Question

For the following logic circuit, the truth table is:

• A. \(\begin{matrix} A &B  &Y \\   0&0  &0 \\   0&1  &1 \\   1&0  &1 \\   1&1  &1  \end{matrix}\)
• B. \(\begin{matrix} A &B  &Y \\   0&0  &1 \\   0&1  &0 \\   1&0  &1 \\   1&1  &0  \end{matrix}\)
• C. \(\begin{matrix} A &B  &Y \\   0&0  &0 \\   0&1  &0 \\   1&0  &0 \\   1&1  &1  \end{matrix}\)
• D. \(\begin{matrix} A &B  &Y \\   0&0  &1 \\   0&1  &1 \\   1&0  &1 \\   1&1  &0  \end{matrix}\)

Answer 1

The Correct Option is C

To determine the truth table for the given logic circuit, we need to follow the circuit logic step by step. The circuit consists of two NOT gates and one AND gate.

Let's analyze the circuit: 

1. Inputs: The inputs are \( A \) and \( B \).
2. NOT Gates: There are two NOT gates:
   • The first NOT gate takes input \( A \) and outputs \( \overline{A} \).
   • The second NOT gate takes input \( B \) and outputs \( \overline{B} \).
3. AND Gate: The AND gate receives inputs \( \overline{A} \) and \( \overline{B} \), giving the output \( Y = \overline{A} \cdot \overline{B} \).

The operation of the AND gate is such that it outputs 1 only when both inputs are 1. To achieve that, here the input must be the inverted outputs, meaning:

• If both \( A \) and \( B \) are 1, both \( \overline{A} \) and \( \overline{B} \) will be 0, making \( Y = 1 \cdot 1 = 1 \).
• Any other combination of \( A \) and \( B \) will result in 0 because one or both of \( \overline{A} \) or \( \overline{B} \) will be 1 (thus after NOT, they are 0).

The truth table can now be created:

A	B	Y
0	0	0
0	1	0
1	0	0
1	1	1

Therefore, the correct option is the truth table:

\(\begin{matrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{matrix}\)