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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2021 · 31 August · Morning Shift · Physics · Semiconductors

In the following logic circuit the sequence of the inputs A, B are...

In the following logic circuit the sequence of the inputs A,BA, B are (0,0),(0,1),(1,0)( 0, 0 ) , ( 0, 1 ) , ( 1, 0 ) and (1,1).( 1, 1 ) . The output YY for this sequence will be :
In the following logic circuit the sequence of the inputs A, B are ( 0, 0 ) , ( 0, 1 ) , ( 1, 0 ) and ( 1, 1 ) . The

Options

  1. A0,1,0,10, 1, 0, 1
  2. B0,0,1,10, 0, 1, 1
  3. C1,1,1,01, 1, 1, 0
  4. D1,0,1,01, 0, 1, 0

Correct answer

C. 1,1,1,01, 1, 1, 0

Step-by-step solution

Look at the above circuit, the value of PP can be found by the idea of AND gate, P=A⋅BP = A \cdot B and the value of QQ, can be calculated by the idea of OR gate, Q=A+BQ = A + B. Then expression for YY can be taken by the idea of NAND gate, ⇒Y=P⋅Q¯⇒ Y = \overset{¯}{P \cdot Q} by the idea of Demorgan's theorem, Y=P¯+Q¯=A¯+B¯+A¯⋅B¯Y = \overset{¯}{P} + \overset{¯}{Q} = \overset{¯}{A} + \overset{¯}{B} + \overset{¯}{A} \cdot \overset{¯}{B}
In the following logic circuit the sequence of the inputs A, B are ( 0, 0 ) , ( 0, 1 ) , ( 1, 0 ) and ( 1, 1 ) . The

{ABY=A¯+B¯+A¯⋅B¯001011101110\begin{cases}A & B & Y = \overset{¯}{A} + \overset{¯}{B} + \overset{¯}{A} \cdot \overset{¯}{B} \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{cases}

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