JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    Truth table for system of four NAND gates as shown in figure is :                AIEEE  Solved  Paper-2012

    A)
    A B Y
    0 0 0
    0 1 1
    1 0 1
    1 1 0

    B)  
    A B Y
    0 0 0
    0 1 0
    1 0 1
    1 1 1

    C)  
    A B Y
    0 0 1
    0 1 1
    1 0 0
    1 1 0

    D)  
    A B Y
    0 0 1
    0 1 0
    1 0 0
    1 1 1

    Correct Answer: A

    Solution :

                 \[y=\overline{\left( \overline{A.\overline{\,A.\,B}} \right).\left( \overline{B.\,\overline{A.\,B}} \right)}\] \[=\left( \overline{\overline{A.\overline{\,A.\,B}}} \right).\left( \overline{\overline{B.\,\overline{A.\,B}}} \right)\] \[=A.\,\left( \overline{A}+\overline{B} \right)+B.\left( \overline{A}+\overline{B} \right)\] \[=A.\,\overline{A}+A.\,\overline{B}+B.\,\overline{A}+B.\,\overline{B}\] \[y=0+A.\,\overline{B}+B.\,\overline{A}+0\]


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