Discussion
Consider the circuit of the figure, the voltage Vc , in Volts, is:
Answer: D
\(V_C\) is 10 V.
For an npn transistor to conduct, its base-emitter junction must be forward biased by about 0.7 V, with the base above the emitter. Here the base is tied to ground (0 V) and the emitter returns to ground through \(R_E\), with no negative supply. The emitter can never be pulled below 0 V, so \(V_{BE} \le 0\) and the transistor is cut off.
With no collector current there is no drop across \(R_C\):
\[ V_C = 10 - I_C R_C = 10 - 0 = 10\ \text{V} \]
0 V would describe a short to ground, and 9.8 V or 1.25 V would need a collector current to flow, which assumes the transistor is biased on. The values of \(R_C\) and \(R_E\) do not enter at all.
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