Animated Solution for Physics - Semiconductors: In the following logic circuit, the sequence of the inputs $A,B$ are $(0,0),(0,1),(1,0)$ and $(1,1)$. The output $Y$ for this sequence will be

The world of digital electronics is built on simple logical rules that combine to form complex behaviors. In this problem, we are presented with a logic circuit containing an AND gate, an OR gate, and a NAND gate. Our goal is to determine the output sequence for a given set of input pairs.

Let's embark on this logical journey and decode the circuit step by step!

Imagine the electrical signals flowing through the wires. We have two primary inputs, $A$ and $B$. These inputs are fed simultaneously into two different gates.

The top gate is an AND gate. It takes $A$ and $B$ and produces an intermediate output, which we will call $P$.

The bottom gate is an OR gate. It also takes $A$ and $B$ as inputs, producing another intermediate output, $Q$.

Finally, these two intermediate signals, $P$ and $Q$, are routed into a NAND gate to produce our final output, $Y$.

To solve this mathematically, we need to translate the physical gates into Boolean algebra.

For the AND gate, the output is true only if both inputs are true. Therefore, we can write:

For the OR gate, the output is true if at least one input is true. This gives us:

The final gate is a NAND gate, which is an AND gate followed by a NOT gate. It takes the product of its inputs and inverts it. Thus, our final output equation becomes:

Substituting our expressions for $P$ and $Q$, we get the complete Boolean equation for the circuit:

At first glance, this equation looks intimidating. You might be tempted to immediately apply De Morgan's theorem to break the long bar. However, there is a much more elegant logical shortcut!

Look closely at the expression inside the inversion bar: $(A\cdot B)\cdot (A+B)$.

This expression demands that both $(A\cdot B)$ AND $(A+B)$ must be true. But think about the conditions. For $(A\cdot B)$ to be true, both $A$ and $B$ must be 1.

If both $A$ and $B$ are 1, then $(A+B)$ is automatically 1! Therefore, the condition $(A+B)$ is redundant when $(A\cdot B)$ is already enforced.

Mathematically, their intersection is simply the stricter condition:

Substituting this beautiful simplification back into our main equation, we get:

This reveals a stunning truth: our entire complex circuit of three gates is functionally identical to a single, simple NAND gate!

Now that we know our circuit is just a disguised NAND gate, finding the output sequence is a breeze.

A NAND gate outputs 0 only when both of its inputs are 1. For all other input combinations, it outputs 1.

Let's evaluate the given sequence of inputs: For $(0,0)$, the output is $1$. For $(0,1)$, the output is $1$. For $(1,0)$, the output is $1$. For $(1,1)$, the output is $0$.

Therefore, the final output sequence is 1, 1, 1, 0. This perfectly matches option (c).

By combining careful tracing with the power of Boolean algebra, we turned a complex web of logic gates into a trivial calculation!