The MOSFET switches shown in the circuit are ideal. Which of the following is the correct option for Boolean logical expression of the output (OUT), and the maximum possible power (P) consumed by the circuit?

The correct answer is:
$OUT=\overline{AB}C, P=7.5\text{ mW}$

Step-by-step Explanation:

1. Logic Analysis of the Circuit:
Let us analyze the behavior of the ideal MOSFET switches in the circuit diagram shown in the first image:
- The NMOS transistor controlled by input $C$ has its drain connected to node $X$ and its source connected to ground. Node $X$ is also pulled up to $5\text{V}$ via a $10\text{kΩ}$ resistor.
- When $C=1$, this NMOS transistor turns ON and pulls node $X$ to ground ($0\text{V}$).
- When $C=0$, the transistor is OFF, so node $X$ is pulled up to $5\text{V}$. Thus, the logical state of node $X$ is:
$X=\overline{C}$

Next, let us look at the pull-down paths for the output node $OUT$:
- The middle NMOS transistor has its gate connected to node $X$. It turns ON when $X=1$ (which corresponds to $C=0$), pulling $OUT$ to ground ($0\text{V}$).
- The two NMOS transistors controlled by $A$ and $B$ are connected in series. They form a pull-down path that is active when both $A=1$ and $B=1$.
- The output node $OUT$ is connected to $5\text{V}$ via a $5\text{kΩ}$ pull-up resistor. Therefore, the output is low ($0$) if any pull-down path is ON. This gives the logical expression for the low output as:
$\overline{OUT}=(A\cdot B)+\overline{C}$

Applying De Morgan's laws to find the expression for $OUT$:
$OUT=\overline{(A\cdot B)+\overline{C}}=\overline{A\cdot B}\cdot \overline{\overline{C}}=\overline{AB}C$

2. Maximum Power Consumption Analysis:
The power consumed by the circuit depends on the current drawn from the $5\text{V}$ source. There are two paths connected to the supply voltage:
- Path 1 (via $10\text{kΩ}$ resistor): When transistor $C$ is ON ($C=1$), node $X$ is connected directly to ground. The current flowing through this branch is:
$I_{10\text{k}}=\frac{5\text{V}-0\text{V}}{10\text{kΩ}}$
- Path 2 (via $5\text{kΩ}$ resistor): When $OUT$ is pulled down to $0\text{V}$, current flows through the $5\text{kΩ}$ resistor:
$I_{5\text{k}}=\frac{5\text{V}-0\text{V}}{5\text{kΩ}}$

To maximize power consumption, both paths must draw maximum current simultaneously. This happens when:
- $C=1$ (so that $I_{10\text{k}}=0.5\text{mA}$). Node $X$ is $0\text{V}$, and the middle switch is OFF.
- $A=1$ and $B=1$ (so that the series branch pulls $OUT$ to ground, making $I_{5\text{k}}=1.0\text{mA}$).

Under these conditions ($A=1, B=1, C=1$), the total maximum current is:
$I_{\text{max}}=I_{10\text{k}}+I_{5\text{k}}=0.5\text{mA}+1.0\text{mA}=1.5\text{mA}$

The maximum power consumed by the circuit is:
$P_{\text{max}}=V_{DD}\cdot I_{\text{max}}=5\text{V}\times 1.5\text{mA}=7.5\text{mW}$