In the given circuit if the power rating of Zener diode is 10~mW, the value of series resistance Rs to regulate the…

Solution & Explanation

Related Formula

Voltage drop across series resistor:

Vₛ = Vᵢₙ - VZ

Load current:

IL = (VZ)/(RL)

Maximum Zener current:

IZmax = (PZ)/(VZ)

Core Logic

Given values: $Vᵢₙ = 8~V$ $V_{\text{in}} = 8\mathrm{~V}$, $VZ = 5~V$ $V_Z = 5\mathrm{~V}$, $RL = 1~kΩ$ $R_L = 1\mathrm{~k}\Omega$, $PZ = 10~mW$ $P_Z = 10\mathrm{~mW}$.

Voltage across $Rₛ$ $R_s$:

VRₛ = 8 - 5 = 3~V

Current through the load resistor:

IL = (5)/(1 \times 10³) = 5~mA

Maximum current allowed through the Zener diode:

IZmax = 10 \times 10⁻³5 = 2~mA

Step 1: Determine the Range of Resistance

Total current through the series loop: $Iₛ = IZ + IL$ $I_s = I_Z + I_L$

For maximum safety configuration (Zener operating at peak current):

Ismax = IZmax + IL = 2~mA + 5~mA = 7~mA

Rsmin = VRₛIsmax = 3~V7~mA = (3)/(7)~kΩ \approx 428.6~Ω

For minimum Zener current requirement ( $IZ \to 0$ $I_Z \to 0$):

Ismin = 0 + 5~mA = 5~mA

Rsmax = VRₛIsmin = 3~V5~mA = (3)/(5)~kΩ = 600~Ω

Therefore, the required window for regulation is:

(3)/(7)~kΩ < Rₛ < (3)/(5)~kΩ

Step 2: Note on Official Key

None of the given multiple-choice options fall strictly within the stable bounds $[428.6~Ω, 600~Ω]$ . Officially, the answer key evaluates option (3) as correct, though the problem functions fundamentally as a bonus candidate under rigorous design tolerances.

Pattern Recognition

Always solve the current constraints at both boundaries ( $IZ = 0$ $I_{Z} = 0$ and $IZ = Imax$ $I_{Z} = I_{\text{max}}$) to bracket the allowable series resistor zone.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Detailed currents in Zener stabilizer solution path for Q33 — The diagram illustrates a Zener diode stabilizer network with an unregulated input supply of 8V, a Zener voltage of 5V, a series resistor Rs, and a load resistor RL of 1 kOhm.

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Determine the Range of Resistance

Step 2: Note on Official Key

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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Solution & Explanation

Related Formula

Core Logic

Step 1: Determine the Range of Resistance

Step 2: Note on Official Key

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_2024_01_february_morning

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