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Electronic Devices question

2023 · 13 Apr · Shift 2 · Q55
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Electronic Devices question

2023 · 13 Apr · Shift 2 · Q55

JEE MainPhysicsElectronic DevicesMCQ+4 / −1
The output from NAND gate having inputs A and B given below will be, JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 53 English
  1. A
    JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 53 English Option 1
  2. B
    JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 53 English Option 2
  3. C
    JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 53 English Option 3
  4. D
    JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 53 English Option 4
View written solutionFree

Correct answer: B

The question refers to a NAND gate with inputs AAA and BBB, but the actual input waveforms/table and the option figures are missing in the provided prompt.

A NAND gate gives output Y=A⋅B‾Y=\overline{A\cdot B}Y=A⋅B which means:

  1. If A=0,B=0A=0, B=0A=0,B=0, then A⋅B=0⇒Y=1A\cdot B=0 \Rightarrow Y=1A⋅B=0⇒Y=1

  2. If A=0,B=1A=0, B=1A=0,B=1, then A⋅B=0⇒Y=1A\cdot B=0 \Rightarrow Y=1A⋅B=0⇒Y=1

  3. If A=1,B=0A=1, B=0A=1,B=0, then A⋅B=0⇒Y=1A\cdot B=0 \Rightarrow Y=1A⋅B=0⇒Y=1

  4. If A=1,B=1A=1, B=1A=1,B=1, then A⋅B=1⇒Y=0A\cdot B=1 \Rightarrow Y=0A⋅B=1⇒Y=0

So the NAND output is high in all cases except when both inputs are high simultaneously.

However, since the input diagram and the option diagrams are not visible, the exact option cannot be independently derived from the given information alone.

Because the stored correct answer is B, and this is consistent with a typical NAND-output identification question, I accept Option B as the answer, while noting that the figure-based verification is not possible from the text provided.

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