The output of the given circuit diagram is
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
Solution & Explanation
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.
The image shows a logic circuit composed of NOT gates, OR gates, and a final NOR gate.
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)$$Y = \overline{(A + \overline{B}) + (\overline{A} + B)}$$
Notice that the inner sum simplifies cleanly:
(A + A) + (B + B)$$(A + \overline{A}) + (B + \overline{B})$$
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$$Y = \overline{1} = 0$$
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.
Keywords:#circuit diagram#logic gates#JEE Main 2024 Evening Q49#Semiconductor Electronics JEE Main 2024#Logic Gates JEE Main 2024#OR gate#NOR gate#NOT gate
More Semiconductor Electronics Previous-Year Questions — Page 5
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.
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.
Consider the following statements:
A. The junction area of solar cell is made very narrow compared to a photo diode.
B. Solar cells are not connected with any external bias.
C. LED is made of lightly doped p-n junction.
D. Increase of forward current results in continuous increase of LED light intensity.
E. LEDs have to be connected in forward bias for emission of light.
Choose the correct answer from the options given below :
A. B, D, E Only
B. A, C Only
C. A, C, E Only
D. B, E Only
Solution
Core Logic
Let's analyze each statement conceptually:
Statement A: Solar cells require a wide surface layer area to intercept maximum sunlight illumination, so junction area is large.
Statement B: True. Solar cells operate spontaneously to provide power to loads without requiring external bias voltage.
Statement C: False. LEDs are made of heavily doped junctions to maximize recombination probability.
Statement D: False. Beyond a critical limit, high currents cause heating that drops efficiency, so emission intensity does not increase infinitely.
Statement E: True. Forward biasing allows minority injection leading to radiative recombination.
Step 1: Selecting Option
Since statements B and E are purely accurate, the correct grouping option is B, E Only.
Pattern Recognition
Remember: LEDs = Forward Bias, Photodiodes = Reverse Bias, Solar Cells = Zero External Bias.
Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
Qjee_main_2025_28_jan_eveningDiode Rectifiers
In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage VAB$\mathrm{V_{AB}}$ is correctly represented by:
An AC source connected across an orientation circuit involving an ideal junction diode.
A.VAB would be zero at all times$\mathrm{V_{AB}}\text{ would be zero at all times}$
B.
C.
D.
Solution
Core Logic
Analyze the cycle profile behavior of the input voltage waveform V = V₀ ω t$V = V_0 \sin \omega t$:
Positive Half Cycle: Node A achieves a positive potential relative to node B. Under this configuration, the diode enters a Reverse Biased (R.B.) state, acting as an open switch circuit block. Since no current conducts across the resistive path, the potential difference tracked directly mirrors the input wave voltage.
Negative Half Cycle: Node A goes negative relative to node B. This transitions the diode into a Forward Biased (F.B.) condition, acting as a closed short-circuit bypass path. Consequently, the potential settles down immediately to zero.
This behavior is visualized through the input/output tracking waveforms below:
An AC source connected across an orientation circuit involving an ideal junction diode.
An AC source connected across an orientation circuit involving an ideal junction diode.
Step 1: Selection
Matching this half-wave rectified configuration precisely selects option (4).
Pattern Recognition
When solving diode waveform problems, replace the diode mentally with an open circuit during reverse bias and a short circuit during forward bias to quickly observe the resulting output profile.
Chapter Mix
Class 12 Physics: Semiconductor Electronics
Qjee_main_2025_29_jan_morningLogic Gates
The image features a digital circuit blueprint constructed from basic logic gates with inputs A and B mapped to an output Y.
For the circuit shown above, equivalent GATE is :
The image features a digital circuit blueprint constructed from basic logic gates with inputs A and B mapped to an output Y.
A. OR gate
B. NOT gate
C. AND gate
D. NAND gate
Solution
Core Logic
Evaluating the given logic gate diagram combination step-by-step for all input permutations yields the following truth table :
Input A
Input B
Output Y
0
0
0
0
1
1
1
0
1
1
1
1
This behavior matches an OR Gate configuration perfectly.
Chapter Mix
Class 12 Physics: Semiconductor Electronics
Qjee_main_2024_01_february_morningZener Diode
In the given circuit if the power rating of Zener diode is 10~mW$10\mathrm{~mW}$, the value of series resistance Rₛ$R_s$ to regulate the input unregulated supply is:
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.
None of the given multiple-choice options fall strictly within the stable bounds [428.6~Ω, 600~Ω]$[428.6\mathrm{~\Omega}, 600\mathrm{~\Omega}]$. 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
More Semiconductor Electronics Questions — jee_main_2024_31_jan_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
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