The given circuit works as : — jee_main_2026_21_jan

Q: The given circuit works as :

### Related Formula De Morgan's Laws: ### Core Logic Analyzing the circuit diagram: - Top branch has a NOT gate on A, so P = A - Bottom branch has a NOT gate on B, so Q = B - They enter a NOR gate, giving output R = A + B - Finally, R passes through a NOT gate to give S = R ### Step 1: Boolean Simplification The expression AB is the Boolean expression for a NAND gate. ### Pattern Recognition Two NOTs feeding into a NOR equals an AND gate (AB). Adding a final NOT gate turns the AND into a NAND. #

Solution & Explanation

Related Formula

De Morgan's Laws:

A + B = A \cdot B

A = A

Core Logic

Analyzing the circuit diagram:

Logic Gates solution diagram for Q42 - JEE Main 2026 Morning — Circuit containing multiple interconnected logic gates leading to a final output.

Top branch has a NOT gate on A, so $P = A$ $P = \overline{A}$
Bottom branch has a NOT gate on B, so $Q = B$ $Q = \overline{B}$
They enter a NOR gate, giving output $R = A + B$
Finally, R passes through a NOT gate to give $S = R$ $S = \overline{R}$

Step 1: Boolean Simplification

R = A + B = AB = AB

S = R = AB

The expression $AB$ $\overline{AB}$ is the Boolean expression for a NAND gate.

Pattern Recognition

Two NOTs feeding into a NOR equals an AND gate ( $AB$ $AB$). Adding a final NOT gate turns the AND into a NAND.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

More Semiconductor Electronics 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 given circuit works as :
Circuit containing multiple interconnected logic gates leading to a final output.

Solution & Explanation

Related Formula

Core Logic

Step 1: Boolean Simplification

Pattern Recognition

Chapter Mix

More Semiconductor Electronics 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

Chapter Mix

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The given circuit works as : Circuit containing multiple interconnected logic gates leading to a final output.

Solution & Explanation

Related Formula

Core Logic

Step 1: Boolean Simplification

Pattern Recognition

Chapter Mix

More Semiconductor Electronics 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

Chapter Mix

More Semiconductor Electronics Questions — jee_main_2026_21_jan_morning

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The given circuit works as :
Circuit containing multiple interconnected logic gates leading to a final output.