In the digital circuit shown in the figure, for the given inputs the P and Q values are :
The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.
### Related Formula
Truth relations of basic logic operations:
- NAND operation: Y = overlineA cdot B$Y = \overline{A \cdot B}$
- NOR operation: Y = overlineA + B$Y = \overline{A + B}$
- NOT operation: Y = overlineA$Y = \overline{A}$
- OR operation: Y = A + B$Y = A + B$
### Core Logic
The inputs are:
- Top input = 1$1$
- Bottom input = 1$1$
Let's analyze step-by-step from left to right:
1. **First Gate (NAND gate at the top-left):**
- Inputs are 1$1$ and 1$1$.
- Output = overline1 cdot 1 = 0$\overline{1 \cdot 1} = 0$.
2. **Bottom-left path with NOT gates:**
- Top input (1$1$) goes to a NOT gate, producing 0$0$.
- Bottom input (1$1$) goes to a NOT gate, producing 0$0$.
- These two 0$0$ values feed into the OR gate:
- Output = 0 + 0 = 0$0 + 0 = 0$.
### Step 1: Calculate output P
Now trace the path to P$P$:
- The inputs to the top-right AND gate are:
- Output of the top-left NAND gate = 0$0$
- Output of the bottom-left OR gate = 0$0$
- Therefore, output P$P$ is:
P = 0 cdot 0 = 0$$P = 0 \cdot 0 = 0$$
### Step 2: Calculate output Q
Now trace the path to Q$Q$:
- The gate at the bottom-right is a NOR gate with two inputs:
- Input 1: Output of the top-left NAND gate (0$0$) inverted by a NOT gate = overline0 = 1$\overline{0} = 1$.
- Input 2: Output of the bottom-left OR gate (0$0$).
- Passing these inputs (1$1$ and 0$0$) through the final NOR gate:
Q = overline1 + 0 = overline1 = 0$$Q = \overline{1 + 0} = \overline{1} = 0$$
Thus, both P = 0$P = 0$ and Q = 0$Q = 0$.
### Pattern Recognition
Sees: Combinational trace with inverted nodes.
Trap: Missing bubbles (NOT gates) representing inversion on internal circuit paths.
Shortcut: The first NAND gate output is 0$0$ (since both inputs are 1). This 0$0$ directly goes to the upper AND gate, immediately guaranteeing output P = 0$P = 0$ (eliminates options 1 and 4). Now, you only need to evaluate Q$Q$ to choose between options 2 and 3.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.
Keywords:#logic gates digital circuit diagram#JEE Main 2025 Evening Q13#trace combinational logic circuit#Semiconductor Electronics logic gates#NAND gate#NOR gate#NOT gate#digital logic circuit
More Semiconductor Electronics: Materials, Devices and Simple Circuits 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
Q11jee_main_2025_07_april_morningZener Diode
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.
A.24mathrmmA$24\mathrm{mA}$
B.80mathrmmA$80\mathrm{mA}$
C.10mathrmmA$10\mathrm{mA}$
D.60mathrmmA$60\mathrm{mA}$
Solution
### Related Formula
Voltage division across load R_L$R_L$ with series resistance R_s$R_s$ without Zener regulation:
V_L = V_textin left( fracR_LR_s + R_L right)$$V_L = V_{\text{in}} \left( \frac{R_L}{R_s + R_L} \right)$$
If V_L > V_z$V_L > V_z$, the Zener diode enters breakdown, and potential across the parallel load is clamped at V_L = V_z$V_L = V_z$.
### Core Logic
Verify if Zener diode operates in the breakdown region:
- V_textin = 12 mathrm~V$V_{\text{in}} = 12 \mathrm{~V}$
- R_s = 100 Omega$R_s = 100 \Omega$
- R_L = 400 Omega$R_L = 400 \Omega$
Calculate the unregulated voltage:
V_1 = 12 times left( frac400100 + 400 right) = 12 times frac45 = 9.6 mathrm~V$$V_1 = 12 \times \left( \frac{400}{100 + 400} \right) = 12 \times \frac{4}{5} = 9.6 \mathrm{~V}$$
Since V_1 > V_z$V_1 > V_z$ (9.6 mathrm~V > 4 mathrm~V$9.6 \mathrm{~V} > 4 \mathrm{~V}$), breakdown occurs, and the parallel branch voltage is fixed at V_z = 4 mathrm~V$V_z = 4 \mathrm{~V}$.
### Step 1: Calculate Branch Current
The voltage across the 400 Omega$400 \Omega$ load resistor (which is in series with the ammeter) is locked at 4 mathrm~V$4 \mathrm{~V}$.
The current I$I$ through the ammeter is:
I = fracV_zR_L = frac4 mathrm~V400 Omega = 10^-2 mathrm~A = 10 mathrm~mA$$I = \frac{V_z}{R_L} = \frac{4 \mathrm{~V}}{400 \Omega} = 10^{-2} \mathrm{~A} = 10 \mathrm{~mA}$$
### Pattern Recognition
Sees: Parallel Zener diode configuration.
Shortcut: Always calculate the open-circuit load voltage first. If it exceeds V_z$V_z$, use V_z$V_z$ as the branch potential. The branch current is simply V_z / R_L$V_z / R_L$. Here, 4 / 400 = 10 mathrm~mA$4 / 400 = 10 \mathrm{~mA}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
Q8jee_main_2025_08_april_eveningDiodes
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
### 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
Q18jee_main_2025_29_jan_eveningLogic Gates
The truth table for the circuit given below is :
The image shows a digital logic circuit containing multiple interconnected logic gates with input terminals A and B and output terminal Y.
### Related Formula
Y = A cdot barB + barA cdot B = A oplus B$$Y = A \cdot \bar{B} + \bar{A} \cdot B = A \oplus B$$
### Core Logic
Analyzing the circuit layout:
1. The top AND gate receives inputs A$A$ and barB$\bar{B}$, yielding output term AbarB$A\bar{B}$.
2. The bottom AND gate receives inputs barA$\bar{A}$ and B$B$, yielding output term barAB$\bar{A}B$.
3. These terms pass into a terminal OR gate, producing:
Y = AbarB + barAB$$Y = A\bar{B} + \bar{A}B$$The image shows a digital logic circuit containing multiple interconnected logic gates with input terminals A and B and output terminal Y.The image shows a digital logic circuit containing multiple interconnected logic gates with input terminals A and B and output terminal Y.
This is the precise expression for an **XOR (Exclusive OR) gate**. The corresponding truth table gives an output of 1$1$ only when inputs are mismatched (0,1$0,1$ or 1,0$1,0$), and 0$0$ otherwise. This aligns exactly with Option 1.
### Pattern Recognition
Recognize the symmetric cross-inversion network of gates: (A cdot barB) + (barA cdot B)$(A \cdot \bar{B}) + (\bar{A} \cdot B)$. This combination structurally builds an XOR logic function.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics
More Semiconductor Electronics: Materials, Devices and Simple Circuits Questions — jee_main_2025_02_april_evening
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