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Semiconductor Electronics · Question 20

JEE Main 2026 (Online) 21st January Morning ShiftMCQEasy📈 Topic in 58% papers+4−1
1

The given circuit works as : [figure]

  1. NOR gate
  2. OR gate
  3. AND gate
  4. NAND gate

Answer

(4) NAND gate

Solution

All four gates in the circuit are NOR gates.

A NOR gate with both inputs joined acts as a NOT gate: A + A = A

Outputs of the first two gates are A and B.

Third gate (NOR): Y1 = A + B = A B (De Morgan's theorem)

Last gate (inputs joined, acts as NOT): Y = Y1 = A B

The circuit works as a NAND gate.

Answer: (4)

JEE Main 2026 (Online) 21st January Morning Shift

Keep practising — Semiconductor Electronics

2

The charge stored by the capacitor C in the given circuit in the steady state is . [figure]

  1. 7.5
  2. 5
  3. 10
  4. 12.5
JEE Main 2026 (Online) 21st January Evening Shift
3

The output voltage in the following circuit is (Consider ideal diode case) [figure]

  1. 0 V
  2. +5 V
  3. −5 V
  4. 10 V
JEE Main 2025 (Online) 8th April Evening Shift
4

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

  1. 60 mA
  2. 80 mA
  3. 10 mA
  4. 24 mA
JEE Main 2025 (Online) 7th April Morning Shift
5

Consider the following logic circuit. [figure] The output is Y = 0 when :

  1. A = 1 and B = 0
  2. A = 0 and B = 0
  3. A = 1 and B = 1
  4. A = 0 and B = 1
JEE Main 2025 (Online) 7th April Evening Shift
6

The Boolean expression Y = A BC + A C can be realised with which of the following gate configurations. Choose the correct answer from the options given below:

  1. B, C Only
  2. A, B, C Only
  3. A, B Only
  4. A, C Only
JEE Main 2025 (Online) 4th April Morning Shift
7

Consider an n-type semiconductor in which ne and nh are number of electrons and holes, respectively. Choose the correct answer from the options given below :

  1. (A), (B), (E) only
  2. (A), (C), (E) only
  3. (A), (B), (C) only
  4. (A), (C), (D) only
JEE Main 2025 (Online) 4th April Evening Shift
8

Choose the correct logic circuit for the given truth table having inputs A and B. [figure]

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 3rd April Morning Shift
9

The truth table corresponding to the circuit given below is: [figure]

  1. arrayccc A & B & C \ 0 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & 1 \ 1 & 1 & 1 array
  2. arrayccc A & B & C \ 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 1 & 0 array
  3. arrayccc A & B & C \ 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \ 1 & 1 & 0 array
  4. arrayccc A & B & C \ 0 & 0 & 0 \ 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 1 & 1 array
JEE Main 2025 (Online) 3rd April Evening Shift
10

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 resistor connected in series, the zener current is found to be 4 times load current. The load current (IL) and load resistance (RL) are :

  1. IL = 0.02 mA; RL = 250
  2. IL = 10 A; RL = 0.5
  3. IL = 10 mA; RL = 500
  4. IL = 20 mA; RL = 250
JEE Main 2025 (Online) 2nd April Morning Shift
11

In the digital circuit shown in the figure, for the given inputs the P and Q values are : [figure]

  1. P = 0, Q = 1
  2. P = 1, Q = 0
  3. P = 0, Q = 0
  4. P = 1, Q = 1
JEE Main 2025 (Online) 2nd April Evening Shift

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