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(B) The circuit consists of a $NOR$ gate followed by a $NOT$ gate, which together form an $OR$ gate for the inputs $A$ and $B$. Let this output be $Y_

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$1, 0, 0$

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(B) The circuit consists of a $NOR$ gate followed by a $NOT$ gate, which together form an $OR$ gate for the inputs $A$ and $B$. Let this output be $Y_1$. Thus, $Y_1 = A + B$.The inputs $C$ and $D$ are connected to an $AND$ gate, so $Y_2 = C \cdot D$.The outputs $Y_1$ and $Y_2$ are then fed into a $NOR$ gate to produce $Y_3$. Thus, $Y_3 = \overline{Y_1 + Y_2}$.Given inputs: $A = 0, B = 1, C = 0, D = 1$.Calculating $Y_1$: $Y_1 = A + B = 0 + 1 = 1$.Calculating $Y_2$: $Y_2 = C \cdot D = 0 \cdot 1 = 0$.Calculating $Y_3$: $Y_3 = \overline{Y_1 + Y_2} = \overline{1 + 0} = \overline{1} = 0$.Therefore, the new outputs are $(1, 0, 0)$.

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

In the logic circuit diagram, when all the four inputs, $A, B, C, D$ are 'one' the outputs $Y_1, Y_2, Y_3$ are respectively $(1, 1, 0)$. When the inputs $A$ and $C$ are changed to zero and $B$ and $D$ are still 'one',then the outputs $Y_1, Y_2, Y_3$ are respectively change to

Similar Questions

$A$ combination of logic gates is as shown in the figure. $P$,$Q$,$R$,and $S$ are the inputs and $X$,$Y$,and $Z$ are outputs. When inputs $P$ and $R$ are 'low' $(0)$ and $Q$ and $S$ are 'high' $(1)$,the outputs $X$,$Y$,and $Z$ are respectively:

The combination of gates shown below yields:

To obtain the truth-table shown, from the following logic circuit, the gate $G$ should be
$A$ $B$ $Y$
$0$ $0$ $1$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $1$

The given truth table is for:
$A$ $B$ $Y$
$0$ $0$ $1$
$0$ $1$ $1$
$1$ $0$ $1$
$1$ $1$ $0$

Which one of the following is the Boolean expression for $NOR$ gate?

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