Discussion
A signal of \( 5 \ \mathrm{mV} \) is applied to a stage of gain \( -180 \). The output is
Answer: A
Multiplying amplitude by gain and noting the sign:
\[ v_{o} = -180 \times 5 \ \mathrm{mV} = -0.9 \ \mathrm{V} \]
so 0.9 volts, inverted. The magnitude alone is not the whole answer in a common-emitter stage, where the inversion matters as soon as the stage is placed in a feedback loop or cascaded. Carrying the sign through rather than taking magnitudes is a habit worth keeping.
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