Animated Solution for Physics - Semiconductors: The truth table for the circuit given in the figure is

Logic gates are the fundamental building blocks of digital electronics, and mastering them is like learning the alphabet of computers. In this problem, we are presented with a combination of an OR gate and a NAND gate and asked to find the corresponding truth table. While we could simply plug in all possible combinations of 0s and 1s, using Boolean algebra offers a much more elegant and insightful approach.

Let's start by carefully tracing the paths of our inputs, $A$ and $B$. Input $A$ splits into two paths: one goes directly into the top input of the NAND gate, and the other goes into the top input of the OR gate. Input $B$ goes straight into the bottom input of the OR gate.

To make our analysis systematic, let's introduce an intermediate variable. Let's call the output of the OR gate $X$. Since the inputs to this OR gate are $A$ and $B$, we can immediately write down the Boolean expression for $X$:

Now, let's look at the final stage of our circuit: the NAND gate. The inputs feeding into this NAND gate are the original input $A$ and our intermediate signal $X$. A NAND gate performs an AND operation followed by a NOT operation (inversion). Therefore, the final output $Y$ can be expressed as:

Now that we have our expressions, we can combine them. By substituting the expression for $X$ into our equation for $Y$, we get a single Boolean equation that describes the entire circuit:

This expression looks a bit bulky, but this is where the magic of Boolean algebra comes in. We can use standard laws to simplify it. First, let's apply the Distributive Law to expand the term inside the parenthesis:

Next, we apply the Idempotent Law, which states that any variable ANDed with itself is just the variable itself ($A\cdot A=A$). This makes intuitive sense: if a statement is true AND true, it's just true. Applying this simplifies our expression to:

Now, we encounter a beautiful shortcut known as the Absorption Law. This law states that $A+A\cdot B=A$. Why? Because if $A$ is 1 (True), the whole expression is 1 regardless of $B$. If $A$ is 0 (False), the $A\cdot B$ term is also 0, making the whole expression 0. The term $A\cdot B$ is completely "absorbed" by $A$. Applying this law, the entire expression under the bar collapses:

What a remarkable result! Despite having two inputs and two logic gates, the final output $Y$ is simply the inverse of input $A$. The input $B$ has absolutely no effect on the final outcome. It's a "dummy" input in this specific configuration.

Knowing that $Y=\overline{A}$, constructing the truth table is trivial. Whenever $A$ is 0, $Y$ must be 1. Whenever $A$ is 1, $Y$ must be 0. Let's write it out:

Comparing this result with the given options, we can confidently conclude that option (c) is the correct answer. This problem beautifully illustrates how Boolean algebra can reveal the hidden simplicity behind seemingly complex digital circuits.