Question 154 of 209

The output of the given circuit diagram is

ABCD

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Answer: (C)

If $A=0;\bar{A}=1$
$A=1;\bar{A}=0$
$B=0;\bar{B}=1$
$B=1;\bar{B}=0$
Using De Morgan's theorem, we can write
$Y=\bar{\left(A+\bar{B}\right)+\left(\bar{A}+B\right)} =\bar{\left(A+\bar{B}\right)}\cdot \bar{\left(\bar{A}+B\right)} =\left(\bar{A}\cdot B\right)\cdot \left(A\cdot \bar{B}\right) =0$

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