In the circuit shown, the open loop gain of the operational amplifier is \(A_0=105\), with \(R_{in}=\infty\ \Omega\) and \(R_{out}=0\ \Omega\).
The circuit is a standard inverting amplifier: \(V_{in}=100\) mV drives the inverting input through a \(5\ k\Omega\) input resistor, and a \(100\ k\Omega\) resistor feeds back from \(V_{out}\) to the inverting input, with the non-inverting input grounded. What is the voltage gain of the circuit? (Round off to two decimal places)
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Use the finite-gain inverting amplifier formula: gain = -(Rf/Rin) divided by [1 + (1+Rf/Rin)/A0].
Step 1: Do not assume a perfect virtual short.
With a finite gain $A_0$, the inverting input is not exactly at ground. Call the inverting input voltage $v_-$ and the output $v_o$. Since the op-amp obeys $v_o=-A_0v_-$ (non-inverting input grounded), we get
\[ v_-=-\frac{v_o}{A_0} \]
Step 2: Write the current balance at the inverting node.
No current enters the op-amp input, so the current through $R_{in}$ from $V_{in}$ must equal the current through $R_f$ into the output:
\[ \frac{V_{in}-v_-}{R_{in}}=\frac{v_--v_o}{R_f} \]
Step 3: Substitute $v_-=-v_o/A_0$.
\[ \frac{V_{in}+v_o/A_0}{R_{in}}=\frac{-v_o/A_0-v_o}{R_f} \]
Multiply both sides by $R_f R_{in}$:
\[ R_f\left(V_{in}+\frac{v_o}{A_0}\right)=R_{in}\left(-\frac{v_o}{A_0}-v_o\right) \]
Step 5: Plug in numbers.
With $R_{in}=5$k, $R_f=100$k, $A_0=105$:
\[ R_{in}\left(1+\frac{1}{105}\right)=5\times1.00952=5.0476 \]
\[ \frac{R_f}{A_0}=\frac{100}{105}=0.9524 \]
\[ \frac{v_o}{V_{in}}=\frac{-100}{5.0476+0.9524}=\frac{-100}{6.0}=-16.67 \]
Step 6: Conclude.
This matches the standard finite-gain formula and confirms the gain is not the ideal $-20$ but a slightly smaller magnitude, $-16.67$.
\[ \boxed{-16.67} \]
