In the given circuit, value of Y is :

Solution

Physics sir · 4 steps

There is no feedback loop here. Start at the inputs and push 0s and 1s gate by gate until you reach Y.

1. First NOT. The input 1 passes through the first NOT gate and becomes 0. This 0 goes three places: both inputs of the top NAND, and the second NOT gate.
2. Top NAND. Its inputs are 0 and 0, so its output is 0 · 0̅ = 1. This wire runs down the right side and hops over the Y line, so it is not joined to Y. It goes on to the lower input of the bottom NAND.

   Gate	Inputs	Output
   First NOT	1	0
   Top NAND	0, 0	1
   Second NOT	0	1
   Bottom NAND	0 (input), 1 (from top NAND)	1
   Middle NAND	1 (from second NOT), 1 (from bottom NAND)	0 = Y

3. Bottom and middle. The bottom NAND gets 0 and 1, so it gives 1. That 1 feeds the middle NAND. The middle NAND also gets 1 from the second NOT, and a NAND with both inputs 1 gives 0.
4. So. Y = 1 · 1̅ = 0, a fixed value, which is option (d).

The idea to keep. In a gate network, first check whether a wire truly joins or just hops over. If no output feeds back into its own gate, the output is fixed, so trace it from the inputs one gate at a time.

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