The output of the circuit is low (zero) for :
(A) X=0, Y=0
(B) X=0, Y=1
(C) X=1, Y=0
(D) X=1, Y=1
Choose the correct answer from the options given below :

Solution

Physics sir · 3 steps

The second gate's two inputs are joined, so that NOR is just a NOT. OR then NOT makes the whole circuit a NOR.

1. First gate. X and Y go into an OR gate, so the wire leaving it carries X + Y.
2. Second gate. That one wire feeds both inputs of a NOR gate. A NOR with the same signal A on both inputs gives A + A̅ = A̅, so it works as a NOT gate.

   Output = X + Y̅

   Case	X, Y	X + Y	Output X̅+̅Y̅
   (A)	0, 0	0	1 (high)
   (B)	0, 1	1	0 (low)
   (C)	1, 0	1	0 (low)
   (D)	1, 1	1	0 (low)

3. So. The output is low for (B), (C) and (D). It is high only when both inputs are 0, which is option (c).

The idea to keep. When both inputs of a NAND or NOR gate are tied together, the gate becomes a NOT gate. Simplify it first, then the rest of the circuit is easy to read.

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