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QJEEM24Ph6fb6: JEE Main 2024 Physics - Class 12 JEEM24Ph6fb6 Solution: The output of the given circuit diagram is -

JEE Main 2024 (Online) 31st January Evening Shift Physics - Semiconductor Question 27 English

The output of the given circuit diagram is -

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/f82d3a2991645aa4f2bbdcc68eb022f0c08756227d0ddcc28462db435d6f460c.webp
  • B https://ik.imagekit.io/novatra/photos/be7fd49fd56235b6c82993d6694e5bb9520fac8cc0fe39c8e0c359b1bdb0113d.webp
  • C https://ik.imagekit.io/novatra/photos/eeb5a9d67a90ac8845e1574a54f41820463f350f4ed9cb94945189db73796be9.webp
  • D https://ik.imagekit.io/novatra/photos/dd9b7d8a7dc4dcf44a2f9adcd876eb09a5318a83ff93ce11f859eda3288dc26c.webp

Step-by-Step Verified Solution

JEE Main 2024 (Online) 31st January Evening Shift Physics - Semiconductor Question 27 English Explanation

 If Aamp;=0;A‾=1 Aamp;=1;A‾=0 Bamp;=0;B‾=1 Bamp;=1;B‾=0Yamp;=(A+B‾)+(A‾+B)‾=(1+1)‾=0\begin{aligned}\text { If } \mathrm{A} & =0 ; \overline{\mathrm{A}}=1 \\\mathrm{~A} & =1 ; \overline{\mathrm{A}}=0 \\\mathrm{~B} & =0 ; \overline{\mathrm{B}}=1 \\\mathrm{~B} & =1 ; \overline{\mathrm{B}}=0 \\\mathrm{Y} & =\overline{(\mathrm{A}+\overline{\mathrm{B}})+(\overline{\mathrm{A}}+\mathrm{B})}=\overline{(1+1)}=0\end{aligned}

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