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

The truth table for the given logic circuit is :

logic circuit with diode, two NAND gates and an AND-type gate, inputs A,B output C

1. Truth table (A B C): 0 0 0; 0 1 1; 1 0 0; 1 1 1

2. Truth table (A B C): 0 0 0; 0 1 1; 1 0 1; 1 1 0

3. Truth table (A B C): 0 0 1; 0 1 0; 1 0 0; 1 1 1

4. Truth table (A B C): 0 0 1; 0 1 0; 1 0 1; 1 1 0

Doubtly Asked 30 Oct 2026 NEET UG 2022 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 4. Truth table (A B C): 0 0 1; 0 1 0; 1 0 1; 1 1 0

Trace the circuit. The upper NAND gets A and B: \(Y_1=\overline{A\cdot B}\). The lower NAND gets \(\bar A\) (through the NOT gate) and B: \(Y_2=\overline{\bar A\cdot B}\). The AND gate gives \(C=Y_1\cdot Y_2\).

A	B	Y1	Y2	C
0	0	1	1	1
0	1	1	0	0
1	0	1	1	1
1	1	0	1	0

C = 1, 0, 1, 0, which is option (4). (This equals \(\bar B\).)

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

Was this answer helpful?

The truth table for the given logic circuit is :

1 answer

Your answer