Discussion
With \( A = 500 \) and \( \beta = 0.02 \), the gain once the loop is closed comes to
Answer: B
Computing the loop gain first, then applying the equation:
\[ \begin{aligned} A\beta &= 500 \times 0.02 = 10 \\ A_{f} &= \tfrac{500}{11} \approx 45.5 \end{aligned} \]
giving about 45.5, some nine per cent below the ideal fifty. A loop gain of ten is modest, so the departure from the ideal is visible; at a loop gain of a thousand it would be a tenth of a per cent. The size of that departure is itself the measure of how much feedback is present.
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