JEE Main 2019 — Physics solution

JEE Main2019Physics

The truth table for the circuit given in the fig. is:

A
\left| {\begin{array}{*{20}{c}} A\,\,\,\,B\,\,\,\,Y \\0\,\,\,\,0\,\,\,\,1 \\ 0\,\,\,\,1\,\,\,\,0 \\ 1\,\,\,\,0\,\,\,\,0 \\ 1\,\,\,\,1\,\,\,\,0 \end{array} } \right|\,\,
B
\left|{\begin{array}{*{20}{c}}A\,\,\,\,B\,\,\,\,Y\\ 0\,\,\,\,0\,\,\,\,0\\ 0\,\,\,\,1\,\,\,\,0 \\ 1\,\,\,\,0\,\,\,\,1\\1\,\,\,\,1\,\,\,\,1 \end{array} } \right|
C
\left| {\begin{array}{*{20}{c}}A\,\,\,\,B\,\,\,\,Y \\ 0\,\,\,\,0\,\,\,\,1 \\ 0\,\,\,\,1\,\,\,\,1 \\ 1\,\,\,\,0\,\,\,\,1 \\ 1\,\,\,\,1\,\,\,\,1 \end{array} } \right|
D
\left|{\begin{array}{*{20}{c}} A\,\,\,\,B\,\,\,\,Y \\ 0\,\,\,\,0\,\,\,\,1 \\ 0\,\,\,\,1\,\,\,\,1 \\ 1\,\,\,\,0\,\,\,\,0 \\ 1\,\,\,\,1\,\,\,\,0 \end{array} } \right|

Solution

$\Rightarrow y = \overline{A . (A + B)} = \overline{A} + \overline{(A + B)} \Rightarrow y = \overline{A} + \overline{A} . \overline{B} = \overline{A} (1 + \overline{B}) \Rightarrow y = \overline{A}$

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