The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at VZ = 5V and the desired current in load is 5 mA. The unregulated voltage source can supply up to 25V. Considering the Zener diode can withstand four times of the load current, the value of resistor RS (shown in circuit) should be ____ Omega.

The question is dropped by the examination authority, meaning there was an anomaly in data or multiple correct interpretations, likely due to imprecise wording regarding the maximum source voltage and the exact safe current rating. In a nominal calculation, IZ_max = 4 x 5 mA = 20 mA. Total current I = 25 mA. RS = (25 - 5) / (25 x 10-3) = 800Omega, which is not in the options. Question Dropped. Sees: "Question Dropped" → This often occurs when mathematical data maps to none of the provided MCQs. Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at VZ = 5V and the desired current in load is 5 mA. The unregulated voltage source can supply up to 25V. Considering the Zener diode can withstand four times of the load current, the value of resistor RS (shown in circuit) should be ____ Omega.

The question is dropped by the examination authority, meaning there was an anomaly in data or multiple correct interpretations, likely due to imprecise wording regarding the maximum source voltage and the exact safe current rating. In a nominal calculation, IZ_max = 4 x 5 mA = 20 mA. Total current I = 25 mA. RS = (25 - 5) / (25 x 10-3) = 800Omega, which is not in the options. Question Dropped. Sees: "Question Dropped" → This often occurs when mathematical data maps to none of the provided MCQs. Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at VZ = 5V and the desired…

Solution & Explanation

Related Formula

Itotal = IZ + IL

Vᵢₙ - VZ = Itotal RS

Core Logic

The question is dropped by the examination authority, meaning there was an anomaly in data or multiple correct interpretations, likely due to imprecise wording regarding the maximum source voltage and the exact safe current rating. In a nominal calculation, $I_Zmax = 4 \times 5 mA = 20 mA$ . Total current $I = 25 mA$ $I = 25 \text{ mA}$. $RS = (25 - 5) / (25 \times 10⁻³) = 800Ω$ , which is not in the options.

Step 1: Final Conclusion

Question Dropped.

Pattern Recognition

Sees: "Question Dropped" → This often occurs when mathematical data maps to none of the provided MCQs.

Chapter Mix

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

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

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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Step 1: Final Conclusion

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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

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