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JEE Main 202125 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2021 · 25 February · Evening Shift · Physics · Semiconductors

The truth table for the following logic circuit is :

The truth table for the following logic circuit is :

The truth table for the following logic circuit is :

Options

  1. AA uniform electric field E = ((8m)/(e))V m ^ - 1 is created between two parallel plates of length 1m as shown in
  2. BA uniform electric field E = ((8m)/(e))V m ^ - 1 is created between two parallel plates of length 1m as shown in
  3. CA uniform electric field E = ((8m)/(e))V m ^ - 1 is created between two parallel plates of length 1m as shown in
  4. DA uniform electric field E = ((8m)/(e))V m ^ - 1 is created between two parallel plates of length 1m as shown in

Correct answer

A. A uniform electric field E = ((8m)/(e))V m ^ - 1 is created between two parallel plates of length 1m as shown in

Step-by-step solution

y=(AB¯+A¯B)¯y = \overset{¯}{( A\overset{¯}{B} + \overset{¯}{A}B )}

y=AB¯⋅A¯B¯y = \overset{¯}{A B} \cdot \overset{¯}{\overset{¯}{A}B}

y=(A¯+B)⋅(A+B¯)y = ( \overset{¯}{A} + B ) \cdot ( A + \overset{¯}{B} )

y=A¯⋅A+A¯B¯+A⋅B+BB¯y = \overset{¯}{A} \cdot A + \overset{¯}{A}\overset{¯}{B} + A \cdot B + B\overset{¯}{B}

y=AB+A¯B¯y = AB + \overset{¯}{A}\overset{¯}{B}

The truth table for the following logic circuit is :

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