For the following logic circuit, the truth table is: 1. A B Y: 0 0 0; 0 1 0; 1 0 0; 1 1 1 2. A B Y: 0 0 1; 0 1 1; 1 0 1; 1 1 0 3. A B Y: 0 0 0; 0 1 1; 1 0 1; 1 1 1 4. A B Y: 0 0 1; 0 1 0; 1 0 1; 1 1 0

Answer: 3. A B Y: 0 0 0; 0 1 1; 1 0 1; 1 1 1 The inputs A and B pass through NOT gates, and the results go into a NAND gate. Y=bar A·bar B=A+B (De Morgan), which is an OR gate. ABY 000 011 101 111 This is option 3.

Topic: Logic gates (OR, AND, NOT, NAND, NOR)· Electronic Devices

For the following logic circuit, the truth table is:

A and B each through a NOT gate, outputs into a NAND gate giving Y

1. A B Y: 0 0 0; 0 1 0; 1 0 0; 1 1 1

2. A B Y: 0 0 1; 0 1 1; 1 0 1; 1 1 0

3. A B Y: 0 0 0; 0 1 1; 1 0 1; 1 1 1

4. A B Y: 0 0 1; 0 1 0; 1 0 1; 1 1 0

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Answer: 3. A B Y: 0 0 0; 0 1 1; 1 0 1; 1 1 1

The inputs A and B pass through NOT gates, and the results go into a NAND gate.

\(Y=\overline{\bar A\cdot\bar B}=A+B\) (De Morgan), which is an OR gate.

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

This is option 3.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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For the following logic circuit, the truth table is:

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