Discussion
A stage with \( R_{C} = 5 \ \mathrm{k\Omega} \) drives another of \( 5 \ \mathrm{k\Omega} \) input impedance. Its effective load is
Answer: B
The two appear in parallel to the signal:
\[ R_{L} = \frac{5 \times 5}{10} = 2.5 \ \mathrm{k\Omega} \]
so 2.5 kilohms, halving the first stage's gain. Adding them would apply to a series connection, which is not what a coupled load is. Recognising which combination applies is settled by asking whether the two elements share a voltage or a current.
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