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.
### Related Formula
Truth relations of basic logic operations:
- NAND operation: Y = overlineA cdot B$Y = \overline{A \cdot B}$
- NOR operation: Y = overlineA + B$Y = \overline{A + B}$
- NOT operation: Y = overlineA$Y = \overline{A}$
- 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:
1. **First Gate (NAND gate at the top-left):**
- Inputs are 1$1$ and 1$1$.
- Output = overline1 cdot 1 = 0$\overline{1 \cdot 1} = 0$.
2. **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$$P = 0 \cdot 0 = 0$$
### 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 = overline0 = 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 = overline1 + 0 = overline1 = 0$$Q = \overline{1 + 0} = \overline{1} = 0$$
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.
Keywords:#logic gates digital circuit diagram#JEE Main 2025 Evening Q13#trace combinational logic circuit#Semiconductor Electronics logic gates#NAND gate#NOR gate#NOT gate#digital logic circuit
More Semiconductor Electronics: Materials, Devices and Simple Circuits Previous-Year Questions — Page 2
Q13jee_main_2025_28_jan_morningLogic Gates
Which of the following circuits has the same output as that of the given circuit?
A combination gate circuit configuration evaluated for total Boolean output expressions.
A.textCircuit (1)$\text{Circuit (1)}$
B.textCircuit (2)$\text{Circuit (2)}$
C.textCircuit (3)$\text{Circuit (3)}$
D.textCircuit (4)$\text{Circuit (4)}$
Solution
### Core Logic
Let's perform Boolean analysis on the configuration steps mapped below:
mathrmP = mathrmA cdot barmathrmB$$\mathrm{P} = \mathrm{A} \cdot \bar{\mathrm{B}}$$mathrmQ = mathrmA cdot mathrmB$$\mathrm{Q} = \mathrm{A} \cdot \mathrm{B}$$mathrmY = overlinemathrmP + mathrmQ = overlinemathrmA cdot barmathrmB + mathrmA cdot mathrmB$$\mathrm{Y} = \overline{\mathrm{P} + \mathrm{Q}} = \overline{\mathrm{A} \cdot \bar{\mathrm{B}} + \mathrm{A} \cdot \mathrm{B}}$$
Factoring using distributive Boolean rules:
mathrmY = overlinemathrmA cdot (mathrmB + barmathrmB) = overlinemathrmA cdot 1$$\mathrm{Y} = \overline{\mathrm{A} \cdot (\mathrm{B} + \bar{\mathrm{B}})} = \overline{\mathrm{A} \cdot 1}$$mathrmY = barmathrmA$$\mathrm{Y} = \bar{\mathrm{A}}$$
### Step 1: Final Reduction
The expression reduces to a simple inverter (NOT gate) processing input A. This aligns with Circuit (1), matching option (1).
### Pattern Recognition
Identify standard combinations: (mathrmA text AND NOT mathrmB) text OR (mathrmA text AND mathrmB)$(\mathrm{A} \text{ AND NOT } \mathrm{B}) \text{ OR } (\mathrm{A} \text{ AND } \mathrm{B})$ collapses back into simply input A because operand B covers all possible states.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
Q3jee_main_2025_03_april_morningLogic Gates
Choose the correct logic circuit for the given truth table having inputs A and B.
Inputs
Output
A$A$
B$B$
Y$Y$
0
0
0
0
1
0
1
0
1
1
1
1
A. Circuit (1)
B. Circuit (2)
C. Circuit (3)
D. Circuit (4)
Solution
### Related Formula
Let us inspect the Boolean expression for the output Y$Y$ from the truth table.
From the table:
- If A=0$A=0$, Y=0$Y=0$ regardless of B$B$.
- If A=1$A=1$, Y=1$Y=1$ regardless of B$B$.
Thus, the truth table is represented by the simple direct logical equation:
Y = A$Y = A$
### Core Logic
Let's check the Boolean output of the options shown in the question paper:
- **Circuit (1)**: Inputs A$A$ and B$B$ go into an OR gate, outputting (A + B)$(A + B)$. This output and B$B$ then go to an AND gate.
Y = (A + B) cdot B = Acdot B + Bcdot B = B(A + 1) = B$$Y = (A + B) \cdot B = A\cdot B + B\cdot B = B(A + 1) = B$$
This gives Y = B$Y = B$ (Not matching table).
- **Circuit (2)**: Inputs A$A$ and B$B$ go into an OR gate, outputting (A + B)$(A + B)$. This and A$A$ then go into an AND gate.
Y = (A + B) cdot A = Acdot A + Acdot B = A + Acdot B = A(1 + B) = A$$Y = (A + B) \cdot A = A\cdot A + A\cdot B = A + A\cdot B = A(1 + B) = A$$
This gives Y = A$Y = A$ (Perfect match to the truth table where Y$Y$ exactly copies A$A$).
### Step 1: Verification of Circuit (2)
Let's double-check the truth table values for Circuit (2):
- For A=0, B=0$A=0, B=0$: Y = (0 + 0) cdot 0 = 0$Y = (0 + 0) \cdot 0 = 0$.
- For A=0, B=1$A=0, B=1$: Y = (0 + 1) cdot 0 = 0$Y = (0 + 1) \cdot 0 = 0$.
- For A=1, B=0$A=1, B=0$: Y = (1 + 0) cdot 1 = 1$Y = (1 + 0) \cdot 1 = 1$.
- For A=1, B=1$A=1, B=1$: Y = (1 + 1) cdot 1 = 1$Y = (1 + 1) \cdot 1 = 1$.
This perfectly matches the given truth table. Therefore, Circuit (2) is correct.
### Pattern Recognition
Identify the logic expression directly from the truth table first! Notice that Y$Y$ is completely independent of B$B$ and strictly equals A$A$. This immediately points to any Boolean simplification that collapses to A$A$ (such as absorption law: A(A+B) = A$A(A+B) = A$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
Consider a n-type semiconductor in which n_e$n_{e}$ and n_h$n_{h}$ are number of electrons and holes, respectively.
(A) Holes are minority carriers
(B) The dopant is a pentavalent atom
(C) n_en_hne n_i^2$n_{e}n_{h}\ne n_{i}^{2}$ (where n_i$n_{i}$ is number of electrons or holes in semiconductor when it is in intrinsic form)
(D) n_en_hge n_i^2$n_{e}n_{h}\ge n_{i}^{2}$
(E) The holes are not generated due to the donors
Choose the correct answer from the options given below:
A. (A), (C), (D) only
B. (A), (C), (E) only
C. (A), (B), (E) only
D. (A), (B), (C) only
Solution
### Related Formula
Mass Action Law:
n_e cdot n_h = n_i^2$$n_e \cdot n_h = n_i^2$$
### Core Logic
Let's analyze each statement for an n-type semiconductor:
- (A) Holes are minority carriers: True, electrons are the majority carriers.
- (B) The dopant is a pentavalent atom: True (like Phosphorus, Arsenic) which provides extra free electrons.
- (C) and (D) contradict the fundamental mass action law n_e n_h = n_i^2$n_e n_h = n_i^2$, so they are False.
- (E) Holes are generated purely due to thermal excitation, not due to donor atoms: True.
### Step 1: Assemble Correct Set
Statements (A), (B), and (E) are explicitly correct.
### Pattern Recognition
Mass action law (n_e n_h = n_i^2$n_e n_h = n_i^2$) holds uniformly for both doped types at thermal equilibrium. In n-type systems, donors directly inject electrons only; holes emerge solely from thermal breakages of lattice bonds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
Q15jee_main_2025_04_april_morningLogic Gates
The Boolean expression Y=AoverlineBC+overlineAoverlineC$Y=Aoverline{B}C+overline{A}overline{C}$ can be realised with which of the following gate configurations.
A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate
B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate
C. 3-input OR gate, 3 NOT gates and one 2-input AND gate
Choose the correct answer from the options given below
A. B, C Only
B. A, B Only
C. A, B, C Only
D. A, C Only
Solution
### Related Formula
Given logical expression:
Y = AoverlineBC + overlineAoverlineC$$Y = Aoverline{B}C + overline{A}overline{C}$$
By Boolean theorem logic:
overlineAoverlineC = overlineA+C quad text(NOR configuration)
$$overline{A}overline{C} = overline{A+C} \quad \text{(NOR configuration)}
$$
### Core Logic
Let's analyze configuration options A and B:
* **Configuration A:** Creates AoverlineBC$Aoverline{B}C$ via one 3-input AND gate and one NOT gate for B$B$. Creates overlineAoverlineC$overline{A}overline{C}$ via one 2-input AND gate and two separate NOT gates for A$A$ and C$C$. Finally combines them using a 2-input OR gate. This perfectly mirrors the terms. (Total gates: one 3-input AND, one 2-input AND, three NOT, one 2-input OR).
Logic circuit layout A for Q15 - JEE Main 2025 Morning
### Step 1: Verify Configuration B
* **Configuration B:** Creates AoverlineBC$Aoverline{B}C$ via one 3-input AND gate and one NOT gate for B$B$. Re-writes the secondary term overlineAoverlineC$overline{A}overline{C}$ into overlineA+C$overline{A+C}$, which corresponds directly to a single 2-input NOR gate. Merges both sub-tracks using a 2-input OR gate. This is also completely valid.
Logic circuit layout A for Q15 - JEE Main 2025 Morning
### Pattern Recognition
Always utilize De Morgan's identity triggers to dynamically simplify multiple product complements like overlineAoverlineC$overline{A}overline{C}$ into unified NOR formats.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
Q15jee_main_2025_24_jan_eveningLogic Gates
The output of the circuit is low (zero) for : The figure contains a digital combination logic gate configuration with input variables X and Y.
(A) X = 0, Y = 0
(B) X = 0, Y = 1
(C) X = 1, Y = 0
(D) X = 1, Y = 1
Choose the correct answer from the options given below:
A. (A), (C) and (D) only
B. (A), (B) and (C) only
C. (B), (C) and (D) only
D. (A), (B) and (D) only
Solution
### Core Logic
Let us check the gate outputs row-by-row to find the boolean expression or map the truth table values: The figure contains a digital combination logic gate configuration with input variables X and Y.beginarrayccc X & Y & textOutput \hline 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \ 1 & 1 & 0 endarray$$\begin{array}{ccc} X & Y & \text{Output} \\hline 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \ 1 & 1 & 0 \end{array}$$
The output is low (zero) for configurations (B) X=0, Y=1$X=0, Y=1$, (C) X=1, Y=0$X=1, Y=0$, and (D) X=1, Y=1$X=1, Y=1$. Therefore, options (B), (C) and (D) only are correct.
### Pattern Recognition
The truth table profile matches a standard NOR logic configuration where the output is 1 only when all input lines are completely low.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
More Semiconductor Electronics: Materials, Devices and Simple Circuits Questions — jee_main_2025_02_april_evening
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