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An amplifier has an open-loop gain of $100$, an input impedance of $1 \; \mathrm{k \Omega}$, and an output impedance of $100 \; \Omega$. A feedback network with a feedback factor of $0.99$ is connected to the amplifier in a voltage series feedback mode. The new input and output impedances, respectively, are

  1. $10 \; \Omega$ and $1 \; \Omega$
  2. $10 \; \Omega$ and $10 \; k \Omega$
  3. $100 \; \Omega$ and $1 \; \Omega$
  4. $100 \; k \Omega$ and $1 \; k\Omega$

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None of these. Answer is 100k ohm and 1 ohm respectively.

factor = 1 + Ab = 1 + (100)(0.99) = 100

For voltage series, Zif = Zi x factor = 1000 x 100 = 100k ohm

Zof = Zi / factor = 100 / 100 = 1 ohm
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