JEE Main 2023 · 11 April · morning · Single correct · +4 / −1
The logic performed by the circuit shown in figure is equivalent to :
- (a) NAND
- (b) AND
- (c) OR
- (d) NOR
Answer and solution
Answer (b) AND
Solution
Physics sir · 3 steps
A NOR gate with both inputs tied together is a NOT gate, so this circuit is NOR of ā and b̄.
- First two gates. Each input gate is a NOR with both its inputs joined. That gives a + a̅ = ā and b + b̅ = b̄, so each one works as a NOT gate.
- Final gate. The last NOR receives ā and b̄. Apply De Morgan's theorem: the bar over a sum becomes a product of the barred terms.
Y = ā + b̄̅ = ā̄̄ · b̄̄̄ = a · b
a b ā b̄ Y = ā + b̄̅ 0 0 1 1 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 1 - So. The output is 1 only when both inputs are 1, which is the AND gate, option (b).
The idea to keep. NOR with tied inputs gives NOT. Invert both inputs and then NOR them, and you get AND. Invert both inputs and then NAND them, and you get OR.
Answer from the NTA final answer key. Solution by PrepWiser.