The output characteristics of a transistor is shown in the figure. When $$V_{CE}$$ is $$10\,\text{V}$$ and $$I_C = 4.0\,\text{mA}$$, then value of $$\beta_{ac}$$ is___

From the given output characteristics graph at $$V_{CE} = 10\text{ V}$$:

Operating point: $$I_C = 4.0\text{ mA} \implies I_B = 30\ \mu\text{A}$$

Taking adjacent points around the operating point at $$V_{CE} = 10\text{ V}$$:

Upper point: $$I_{C2} = 5.5\text{ mA}$$ at $$I_{B2} = 40\ \mu\text{A}$$

Lower point: $$I_{C1} = 4.0\text{ mA}$$ at $$I_{B1} = 30\ \mu\text{A}$$

$$\Delta I_C = I_{C2} - I_{C1} = 5.5 - 4.0 = 1.5\text{ mA} = 1.5 \times 10^{-3}\text{ A}$$

$$\Delta I_B = I_{B2} - I_{B1} = 40 - 30 = 10\ \mu\text{A} = 10 \times 10^{-6}\text{ A}$$

$$\beta_{ac} = \frac{1.5 \times 10^{-3}}{10 \times 10^{-6}} = 150$$