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Electronic Devices question

2022 · 25 Jul · Shift 1 · Q60
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Electronic Devices question

2022 · 25 Jul · Shift 1 · Q60

JEE MainPhysicsElectronic DevicesMCQ+4 / −1
In the circuit, the logical value of A=1A=1A=1 or B=1B=1B=1 when potential at AAA or BBB is 5 V5 \mathrm{~V}5 V and the logical value of A=0A=0A=0 or B=0B=0B=0 when potential at AAA or BBB is 0 V0 \mathrm{~V}0 V. JEE Main 2022 (Online) 25th July Morning Shift Physics - Semiconductor Question 77 English The truth table of the given circuit will be :
  1. A
    ABY000100010111\begin{matrix} A & B & Y \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix}A0101​B0011​Y0001​
  2. B
    ABY000101011111\begin{matrix} A & B & Y \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix}A0101​B0011​Y0111​
  3. C
    ABY000100010110\begin{matrix} A & B & Y \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end{matrix}A0101​B0011​Y0000​
  4. D
    ABY001101011110\begin{matrix} A & B & Y \\ 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ \end{matrix}A0101​B0011​Y1110​
View written solutionFree

Correct answer: A

  1. Identify the circuit logic

    This standard electronic-devices question corresponds to a diode/transistor arrangement in which the output becomes high only when both inputs are high.

    So we test all input combinations:

  2. Case-wise analysis

    Let logic levels be:

    • 1↔5 V1 \leftrightarrow 5\,\text{V}1↔5V
    • 0↔0 V0 \leftrightarrow 0\,\text{V}0↔0V

    We evaluate output YYY for each pair (A,B)(A,B)(A,B).

    (i) A=0,B=0A=0, B=0A=0,B=0

    • Neither input is high.
    • Output remains low.
    • Hence Y=0Y=0Y=0.

    (ii) A=1,B=0A=1, B=0A=1,B=0

    • Only one input is high.
    • Output is still low.
    • Hence Y=0Y=0Y=0.

    (iii) A=0,B=1A=0, B=1A=0,B=1

    • Only one input is high.
    • Output is still low.
    • Hence Y=0Y=0Y=0.

    (iv) A=1,B=1A=1, B=1A=1,B=1

    • Both inputs are high.
    • Output becomes high.
    • Hence Y=1Y=1Y=1.

  3. Truth table obtained

    A & B & Y \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{matrix}$$

  4. Match with options

    This matches Option A.

  5. Conclusion

    The given circuit behaves as an AND gate.

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