The logic circuit shown above is equivalent to :

Solution

Physics sir · 3 steps

Write the output as one Boolean expression, then let De Morgan's theorem turn the NOR into an AND.

1. Read the circuit. B passes through a NOT gate, so the OR gate receives A and B̅ and gives A + B̅. The final NOT gate inverts this whole sum.

   C = A + B̅̅ = A̅·B̅̅̅ = A̅· B

2. Apply De Morgan. The inverse of a sum is the product of the inverses. The double bar on B cancels, so the circuit is an AND gate fed by A̅ and B.

   Option	Gate and inputs	Output
   (a)	NAND of A̅ and B	A̅̅̅B̅ = A + B̅
   (b)	NOR of A and B	A + B̅ = A̅ B̅
   (c)	NAND of A and B̅	A̅B̅̅̅ = A̅ + B
   (d)	AND of A̅ and B	A̅ B

3. So. Only option (d) gives C = A̅ B. Check one case: with A = 0 and B = 1, the OR gets 0 and 0, gives 0, and the NOT makes C = 1. The AND in (d) also gives 1.

The idea to keep. A NOT after an OR is a NOR, and by De Morgan a NOR is an AND of the inverted inputs. Inverting a signal twice gives back the original signal.

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