Question

In the given circuit $P$ and $Q$ form the inputs. The output $Y$ is

• A. $Y=\bar{P}$
• B. $Y=P\bar{Q}$
• C. $Y=P+Q$
• D. $Y=\bar{Q}$
• E. $Y=\overline{P+Q}$

Answer 1

Answer: (D)

From the given circuit, the output $Y$ is
$Y=\overline{(\bar{P}+Q)(P+Q)}$
$Y=\overline{\bar{P}\cdot P+\bar{P}Q+Q\cdot P+Q\cdot Q}$
$Y=\overline{Q+PQ+\bar{P}Q}(\because \bar{P}\cdot P=0)$
$Y=\overline{Q(1+P+\bar{P})}(\because 1+A=1)$
$Y=\bar{Q}$