Browse Questions

# The Output Y of the figure is :

$(a)\;0 \\ (b)\;1 \\ (c)\;\bar {A} \\ (d)\;\bar {B}$

$Y=\overline {Y'.A}$
$\qquad= \bar {Y'}. \bar {A}$
$\qquad= \overline {\overline {A.B}}. \bar {A}$
$\qquad=A.B. \bar {A}$
$\qquad= B.A. \bar {A}$
$\qquad= B.0$
$\qquad=0$

edited Jan 30, 2014 by meena.p