Animated Solution for Physics - Semiconductors: In the given circuit, value of $Y$ is

A Step-by-Step Trace

Logic circuits can often look like an intimidating maze of wires and symbols. However, they are governed by strict, unbreakable mathematical rules. By patiently tracing a signal from the input to the output, we can solve even the most complex-looking networks. Let's walk through this specific circuit and uncover a common pitfall found in some reference materials.

Analyzing the Setup

Before we start moving signals around, let's take inventory of our circuit. We have a single input signal with a constant value of $1$. This signal feeds into a network composed of three distinct logic gates: 1. A NOT gate (the triangle with a bubble), which inverts its input. 2. An AND gate (the D-shaped symbol), which outputs $1$ only if all its inputs are $1$. 3. A NAND gate (the D-shaped symbol with a bubble), which is the exact opposite of an AND gate.

The First Split and the NOT Gate

Our journey begins at the input. The signal $1$ travels along the wire and immediately hits a junction, splitting into two parallel paths.

Let's follow the top path first. The signal $1$ enters the NOT gate. The operation here is straightforward: the NOT gate inverts the high signal to a low signal.

So, emerging from the NOT gate, we now have a signal of $0$.

The Feedback Loop and the AND Gate

Now, pay close attention to where this $0$ signal goes. It hits another junction! One branch continues forward to the top input of the final NAND gate. The other branch loops downwards to become the top input of the AND gate.

Let's evaluate the AND gate. We need to know both of its inputs: Top input: $0$ (coming from the NOT gate's output). Bottom input: $1$ (coming directly from the main input split).

The Boolean logic for an AND gate dictates that the output is $1$ if and only if both inputs are $1$. Since one of our inputs is $0$, the output is forced to be $0$.

The Final NAND Gate

We are now ready to tackle the final gate in the circuit, the NAND gate. Let's gather its inputs: Top input: $0$ (coming from the NOT gate). Bottom input: $0$ (coming from the AND gate).

A NAND gate performs an AND operation followed by a NOT operation. First, we evaluate $0\text{ AND }0$, which is $0$. Then, we invert that result. The inverse of $0$ is $1$.

By strictly following the rules of Boolean algebra, we have determined that the final output $Y$ is $1$.

It is crucial to address a common discrepancy here. Some textbooks and answer keys incorrectly list the answer to this specific problem as $0$. This error stems from a flawed evaluation at the very last step, incorrectly assuming that a NAND gate with inputs $0$ and $0$ yields an output of $0$.

As a student of physics and mathematics, you must always trust the fundamental axioms over a printed answer key. The truth table for a NAND gate is absolute: $\overline{0\cdot 0}=1$. Therefore, the correct option is (b) 1.