Because of a manufacturing error, the open-loop gain of each op-amp in the circuit is only $A_{o L}=100$. The open-loop input and output resistances are $R_i=10 \mathrm{k} \Omega$ and $R_o=1 \mathrm{k} \Omega$, respectively. Determine the closed-loop parameters: (a) $R_{i f}$, (b) $R_{o f}$, and (c) $A_{C L}=v_{O 2} / v_I$. (d) What is the ratio of the actual closed-loop gain to the ideal value?

For the op-amp in a differential amplifier, the open-loop parameters are: $A_{O L}=10^3, R_i=\infty$, and $R_o=0$. Determine the actual differential voltage gain $A_d=v_O /\left(v_{I 2}-v_{I 1}\right)$. What is the percentage error from the ideal value?

A summing amplifier has an op-amp with open-loop parameters: $A_{O L}=2 \times 10^3, R_i=\infty$, and $R_o=0$. Determine the actual output voltage as a function of $v_{I 1}$ and $v_{I 2}$. What is the percent error from the ideal value?

For the op-amp in a voltage follower circuit, the open-loop parameters are $A_{O L}=5 \times 10^3, R_i=10 \mathrm{k} \Omega$, and $R_o=1 \mathrm{k} \Omega$. (a) Sketch the small-signal equivalent circuit. (b) Determine the (i) closed-loop voltage gain $v_O / v_I$ and (ii) output resistance $R_{o f}$

The noninverting amplifier has an op-amp with open-loop properties: $A_{O L}=10^3, R_i=20 \mathrm{k} \Omega$, and $R_o=0.5 \mathrm{k} \Omega$. (a) Determine the closed-loop values of $A_{C L}=v_O / v_I, R_{i f}$, and $R_{o f}$. (b) If $A_{O L}$ decreases by 10 percent, determine the percentage change in $A_{C L}$.