Identify Logic Gate from Output Conditions Table

 

❓ Question

The output Y = 0 when:

1. A = 1 and B = 1

2. A = 0 and B = 1

3. A = 1 and B = 0

4. A = 0 and B = 0

(Which logic gate does this condition represent?)

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Logic gates
👉 NAND/NOR operations
👉 Boolean algebra.

Main idea:

Break the circuit gate-by-gate and simplify the Boolean expression.

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🔷 Step 1 — Identify First Gate 💯

Top gate is:

$$\text{AND gate}$$

Inputs:

$$A,\ B$$

Thus output:

$$X=AB$$

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🔷 Step 2 — Bottom Part

Input $B$ first passes through NOT gate:

$$\overline{B}$$

Then enters OR gate with $B$.

Thus OR output:

$$B+\overline{B}$$

Using Boolean identity:

$$\boxed{ B+\overline{B}=1 }$$

So lower branch output is:

$$1$$

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🔷 Step 3 — Final Gate

Final gate is NAND gate.

Inputs are:

$$AB \quad \text{and} \quad 1$$

Thus:

$$Y=\overline{(AB)(1)}$$
$$Y=\overline{AB}$$

This is NAND operation.

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🔷 Step 4 — Find When $Y=0$

For NAND gate:

$$Y=0$$

only when:

$$AB=1$$

That means:

$$A=1,\quad B=1$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Final gate ko NOR samajh lena

❌ $B+\overline B$ simplify na kar pana

❌ Bubble notation ignore kar dena

Remember:

$$\boxed{ X+\overline X=1 }$$

and:

$$\boxed{ \overline{A\cdot B}=\text{NAND} }$$

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✅ Final Answer

$$\boxed{ A=1,\ B=1 }$$

(Option 1)

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