The output voltage in the following circuit is (Consider ideal diode case)
A schematic of a circuit showing an input voltage, two diodes D1 and D2 connected in parallel paths, a resistor, and the output node V_out.
A.10mathrm~V$10\mathrm{~V}$
B.0mathrm~V$0\mathrm{~V}$
C.+5mathrm~V$+5\mathrm{~V}$
D.-5mathrm~V$-5\mathrm{~V}$
Solution & Explanation
### Related Formula
For ideal diodes:
- **Forward Bias**: Acts as a closed switch (zero resistance, short circuit).
- **Reverse Bias**: Acts as an open switch (infinite resistance, open circuit).
### Core Logic
Analyzing the bias condition of the diodes based on the applied potential in the schematic:
- Diode D_1$D_1$ is oriented such that its cathode faces the positive terminal (+5mathrm~V$+5\mathrm{~V}$), making it **Reverse Biased** (no current flows through this branch).
- Diode D_2$D_2$ is oriented such that its anode connects to the +5mathrm~V$+5\mathrm{~V}$ path, making it **Forward Biased**.
Since D_2$D_2$ is forward-biased and ideal, it acts as a short circuit (resistance R_D = 0$R_D = 0$). Current flows through D_2$D_2$ and through the series resistor.
### Step 1: Calculating output node potential
Because the forward-biased ideal diode D_2$D_2$ connects the node directly to the low-resistance ground loop or the reference resistor drop, the entire 5mathrm~V$5\mathrm{~V}$ potential drops across the resistor:
V_textout = 0mathrm~V$$V_{\text{out}} = 0\mathrm{~V}$$
### Pattern Recognition
Sees: Parallel diode configuration with opposite polarities.
Shortcut: Check polarity. D_1$D_1$ is reverse-biased (open), D_2$D_2$ is forward-biased (short). The output terminal is pulled down to the reference ground, leading directly to 0mathrm~V$0\mathrm{~V}$. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
Keywords:#Ideal diode circuit output voltage#JEE Main 2025 Evening Q8#Semiconductor Electronics diodes#Forward and reverse biased ideal diode#diode#ideal diode#reverse biased#forward biased#output voltage
More Semiconductor Electronics Previous-Year Questions
Qjee_main_2026_21_jan_morningLogic Gates
The given circuit works as :
Circuit containing multiple interconnected logic gates leading to a final output.
A. AND gate
B. NOR gate
C. NAND gate
D. OR gate
Solution
### Related Formula
De Morgan's Laws:
overlineA + overlineB = overlineA cdot B$$\overline{A} + \overline{B} = \overline{A \cdot B}$$overlineoverlineA = A$$\overline{\overline{A}} = A$$
### Core Logic
Analyzing the circuit diagram:
Circuit containing multiple interconnected logic gates leading to a final output.
- Top branch has a NOT gate on A, so P = overlineA$P = \overline{A}$
- Bottom branch has a NOT gate on B, so Q = overlineB$Q = \overline{B}$
- They enter a NOR gate, giving output R = overlineoverlineA + overlineB$R = \overline{\overline{A} + \overline{B}}$
- Finally, R passes through a NOT gate to give S = overlineR$S = \overline{R}$
### Step 1: Boolean Simplification
R = overlineoverlineA + overlineB = overlineoverlineAB = AB$$R = \overline{\overline{A} + \overline{B}} = \overline{\overline{AB}} = AB$$S = overlineR = overlineAB$$S = \overline{R} = \overline{AB}$$
The expression overlineAB$\overline{AB}$ is the Boolean expression for a NAND gate.
### Pattern Recognition
Two NOTs feeding into a NOR equals an AND gate (AB$AB$). Adding a final NOT gate turns the AND into a NAND.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
### Related Formula
textNOR Gate: overlineA+B, quad textNAND Gate: overlineA cdot B$$\text{NOR Gate: } \overline{A+B}, \quad \text{NAND Gate: } \overline{A \cdot B}$$
### Core Logic
Logic gate circuit diagram connected to an LED.
LED will glow in forward biasing when point P is at higher potential (1) and point Q is at lower potential (0).
Testing option (4) [1101]:
- Inputs A=1, B=1 through NOR and NAND gates yield P = 1.
- Inputs C=1, D=0 through NOR gate yields Q = 0.
- Forward bias established, LED glows.
### Pattern Recognition
Sees: Logic gates combination with LED forward biasing condition.
Shortcut: Check forward bias requirement (P=1, Q=0) for each option combination.
Check: Option (4) satisfies the condition. ✓
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
Q35jee_main_2026_22_january_eveningLogic Gates and Truth Tables
The correct truth table for the given input data of the following logic gate is :
The figure illustrates a combined logic circuit with inputs A, B, C, D feeding into AND, OR, and NOT gates to produce output Y.
A.
Inputs
Output
A
B
C
D
Y
1
1
0
1
1
0
0
1
1
0
1
0
1
0
1
1
1
1
1
0
B.
Inputs
Output
A
B
C
D
Y
1
1
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
1
1
C.
Inputs
Output
A
B
C
D
Y
1
1
0
1
0
0
0
1
1
0
1
0
1
0
1
1
1
1
1
1
D.
Inputs
Output
A
B
C
D
Y
1
1
0
1
0
0
0
1
1
1
1
0
1
0
1
1
1
1
1
1
Solution
### Related Formula
Y = overline(overlineA cdot B) cdot (C + D)$$Y = \overline{(\overline{A \cdot B}) \cdot (C + D)}$$
Applying De Morgan's Law:
Y = (A cdot B) + (overlineC + D)$$Y = (A \cdot B) + (\overline{C + D})$$
### Core Logic
Analyzing the expression step by step:
1. Upper branch: A$A$ and B$B$ pass through NAND gate rightarrow overlineA cdot B$\rightarrow \overline{A \cdot B}$.
2. Lower branch: C$C$ and D$D$ pass through OR gate rightarrow (C + D)$\rightarrow (C + D)$.
3. Combined in AND gate followed by NOT gate (NAND equivalent output stage):
Y = overline(overlineA cdot B) cdot (C + D) = (A cdot B) + (overlineC + D)$$Y = \overline{(\overline{A \cdot B}) \cdot (C + D)} = (A \cdot B) + (\overline{C + D})$$
Testing conditions for Option (2):
- For A=1, B=1, C=0, D=1 implies (1 cdot 1) + (overline0 + 1) = 1 + 0 = 1$A=1, B=1, C=0, D=1 \implies (1 \cdot 1) + (\overline{0 + 1}) = 1 + 0 = 1$.
- For A=0, B=0, C=1, D=1 implies (0 cdot 0) + (overline1 + 1) = 0 + 0 = 0$A=0, B=0, C=1, D=1 \implies (0 \cdot 0) + (\overline{1 + 1}) = 0 + 0 = 0$.
- For A=1, B=0, C=1, D=0 implies (1 cdot 0) + (overline1 + 0) = 0 + 0 = 0$A=1, B=0, C=1, D=0 \implies (1 \cdot 0) + (\overline{1 + 0}) = 0 + 0 = 0$.
- For A=1, B=1, C=1, D=1 implies (1 cdot 1) + (overline1 + 1) = 1 + 0 = 1$A=1, B=1, C=1, D=1 \implies (1 \cdot 1) + (\overline{1 + 1}) = 1 + 0 = 1$.
The figure illustrates a combined logic circuit with inputs A, B, C, D feeding into AND, OR, and NOT gates to produce output Y.
### Step 1: Final Conclusion
Option (2) presents the correct truth table matching the evaluated output values.
### Pattern Recognition
Boolean reduction: overlinebarX cdot Y = X + barY$\overline{\bar{X} \cdot Y} = X + \bar{Y}$ where X = A cdot B$X = A \cdot B$ and Y = C+D$Y = C+D$. Hence Y = (A cdot B) + (overlineC+D)$Y = (A \cdot B) + (\overline{C+D})$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
Q37jee_main_2026_23_january_morningZener Diode
The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at V_Z = 5V$V_{Z} = 5V$ 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 R_S$R_{S}$ (shown in circuit) should be ____ Omega$\Omega$.
Circuit containing unregulated voltage, series resistor Rs, Zener diode, and load resistor.
A.4000$4000$
B.10$10$
C.100$100$
D.1000$1000$
Solution
### Related Formula
I_texttotal = I_Z + I_L$$I_{\text{total}} = I_{Z} + I_{L}$$V_textin - V_Z = I_texttotal R_S$$V_{\text{in}} - V_{Z} = I_{\text{total}} R_{S}$$
### 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_Z_textmax = 4 times 5 text mA = 20 text mA$I_{Z_{\text{max}}} = 4 \times 5 \text{ mA} = 20 \text{ mA}$. Total current I = 25 text mA$I = 25 \text{ mA}$. R_S = (25 - 5) / (25 times 10^-3) = 800Omega$R_{S} = (25 - 5) / (25 \times 10^{-3}) = 800\Omega$, 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
Q38jee_main_2026_23_january_eveningLogic Gates
For the given logic gate circuit, which of the following is the correct truth table?
Circuit schematic containing an OR gate and a NAND gate.
A.
n
m
z
0
0
1
0
1
0
1
1
0
1
0
0
B.
n
m
z
0
0
0
0
1
1
1
1
0
1
0
1
C.
n
m
z
0
0
1
0
1
0
1
1
1
1
0
0
D.
n
m
z
0
0
1
0
1
1
1
1
0
1
0
0
Solution
### Related Formula
Boolean Algebra rules:
A + AB = A$A + AB = A$A cdot A = A$A \cdot A = A$
NAND Gate: Z = overlineX cdot Y$Z = \overline{X \cdot Y}$
### Core Logic
Gate 1 (Bottom): This is an OR gate with inputs n$n$ and m$m$.
Output = n + m$n + m$
Gate 2 (Top): This is a NAND gate. Its inputs are direct n$n$ and the output of the OR gate (n + m$n + m$).
Output z = overlinen cdot (n + m)$z = \overline{n \cdot (n + m)}$
### Step 1: Simplify the Boolean Expression
First, expand the inner term:
n cdot (n + m) = (n cdot n) + (n cdot m)$$n \cdot (n + m) = (n \cdot n) + (n \cdot m)$$
Since n cdot n = n$n \cdot n = n$, this becomes:
n + n cdot m = n(1 + m) = n cdot 1 = n$$n + n \cdot m = n(1 + m) = n \cdot 1 = n$$
So the overall output is:
z = overlinen cdot (n + m) = overlinen$$z = \overline{n \cdot (n + m)} = \overline{n}$$
### Step 2: Construct the Truth Table
Since z = overlinen$z = \overline{n}$, the output z$z$ only depends on n$n$ and is its exact inverse.
When n=0, m=0 implies z=1$n=0, m=0 \implies z=1$
When n=0, m=1 implies z=1$n=0, m=1 \implies z=1$
When n=1, m=1 implies z=0$n=1, m=1 \implies z=0$
When n=1, m=0 implies z=0$n=1, m=0 \implies z=0$
### Pattern Recognition
Always simplify the Boolean expression using standard absorption laws (like A(A+B) = A$A(A+B) = A$) before manually evaluating every row. It saves significant time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
More Semiconductor Electronics Questions — jee_main_2025_08_april_evening
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