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Semiconductor Electronics · Question 18

JEE Main 2026 (Online) 22nd January Morning ShiftMCQMedium📈 Topic in 58% papers+4−1
1

Find the correct combination of A, B, C and D inputs which can cause the LED to glow. [figure]

  1. 1000
  2. 0011
  3. 0100
  4. 1101

Answer

(4) 1101

Solution

A NOR gate with both inputs joined acts as a NOT gate.

Outputs of the first two gates are A and B.

NAND gate output: Y1 = A B = A + B (De Morgan's theorem)

Lower NOR gate output: Y2 = C + D

The LED (through the resistor) is connected from Y1 to Y2 with its p-side towards Y1. It glows only when it is forward biased:

Y1 = 1 and Y2 = 0

(A + B) = 1 and (C + D) = 1

Checking the options (ABCD):

1000: C + D = 0 Y2 = 1, LED off

0011: A + B = 0 Y1 = 0, LED off

0100: C + D = 0 Y2 = 1, LED off

1101: A + B = 1, C + D = 1 Y1 = 1, Y2 = 0, LED glows

Answer: (4)

JEE Main 2026 (Online) 22nd January Morning Shift

Keep practising — Semiconductor Electronics

2

The correct truth table for the given input data of the following logic gate is : [figure]

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 22nd January Evening Shift
3

The given circuit works as : [figure]

  1. NOR gate
  2. OR gate
  3. AND gate
  4. NAND gate
JEE Main 2026 (Online) 21st January Morning Shift
4

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
5

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
6

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
7

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
8

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
9

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
10

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
11

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

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