Consider an ideal OP-AMP circuit. The resistances $R_1=R_2=R_3=R_4=50k\Omega$. The magnitude of the closed loop gain is $\text{\_\_\_\_}$ (rounded off to two decimal places).

The correct answer is 2.00.

To understand why the closed-loop gain magnitude of the ideal operational amplifier (op-amp) circuit is 2.00, we can analyze the standard non-inverting amplifier configuration.

For a non-inverting operational amplifier circuit:
1. The input voltage is applied directly to the non-inverting terminal (+).
2. A feedback resistor, denoted as $R_f$, is connected between the output terminal and the inverting terminal (-).
3. An input resistor, denoted as $R_{in}$, connects the inverting terminal (-) to ground.

Using the golden rules of an ideal op-amp:
- The current entering both input terminals is zero due to infinite input impedance.
- The voltage at the inverting terminal $V_{-}$ is equal to the voltage at the non-inverting terminal $V_{+}$ due to the virtual short concept ($V_{-}=V_{+}=V_{in}$).

Applying Kirchhoff's Current Law (KCL) at the inverting terminal node:

$\frac{V_{out}-V_{in}}{R_f}=\frac{V_{in}-0}{R_{in}}$

Rearranging the equation to solve for the closed-loop voltage gain ($A_v$):

$A_v=\frac{V_{out}}{V_{in}}=1+\frac{R_f}{R_{in}}$

Given that the resistances are:
$R_1=R_2=R_3=R_4=50\text{k}\Omega$

In a standard non-inverting amplifier layout with these resistors, we associate the feedback resistance $R_f=R_2=50\text{k}\Omega$ and the input resistance $R_{in}=R_1=50\text{k}\Omega$. Substituting these values into the gain formula:

$A_v=1+\frac{50\text{k}\Omega }{50\text{k}\Omega }=1+1=2$

Therefore, the magnitude of the closed-loop gain is:
$\left|A_v\right|=2.00$