JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
The output Y for the inputs A and B of circuit is given by
Truth table of the shown circuit is:
- (a)
- (b)
- (c)
- (d)
Answer and solution
Answer (a)
Solution
Physics sir · 3 steps
Name each gate's output, then push all four input pairs through the circuit one gate at a time.
- Label the outputs. Call the first NAND's output C = A̅B̅. The top NAND gives P = A̅ ̅C̅, the bottom NAND gives Q = B̅ ̅C̅, and the last NAND gives Y = P̅ ̅Q̅.
- Simplify. By De Morgan, Y = P̅ ̅Q̅ = P̅ + Q̅ = AC + BC = (A + B) A̅B̅.
Y = (A + B) A̅B̅ = AB̅ + A̅B = A ⊕ B
A B C = A̅B̅ P = A̅C̅ Q = B̅C̅ Y = P̅Q̅ 0 0 1 1 1 0 0 1 1 1 0 1 1 0 1 0 1 1 1 1 0 1 1 0 - So. The output column reads 0, 1, 1, 0. This is the XOR truth table, option (a).
The idea to keep. Four NAND gates in this shape, one feeding two and those two feeding a last one, always make XOR. Label every intermediate output and the truth table follows row by row.
Answer from the NTA final answer key. Solution by PrepWiser.