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Electronic Devices question

2024 · 5 Apr · Shift 2 · Q75
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Electronic Devices question

2024 · 5 Apr · Shift 2 · Q75

JEE MainPhysicsElectronic DevicesMCQ+4 / −1
The output (Y) of logic circuit given below is 0 only when : JEE Main 2024 (Online) 5th April Evening Shift Physics - Semiconductor Question 21 English
  1. A
    A=0, B=0\mathrm{A}=0, \mathrm{~B}=0A=0, B=0
  2. B
    A=0, B=1\mathrm{A}=0, \mathrm{~B}=1A=0, B=1
  3. C
    A=1, B=0\mathrm{A}=1, \mathrm{~B}=0A=1, B=0
  4. D
    A=1,B=1\mathrm{A=1, B=1}A=1,B=1
View written solutionFree

Correct answer: B, C, D

  1. The circuit shown is the standard OR gate followed by a NOT gate, i.e. a NOR gate.

  2. For an OR gate, the output is: A+BA + BA+B So the output is 1 if either A=1A=1A=1 or B=1B=1B=1.

  3. Since there is inversion at the output, the final output is: Y=A+B‾Y = \overline{A+B}Y=A+B​ This is the Boolean expression of a NOR gate.

  4. Now evaluate all input combinations:

AAABBBA+BA+BA+BY=A+B‾Y=\overline{A+B}Y=A+B​
0001
0110
1010
1110
  1. From the truth table, Y=0Y=0Y=0 when:
  • A=0,B=1A=0, B=1A=0,B=1
  • A=1,B=0A=1, B=0A=1,B=0
  • A=1,B=1A=1, B=1A=1,B=1

So the output is 0 only when at least one input is 1.

Therefore, among the given options, the correct cases are: B, C, D\boxed{B,\ C,\ D}B, C, D​

  1. Comparison with stored answer:
  • Stored correct answer: AAA
  • Derived answer: B,C,DB, C, DB,C,D

Hence, the stored answer does not match the logic of a NOR gate truth table.

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