A = 5 \times 10^{-6} \text{ A}$$

Calculate the small signal current gain: $$\beta_{ac} = \frac{2 \times 10^{-3}}{5 \times 10^{-6}} = \frac{2000}{5} = 400$$

The correct answer is Option B: 400.