Question

The output of the given circuit diagram is

• A. \[\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \\ \hline \end{array}\]
• B. \[\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ \hline \end{array}\]
• C. \[\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \hline \end{array}\]
• D. \[\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ \hline \end{array}\]

Answer 1

The Correct Option is C

The circuit diagram provided is a logic gate combination circuit. Let's analyze the circuit step-by-step to determine the output \( Y \) for different input combinations of \( A \) and \( B \).

The circuit contains:

• An NOT gate inverts the input \( B \).
• Two AND gates and one OR gate are used to process the signals.

We will calculate the output for each input combination in the truth table:

1. Inputs: \( A = 0 \), \( B = 0 \)
   • The NOT gate inverts \( B \) (0 becomes 1).
   • The first AND gate has inputs \( A = 0 \) and NOT \( B = 1 \), so its output is \( 0 \cdot 1 = 0 \).
   • The second AND gate has inputs \( B = 0 \) (inverted becomes 1) and \( A = 0 \), so its output is \( 0 \).
   • The OR gate processes the outputs \( 0 \) and \( 0 \), resulting in \( Y = 0 \).
2. Inputs: \( A = 1 \), \( B = 0 \)
   • The NOT gate inverts \( B \) (0 becomes 1).
   • The first AND gate has inputs \( A = 1 \) and NOT \( B = 1 \), so its output is \( 1 \cdot 1 = 1 \).
   • The second AND gate has inputs \( B = 0 \) (inverted becomes 1) and \( A = 1 \), so its output is \( 0 \).
   • The OR gate processes the outputs \( 1 \) and \( 0 \), resulting in \( Y = 1 \).
3. Inputs: \( A = 0 \), \( B = 1 \)
   • The NOT gate inverts \( B \) (1 becomes 0).
   • The first AND gate has inputs \( A = 0 \) and NOT \( B = 0 \), so its output is \( 0 \).
   • The second AND gate has inputs \( B = 1 \) (inverted becomes 0) and \( A = 0 \), so its output is \( 0 \cdot 0 = 0 \).
   • The OR gate processes the outputs \( 0 \) and \( 0 \), resulting in \( Y = 0 \).
4. Inputs: \( A = 1 \), \( B = 1 \)
   • The NOT gate inverts \( B \) (1 becomes 0).
   • The first AND gate has inputs \( A = 1 \) and NOT \( B = 0 \), so its output is \( 0 \).
   • The second AND gate has inputs \( B = 1 \) (inverted becomes 0) and \( A = 1 \), so its output is \( 1 \cdot 0 = 0 \).
   • The OR gate processes the outputs \( 0 \) and \( 0 \), resulting in \( Y = 0 \).

Based on the above analysis, the correct truth table is:

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

Hence, the correct answer is:

\[\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \hline \end{array}\]

Answer 2

The circuit diagram consists of logic gates. By analyzing each gate’s behavior step-by-step and evaluating the output \( Y \) for each input combination of \( A \) and \( B \), we can determine the output for each case. After constructing the truth table for the circuit, we find that the correct output matches option (3).

Thus, the answer is:

\[ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \\ \hline \end{array} \]