The truth table corresponding to the circuit given below is:

Q: The truth table corresponding to the circuit given below is:

### Related Formula Boolean operators for basic logic gates: - OR gate: Y = A + B - AND gate: Y = A cdot B ### Core Logic Analyze the schematic: 1. Input lines A and B are linked into an OR gate, yielding output: Y = A + B 2. This term A + B and the original input line A form the inputs to a final AND gate. 3. The final output expression C is therefore: ### Step 1: Construct the Truth Table Compute output values for each input combination: - For A=0, B=0: - For A=1, B=0: - For A=0, B=1: - For A=

Year	2026	2025	2024	Total
Questions	9	15	8	32

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Boolean operators for basic logic gates:

OR gate:

$Y = A + B$ $Y = A + B$

AND gate:

$Y = A \cdot B$ $Y = A \cdot B$

Core Logic

Analyze the schematic:

Input lines $A$ $A$ and $B$ $B$ are linked into an OR gate, yielding output:

$Y = A + B$ $Y = A + B$

This term $A + B$ $A + B$ and the original input line $A$ $A$ form the inputs to a final AND gate.
The final output expression $C$ $C$ is therefore:

C = A \cdot (A + B)

Logic Gates and Truth Tables — Diagram showing a combination of logic gates connected to inputs A and B resulting in output C.

Step 1: Construct the Truth Table

Compute output values for each input combination:

For $A=0, B=0$ $A=0, B=0$:

C = 0 \cdot (0 + 0) = 0

For $A=1, B=0$ $A=1, B=0$:

C = 1 \cdot (1 + 0) = 1 \cdot 1 = 1

For $A=0, B=1$ $A=0, B=1$:

C = 0 \cdot (0 + 1) = 0 \cdot 1 = 0

For $A=1, B=1$ $A=1, B=1$:

C = 1 \cdot (1 + 1) = 1 \cdot 1 = 1

This matches Table 2.

Pattern Recognition

Using Boolean algebra:

C = A \cdot (A + B) = A \cdot A + A \cdot B = A + A \cdot B

By absorption law:

A + A \cdot B = A

So the circuit is equivalent to a simple buffer carrying input $A$ $A$. The output $C$ $C$ must match input $A$ $A$ under all conditions. Checking the options, Table 2 is the one where output $C$ $C$ tracks input $A$ $A$ directly ( $0, 1, 0, 1$ $0, 1, 0, 1$).

Chapter Mix

Class 12 Physics: Semiconductors

More Semiconductors Previous-Year Questions — Page 7

Q49 jee_main_2024_31_jan_evening Logic Gates

The output of the given circuit diagram is

Logic Gates diagram for Q49 - JEE Main 2024 Evening — The image shows a logic circuit composed of NOT gates, OR gates, and a final NOR gate.

A.
A B Y
0 0 0
1 0 0
0 1 0
1 1 1
B.
A B Y
0 0 0
1 0 1
0 1 1
1 1 0
C.
A B Y
0 0 0
1 0 0
0 1 0
1 1 0
D.
A B Y
0 0 0
1 0 0
0 1 1
1 1 0

A	B	Y
0	0	0
1	0	0
0	1	0
1	1	1

A	B	Y
0	0	0
1	0	1
0	1	1
1	1	0

A	B	Y
0	0	0
1	0	0
0	1	0
1	1	0

A	B	Y
0	0	0
1	0	0
0	1	1
1	1	0

Solution

Related Formula

Boolean Algebra expressions for logic gates: NOT: $A$ $\overline{A}$ OR: $A + B$ $A + B$ NOR: $A + B$ $\overline{A + B}$

Core Logic

Analyze the paths from inputs A and B to the final output Y.

Step 1: Intermediate Signals

Top OR gate inputs: $A$ $A$ directly, and $B$ $B$ inverted ( $B$ $\overline{B}$). Top OR gate output: $A + B$ $A + \overline{B}$

Bottom OR gate inputs: $A$ $A$ inverted ( $A$ $\overline{A}$), and $B$ $B$ directly. Bottom OR gate output: $A + B$ $\overline{A} + B$

Step 2: Final Gate Evaluation

The final gate is a NOR gate taking the two intermediate outputs as its inputs.

Y = (A + B) + ( A + B)

Notice that the inner sum simplifies cleanly:

(A + A) + (B + B)

Since $A + A = 1$ $A + \overline{A} = 1$ and $B + B = 1$ $B + \overline{B} = 1$, the inner term is $1 + 1 = 1$ $1 + 1 = 1$.

Y = 1 = 0

Step 3: Conclusion

The output Y is always 0 regardless of the inputs A and B. Checking the truth tables, only option 3 satisfies $Y=0$ $Y=0$ for all conditions.

Pattern Recognition

When a Boolean expression groups a variable and its exact complement together in an OR configuration ( $A$ $A$ and $A$ $\overline{A}$), the result instantly hits logic 1. Feeding 1 into any NOR gate guarantees a 0 output universally.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Q jee_main_2024_31_jan_morning Logic Gates

Identify the logic operation performed by the given circuit.

Logic Gates diagram for Q32 - JEE Main 2024 Morning — Two inputs passing through NOT gates before entering a NAND gate.

A. $NAND$ $\text{NAND}$
B. $NOR$ $\text{NOR}$
C. $OR$ $\text{OR}$
D. $AND$ $\text{AND}$

Solution

Related Formula

Y = A \cdot B (NAND)

Y = A + B (De Morgan's)

Core Logic

The inputs $A$ $A$ and $B$ $B$ are first passed through individual NOT gates (made from tied-input NAND gates or standard NOT gates). The outputs become $A$ $\overline{A}$ and $B$ $\overline{B}$.

These are then fed into a NAND gate. The final output $Y$ $Y$ is:

Y = A \cdot B

Applying De-Morgan's Law:

Y = A + B

$Y = A + B$ $Y = A + B$

This represents an OR operation.

Pattern Recognition

Bubbled inputs on a NAND gate convert it directly into an OR gate via De-Morgan's laws. (Bubbled NAND = OR).

Chapter Mix

Class 12 Physics: Semiconductor Electronics

The truth table corresponding to the circuit given below is:
Diagram showing a combination of logic gates connected to inputs A and B resulting in output C.

Solution & Explanation

Related Formula

Core Logic

Step 1: Construct the Truth Table

Pattern Recognition

Chapter Mix

More Semiconductors Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Intermediate Signals

Step 2: Final Gate Evaluation

Step 3: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

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The truth table corresponding to the circuit given below is: Diagram showing a combination of logic gates connected to inputs A and B resulting in output C.

Solution & Explanation

Related Formula

Core Logic

Step 1: Construct the Truth Table

Pattern Recognition

Chapter Mix

More Semiconductors Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Intermediate Signals

Step 2: Final Gate Evaluation

Step 3: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

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The truth table corresponding to the circuit given below is:
Diagram showing a combination of logic gates connected to inputs A and B resulting in output C.