Discussion
Taking \( A = 10^{5} \) and \( \beta = 10^{-3} \) instead, the result settles near
Answer: C
The loop gain is a hundred, so the ideal thousand is approached closely:
\[ A_{f} = \frac{100000}{1 + 100} \approx 990 \]
giving about 990, one per cent short. That one per cent is the residual dependence on the amplifier's own gain, and it halves each time the loop gain doubles. Quoting the ideal figure and the departure from it separately is more informative than either alone.
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