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

Solution & Explanation

Related Formula

Voltage division across load $RL$ $R_L$ with series resistance $Rₛ$ $R_s$ without Zener regulation:

VL = Vᵢₙ ( (RL)/(Rₛ + RL) )

If $VL > Vz$ $V_L > V_z$, the Zener diode enters breakdown, and potential across the parallel load is clamped at $VL = Vz$ $V_L = V_z$.

Core Logic

Verify if Zener diode operates in the breakdown region:

$Vᵢₙ = 12 ~V$ $V_{\text{in}} = 12 \mathrm{~V}$
$Rₛ = 100 Ω$ $R_s = 100 \Omega$
$RL = 400 Ω$ $R_L = 400 \Omega$

Calculate the unregulated voltage:

V₁ = 12 \times ( (400)/(100 + 400) ) = 12 \times (4)/(5) = 9.6 ~V

Since $V₁ > Vz$ $V_1 > V_z$ ( $9.6 ~V > 4 ~V$ $9.6 \mathrm{~V} > 4 \mathrm{~V}$), breakdown occurs, and the parallel branch voltage is fixed at $Vz = 4 ~V$ $V_z = 4 \mathrm{~V}$.

Step 1: Calculate Branch Current

The voltage across the $400 Ω$ $400 \Omega$ load resistor (which is in series with the ammeter) is locked at $4 ~V$ $4 \mathrm{~V}$.

The current $I$ $I$ through the ammeter is:

I = (Vz)/(RL) = 4 ~V400 Ω = 10⁻² ~A = 10 ~mA

Pattern Recognition

Sees: Parallel Zener diode configuration. Shortcut: Always calculate the open-circuit load voltage first. If it exceeds $Vz$ $V_z$, use $Vz$ $V_z$ as the branch potential. The branch current is simply $Vz / RL$ $V_z / R_L$. Here, $4 / 400 = 10 ~mA$ $4 / 400 = 10 \mathrm{~mA}$.

Chapter Mix

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

Phasor and load lines representation for Zener breakdown state — 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.

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 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 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_07_april_morning

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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.

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 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_07_april_morning

Select Your Target Exam

Full Mock Test Hub

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Welcome back

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.