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Electronic Devices question

2016 · 9 Apr · Shift 1 · Q53
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Electronic Devices question

2016 · 9 Apr · Shift 1 · Q53

JEE MainPhysicsElectronic DevicesMCQ+4 / −1
The truth table given in fig. represents :
A B Y
0 0 0
0 1 1
1 0 1
1 1 1
  1. A
    AND - Gate
  2. B
    OR - Gate
  3. C
    NAND - Gate
  4. D
    NOR - Gate
View written solutionFree

Correct answer: B

  1. Write the given truth table
ABY000011101111\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array}A0011​B0101​Y0111​​
  1. Identify when the output is 1

The output Y=1Y=1Y=1 for:

  • 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

The output is 000 only when both inputs are 000.

  1. Compare with standard logic gates
  • AND gate: output is 1 only for (1,1)(1,1)(1,1), so this does not match.
  • OR gate: output is 1 when at least one input is 1, and 0 only for (0,0)(0,0)(0,0). This matches exactly.
  • NAND gate: output is 0 only for (1,1)(1,1)(1,1), so this does not match.
  • NOR gate: output is 1 only for (0,0)(0,0)(0,0), so this does not match.
  1. Conclusion

The given truth table represents an OR gate.

So, the correct option is:

B: OR-Gate\boxed{\text{B: OR-Gate}}B: OR-Gate​

  1. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They match.

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