Discussion
With a base current giving \( I_{C} = 2 \ \mathrm{mA} \), that stage's collector-emitter voltage is
Answer: C
Subtracting both resistive drops from the supply:
\[ \begin{aligned} V_{CE} &= 15 - 2\mathrm{m}(3000+1000) \\ &= 15 - 8 = 7 \ \mathrm{V} \end{aligned} \]
so 7 volts, close enough to mid-supply for a good swing. Nine volts counts only the collector resistor. Because the same current flows in both resistors, they can be added before multiplying, which halves the arithmetic and the chances of an error.
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