Question:
The output Y for the inputs A and B of circuit is given by Truth table of the shown circuit is:
The output Y for the inputs A and B of circuit is given by Truth table of the shown circuit is:
Updated On: Sep 8, 2025
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The Correct Option is D
Approach Solution - 1
Given circuit represent XOR.
The correct answer is (D) : 
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Approach Solution -2
To solve the given circuit and determine the output \( Y \) for the inputs \( A \) and \( B \), we need to analyze the logic gates used in the circuit.
Circuit Analysis:
The circuit provided in the image represents an XOR gate. Let's verify this by constructing the truth table:
1. NAND Gates:
- The first two NAND gates receive inputs \( A \) and \( B \).
- The output of the first NAND gate, \( \text{NAND1} \), is \( \overline{A \cdot B} \).
- The output of the second NAND gate, \( \text{NAND2} \), is \( \overline{A \cdot B} \).
2. Connecting Outputs:
- The output of \( \text{NAND1} \) is one of the inputs to \( \text{NAND3} \), and the input \( A \) is the other input.
- The output of \( \text{NAND2} \) is one of the inputs to \( \text{NAND4} \), and the input \( B \) is the other input.
3. Third and Fourth NAND Gates:
- The output of \( \text{NAND3} \) is \( \overline{A \cdot \overline{A \cdot B}} \).
- The output of \( \text{NAND4} \) is \( \overline{B \cdot \overline{A \cdot B}} \).
4. Final NAND Gate:
- The outputs of \( \text{NAND3} \) and \( \text{NAND4} \) are inputs to the final NAND gate, \( \text{NAND5} \).
- The output of \( \text{NAND5} \) is \( Y = \overline{\left( \overline{A \cdot \overline{A \cdot B}} \cdot \overline{B \cdot \overline{A \cdot B}} \right)} \).
Now, let’s look into the truth table:
Truth Table Explanation:
- When \( A = 0 \) and \( B = 0 \), \( Y = 0 \).
- When \( A = 0 \) and \( B = 1 \), \( Y = 1 \).
- When \( A = 1 \) and \( B = 0 \), \( Y = 1 \).
- When \( A = 1 \) and \( B = 1 \), \( Y = 0 \).
This matches the truth table for an XOR gate. Thus, the correct option is:
Option 4:
\[\begin{array}{c|c|c}A & B & Y \\\hline0 & 0 & 0 \\0 & 1 & 1 \\1 & 0 & 1 \\1 & 1 & 0 \\\end{array}\]
Conclusion:
The given circuit represents an XOR gate, and the output truth table is correctly represented by Option 4.
Circuit Analysis:
The circuit provided in the image represents an XOR gate. Let's verify this by constructing the truth table:
1. NAND Gates:
- The first two NAND gates receive inputs \( A \) and \( B \).
- The output of the first NAND gate, \( \text{NAND1} \), is \( \overline{A \cdot B} \).
- The output of the second NAND gate, \( \text{NAND2} \), is \( \overline{A \cdot B} \).
2. Connecting Outputs:
- The output of \( \text{NAND1} \) is one of the inputs to \( \text{NAND3} \), and the input \( A \) is the other input.
- The output of \( \text{NAND2} \) is one of the inputs to \( \text{NAND4} \), and the input \( B \) is the other input.
3. Third and Fourth NAND Gates:
- The output of \( \text{NAND3} \) is \( \overline{A \cdot \overline{A \cdot B}} \).
- The output of \( \text{NAND4} \) is \( \overline{B \cdot \overline{A \cdot B}} \).
4. Final NAND Gate:
- The outputs of \( \text{NAND3} \) and \( \text{NAND4} \) are inputs to the final NAND gate, \( \text{NAND5} \).
- The output of \( \text{NAND5} \) is \( Y = \overline{\left( \overline{A \cdot \overline{A \cdot B}} \cdot \overline{B \cdot \overline{A \cdot B}} \right)} \).
Now, let’s look into the truth table:
Truth Table Explanation:
- When \( A = 0 \) and \( B = 0 \), \( Y = 0 \).
- When \( A = 0 \) and \( B = 1 \), \( Y = 1 \).
- When \( A = 1 \) and \( B = 0 \), \( Y = 1 \).
- When \( A = 1 \) and \( B = 1 \), \( Y = 0 \).
This matches the truth table for an XOR gate. Thus, the correct option is:
Option 4:
\[\begin{array}{c|c|c}A & B & Y \\\hline0 & 0 & 0 \\0 & 1 & 1 \\1 & 0 & 1 \\1 & 1 & 0 \\\end{array}\]
Conclusion:
The given circuit represents an XOR gate, and the output truth table is correctly represented by Option 4.
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