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QJEEM23Phce8f: JEE Main 2023 Physics - Class 12 JEEM23Phce8f Solution: The output for the inputs and of circuit is given by Truth…

The output YY for the inputs AA and BB of circuit is given by

JEE Main 2023 (Online) 30th January Evening Shift Physics - Semiconductor Question 53 English

Truth table of the shown circuit is:

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/0c5c07fe523c26f50ad62efc99154bd1b63df8d8487cb0d775fe79bcba101fe3.webp
  • B https://ik.imagekit.io/novatra/photos/4ccdcb447df09d648f9f0b67693880324a56f2732041c572215fe0647ecfef78.webp
  • C https://ik.imagekit.io/novatra/photos/6526cbaedc7aad437139c2cb1fb0dae5882f4947f640a7a36aeec39b3ee35f05.webp
  • D https://ik.imagekit.io/novatra/photos/4a09b82747cb2be4cd1e09ae77af335b53d57d590cc664f15d8aa3844d64e5be.webp

Step-by-Step Verified Solution

JEE Main 2023 (Online) 30th January Evening Shift Physics - Semiconductor Question 53 English Explanation 1

Y=(A‾+AB)(B‾+AB)‾=(A+B)(A‾+B‾)Y = \overline {\left( {\overline A + AB} \right)\left( {\overline B + AB} \right)} = \left( {A + B} \right)\left( {\overline A + \overline B } \right)

=AB‾+BA‾= A\overline B + B\overline A (XOR gate)

JEE Main 2023 (Online) 30th January Evening Shift Physics - Semiconductor Question 53 English Explanation 2

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