Which of the following circuits has the same output as that of the given circuit?
A combination gate circuit configuration evaluated for total Boolean output expressions.
A.
A combination gate circuit configuration evaluated for total Boolean output expressions.
B.
A combination gate circuit configuration evaluated for total Boolean output expressions.
C.
A combination gate circuit configuration evaluated for total Boolean output expressions.
D.
A combination gate circuit configuration evaluated for total Boolean output expressions.
Solution & Explanation
Core Logic
Let's perform Boolean analysis on the configuration steps mapped below:
P = A · B$$\mathrm{P} = \mathrm{A} \cdot \bar{\mathrm{B}}$$Q = A · B$$\mathrm{Q} = \mathrm{A} \cdot \mathrm{B}$$Y = P + Q = A · B + A · B$$\mathrm{Y} = \overline{\mathrm{P} + \mathrm{Q}} = \overline{\mathrm{A} \cdot \bar{\mathrm{B}} + \mathrm{A} \cdot \mathrm{B}}$$
Factoring using distributive Boolean rules:
Y = A · (B + B) = A · 1$$\mathrm{Y} = \overline{\mathrm{A} \cdot (\mathrm{B} + \bar{\mathrm{B}})} = \overline{\mathrm{A} \cdot 1}$$Y = A$$\mathrm{Y} = \bar{\mathrm{A}}$$
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)$(\mathrm{A} \text{ AND NOT } \mathrm{B}) \text{ OR } (\mathrm{A} \text{ AND } \mathrm{B})$ 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
A combination gate circuit configuration evaluated for total Boolean output expressions.
More Semiconductor Electronics: Materials, Devices and Simple Circuits Previous-Year Questions — Page 6
Q44jee_main_2024_30_jan_morningZener Diode as a Voltage Regulator
A Zener diode of breakdown voltage 10V$10\mathrm{V}$ is used as a voltage regulator as shown in the figure. The current through the Zener diode is
A Zener diode regulator circuit with a 20V source and two resistors.
A.50 ~mA$50 \mathrm{~mA}$
B.0$0$
C.30 ~mA$30 \mathrm{~mA}$
D.20 ~mA$20 \mathrm{~mA}$
Solution
Related Formula
Itotal = Iz + IL$$I_{\text{total}} = I_z + I_L$$Vload = Vz (if in breakdown)$$V_{\text{load}} = V_z \quad (\text{if in breakdown})$$
Core Logic
A Zener diode regulator circuit with a 20V source and two resistors.
The Zener is in the breakdown region because the open-circuit voltage across it without the Zener (20 × (500)/(700) = 14.28V$20 \times \frac{500}{700} = 14.28\mathrm{V}$) is greater than Vz = 10V$V_z = 10\mathrm{V}$. Therefore, it locks the voltage across the load resistor (500 Ω$500 \,\Omega$) at 10 V$10 \mathrm{V}$.
Step 1: Calculate Currents
Current across the load resistor (500 Ω$500 \,\Omega$):
Always perform the unregulated voltage check first. If Vᵢₙ (RL / (RL + RS)) > VZ$V_{in} (R_L / (R_L + R_S)) > V_Z$, the diode behaves like a constant VZ$V_Z$ battery. Apply nodal analysis.
Chapter Mix
Class 12 Physics: Semiconductor Electronics
Q49jee_main_2024_31_jan_eveningLogic Gates
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
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.
Chapter Mix
Class 12 Physics: Semiconductor Electronics
Qjee_main_2024_31_jan_morningLogic Gates
Identify the logic operation performed by the given circuit.
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 · B (NAND)$$Y = \overline{A \cdot B} \quad \text{(NAND)}$$Y = A + B (De Morgan's)$$Y = \overline{A} + \overline{B} \quad \text{(De Morgan's)}$$
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 · B$$Y = \overline{\overline{A} \cdot \overline{B}}$$
Applying De-Morgan's Law:
Y = A + B$$Y = \overline{\overline{A}} + \overline{\overline{B}}$$
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
More Semiconductor Electronics: Materials, Devices and Simple Circuits Questions — jee_main_2025_28_jan_morning
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.
