The ideal OP-AMP circuit shown in the Figure produces output voltage V_o = x when the Switch, S, is open. Which of the options represents the output voltage when S is closed?

The correct option is:
$\frac{3}{4}x$

Analysis of the Circuit Diagram:
Based on the provided circuit schematic:
1. An ideal operational amplifier (OP-AMP) is configured in a non-inverting feedback configuration.
2. The input voltage at the non-inverting terminal (+) is labeled as $V_i=1\text{V}$.
3. The inverting terminal (-) is connected to:
- A resistor of resistance $1\text{k}\Omega$ connected to ground.
- A feedback branch consisting of a $2\text{k}\Omega$ resistor connected in series with a parallel combination of a $1\text{k}\Omega$ resistor and switch $S$.
4. The output voltage is labeled as $V_o$.

Step-by-Step Derivation:
For an ideal operational amplifier with negative feedback, the virtual short concept applies:
$V_{-}=V_{+}=V_i=1\text{V}$
The current $I_1$ flowing from the inverting terminal node through the grounded $1\text{k}\Omega$ resistor to ground is:
$I_1=\frac{V_{-}-0}{1\text{k}\Omega }=\frac{1\text{V}}{1\text{k}\Omega }=1\text{mA}$
Since no current enters the input terminals of an ideal OP-AMP, the feedback current $I_f$ flowing from the output through the feedback network to the inverting terminal is equal to $I_1$:
$I_f=I_1=1\text{mA}$
Thus, the output voltage is given by:
$V_o=V_{-}+I_f\times R_f$
where $R_f$ is the equivalent resistance of the feedback network.

Case 1: Switch $S$ is Open
When the switch is open, the current must flow through both the $2\text{k}\Omega$ and the $1\text{k}\Omega$ feedback resistors in series:
$R_f=2\text{k}\Omega +1\text{k}\Omega =3\text{k}\Omega$
The output voltage is:
$V_o=1\text{V}+(1\text{mA}\times 3\text{k}\Omega )=1\text{V}+3\text{V}=4\text{V}$
According to the problem statement, this open-switch output voltage is defined as $x$:
$x=4\text{V}$

Case 2: Switch $S$ is Closed
When the switch is closed, it acts as a short circuit (zero resistance) across the $1\text{k}\Omega$ feedback resistor. Therefore, the feedback network's equivalent resistance becomes:
$R_f=2\text{k}\Omega +0=2\text{k}\Omega$
The new output voltage is:
$V_o=1\text{V}+(1\text{mA}\times 2\text{k}\Omega )=1\text{V}+2\text{V}=3\text{V}$

Expressing the Output Voltage in terms of $x$:
To write the closed-switch output voltage in terms of $x$:
$V_{o,\text{closed}}=3\text{V}=\frac{3}{4}\times \left(4\text{V}\right)=\frac{3}{4}x$
Thus, the output voltage when the switch is closed is indeed $\frac{3}{4}x$.