Identify the logic gate given in the circuit :

Solution

Physics sir · 3 steps

Push both NOT gates into the NAND with De Morgan's law, and the double bars cancel to leave A + B.

1. Read the circuit. Input A goes through a NOT gate and becomes Ā. Input B goes through a NOT gate and becomes B̄. These two signals feed a NAND gate.

   Y = Ā·B̄̅ = Ā̄̄ + B̄̄̄ = A + B

2. Apply De Morgan. De Morgan's law turns the bar over a product into a sum of barred terms: X̅Y̅ = X̄ + Ȳ. Two bars on the same variable cancel, so Ā̄̄ = A.

   A	B	Ā	B̄	Y = Ā·B̄̅
   0	0	1	1	0
   0	1	1	0	1
   1	0	0	1	1
   1	1	0	0	1

3. So. The output is 0 only when both inputs are 0, and 1 otherwise. That is the OR gate truth table, so the answer is (a).

The idea to keep. Inverting both inputs of a NAND gives OR, and inverting both inputs of a NOR gives AND. Use De Morgan's law and cancel the double bars.

Answer from the NTA final answer key. Solution by PrepWiser.