In the digital circuit shown in the figure, for the given inputs the P and Q values are :

Q: In the digital circuit shown in the figure, for the given inputs the P and Q values are :

Truth relations of basic logic operations: - NAND operation: Y = A cdot B - NOR operation: Y = A + B - NOT operation: Y = A - OR operation: Y = A + B The inputs are: - Top input = 1 - Bottom input = 1 Let's analyze step-by-step from left to right: 1. **First Gate (NAND gate at the top-left):** - Inputs are 1 and 1. - Output = 1 cdot 1 = 0. 2. **Bottom-left path with NOT gates:** - Top input (1) goes to a NOT gate, producing 0. - Bottom input (1) goes to a NOT gate, producing 0. - These two 0 val

Solution & Explanation

Related Formula

Truth relations of basic logic operations:

NAND operation: $Y = A \cdot B$ $Y = \overline{A \cdot B}$
NOR operation: $Y = A + B$ $Y = \overline{A + B}$
NOT operation: $Y = A$ $Y = \overline{A}$
OR operation: $Y = A + B$ $Y = A + B$

Core Logic

The inputs are:

Top input = $1$ $1$
Bottom input = $1$ $1$

Let's analyze step-by-step from left to right:

First Gate (NAND gate at the top-left):
Inputs are $1$ $1$ and $1$ $1$.
Output = $1 \cdot 1 = 0$ $\overline{1 \cdot 1} = 0$.
Bottom-left path with NOT gates:
Top input ( $1$ $1$) goes to a NOT gate, producing $0$ $0$.
Bottom input ( $1$ $1$) goes to a NOT gate, producing $0$ $0$.
These two $0$ $0$ values feed into the OR gate:
Output = $0 + 0 = 0$ $0 + 0 = 0$.

Step 1: Calculate output P

Now trace the path to $P$ $P$:

The inputs to the top-right AND gate are:
Output of the top-left NAND gate = $0$ $0$
Output of the bottom-left OR gate = $0$ $0$
Therefore, output $P$ $P$ is:

P = 0 \cdot 0 = 0

Step 2: Calculate output Q

Now trace the path to $Q$ $Q$:

The gate at the bottom-right is a NOR gate with two inputs:
Input 1: Output of the top-left NAND gate ( $0$ $0$) inverted by a NOT gate = $0 = 1$ $\overline{0} = 1$.
Input 2: Output of the bottom-left OR gate ( $0$ $0$).
Passing these inputs ( $1$ $1$ and $0$ $0$) through the final NOR gate:

Q = 1 + 0 = 1 = 0

Thus, both $P = 0$ $P = 0$ and $Q = 0$ $Q = 0$.

Pattern Recognition

Sees: Combinational trace with inverted nodes. Trap: Missing bubbles (NOT gates) representing inversion on internal circuit paths. Shortcut: The first NAND gate output is $0$ $0$ (since both inputs are 1). This $0$ $0$ directly goes to the upper AND gate, immediately guaranteeing output $P = 0$ $P = 0$ (eliminates options 1 and 4). Now, you only need to evaluate $Q$ $Q$ to choose between options 2 and 3.

Chapter Mix

Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

Step-by-step digital logic state values annotated on the circuit diagram — The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.

More Semiconductor Electronics: Materials, Devices and Simple Circuits 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

In the digital circuit shown in the figure, for the given inputs the P and Q values are :
The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate output P

Step 2: Calculate output Q

Pattern Recognition

Chapter Mix

Reference Study Guides

More Semiconductor Electronics: Materials, Devices and Simple Circuits 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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In the digital circuit shown in the figure, for the given inputs the P and Q values are : The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate output P

Step 2: Calculate output Q

Pattern Recognition

Chapter Mix

Reference Study Guides

More Semiconductor Electronics: Materials, Devices and Simple Circuits 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: Materials, Devices and Simple Circuits Questions — jee_main_2025_02_april_evening

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In the digital circuit shown in the figure, for the given inputs the P and Q values are :
The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.