In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage = 4 V)

Q jee_main_2025_28_jan_morning Logic Gates

Which of the following circuits has the same output as that of the given circuit?

Logic Gates diagram for Q13 - JEE Main 2025 Morning — A combination gate circuit configuration evaluated for total Boolean output expressions.

A.
B.
C.
D.

Solution

Core Logic

Let's perform Boolean analysis on the configuration steps mapped below:

P = A \cdot B

Q = A \cdot B

Y = P + Q = A \cdot B + A \cdot B

Factoring using distributive Boolean rules:

Y = A \cdot (B + B) = A \cdot 1

Y = 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: $(A AND NOT B) OR (A AND B)$ collapses back into simply input A because operand B covers all possible states.

Chapter Mix

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

Q jee_main_2025_03_april_morning Logic 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.
B.
C.
D.

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 = 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 = 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: $A(A+B) = A$ $A(A+B) = A$).

Chapter Mix

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

Q16 jee_main_2025_04_april_evening Extrinsic Semiconductors

Consider a n-type semiconductor in which

nₑ

and

nh

are number of electrons and holes, respectively. (A) Holes are minority carriers (B) The dopant is a pentavalent atom (C)

nₑnh\neq nᵢ²

(where

nᵢ

is number of electrons or holes in semiconductor when it is in intrinsic form) (D)

nₑnh\geq nᵢ²

(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:

nₑ \cdot nh = nᵢ²

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 $nₑ nh = nᵢ²$ $n_e n_h = n_i^2$, 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 ( $nₑ nh = nᵢ²$ $n_e n_h = n_i^2$) 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.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Q jee_main_2025_04_april_morning Logic Gates

The Boolean expression

Y=AoverlineBC+overlineAoverlineC

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 = A BC + A C

By De Morgan's laws:

A\cdot C = A+C (NOR configuration)

Core Logic

Let's analyze configurations A and B:

Configuration A: Generates $A BC$ $A\overline{B}C$ using one 3-input AND gate and one NOT gate for input $B$ $B$. Generates $A C$ $\overline{A}\overline{C}$ using one 2-input AND gate and two separate NOT gates for inputs $A$ $A$ and $C$ $C$. Combines both terms using a 2-input OR gate.

(Total: one 3-input AND, one 2-input AND, three NOT gates, one 2-input OR gate).

Logic circuit layout A for Q15 - JEE Main 2025 Morning

Step 1: Verify Configuration B

Configuration B: Generates $A BC$ $A\overline{B}C$ using one 3-input AND gate and one NOT gate for input $B$ $B$. Simplifies the second term $A C$ $\overline{A}\overline{C}$ into $A+C$ $\overline{A+C}$, realized directly with a single 2-input NOR gate. Combines both sub-circuits using a 2-input OR gate.

(Total: one 3-input AND, one 2-input NOR, one NOT gate, one 2-input OR gate).

Step 2: Verify Configuration C

Configuration C: Specifies a 3-input OR gate and a 2-input AND gate at the output, which implements a product-of-sums form rather than the required sum-of-products expression. Hence, Configuration C is invalid.

Both configurations A and B correctly realize the logic function.

Pattern Recognition

Apply De Morgan's theorem ( $A\cdot B = A+B$ ) to convert negated AND products into standard NOR gate structures, reducing the total gate count.

Evaluation Rubric / Model Answer

Option B: A, B Only

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Q11 jee_main_2025_07_april_evening Logic Gates

Consider the following logic circuit.

Logic Gates diagram for Q11 - JEE Main 2025 Evening — The diagram illustrates a combination of an AND gate, an inverter, an OR gate, and a terminal NAND gate with inputs A and B.

The output is

Y=0

when : [cite: 64, 89]

A. A=1 and B=1 [cite: 90]
B. A=0 and B=1 [cite: 99]
C. A=1 and B=0 [cite: 91]
D. A=0 and B=0 [cite: 100]

Solution

Core Logic

Let the intermediate outputs of the first layers be $Y₁$ $Y_1$ and $Y₂$ $Y_2$ [cite: 747]:

Top gate is an AND gate with inputs $A$ $A$ and $B$ $B$, so $Y₁ = A \cdot B$ $Y_1 = A \cdot B$ [cite: 747].
Bottom gate is an OR gate where one input is $B$ $B$ and the other is $A$ $\bar{A}$ via a NOT gate, so $Y₂ = A + B$ $Y_2 = \bar{A} + B$ [cite: 747].
The final layer is a NAND gate with inputs $Y₁$ $Y_1$ and $Y₂$ $Y_2$, so $Y = Y₁ \cdot Y₂$ $Y = \overline{Y_1 \cdot Y_2}$[cite: 748].

Step 1: Constructing the Truth Table

Let's compute the output $Y$ $Y$ for all binary input pairs $(A, B)$ $(A, B)$ [cite: 757]:

For $A=0, B=0 Y₁ = 0, Y₂ = 1 Y = 0 \cdot 1 = 1$ [cite: 757].
For $A=1, B=0 Y₁ = 0, Y₂ = 0 Y = 0 \cdot 0 = 1$ [cite: 757].
For $A=0, B=1 Y₁ = 0, Y₂ = 1 Y = 0 \cdot 1 = 1$ [cite: 757].
For $A=1, B=1 Y₁ = 1, Y₂ = 1 Y = 1 \cdot 1 = 0$ [cite: 757].

Thus, $Y=0$ $Y=0$ uniquely when $A=1$ $A=1$ and $B=1$ $B=1$[cite: 89, 90, 757].

Pattern Recognition

A NAND gate produces an output of $0$ $0$ if and only if all its inputs are $1$ $1$. Working backward, this instantly sets $Y₁=1$ $Y_1=1$ and $Y₂=1$ $Y_2=1$. For $Y₁ = A \cdot B = 1$ $Y_1 = A \cdot B = 1$, we must have $A=1$ $A=1$ and $B=1$ $B=1$ simultaneously.

Chapter Mix

Class 12 Physics: Semiconductors

In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage = 4 V) A schematic showing a 12V supply connected to a series 100 ohm resistor, which then branches into a parallel Zener diode and a 400 ohm resistor in series with an ammeter.

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Branch Current

Pattern Recognition

Chapter Mix

More Semiconductor Electronics: Materials, Devices and Simple Circuits Previous-Year Questions — Page 4

Solution

Core Logic

Step 1: Final Reduction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Verification of Circuit (2)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Assemble Correct Set

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Verify Configuration B

Step 2: Verify Configuration C

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Core Logic

Step 1: Constructing the Truth Table

Pattern Recognition

Chapter Mix

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In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage = 4 V)
A schematic showing a 12V supply connected to a series 100 ohm resistor, which then branches into a parallel Zener diode and a 400 ohm resistor in series with an ammeter.