The output of an OR gate is connected to both | vedclass

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(C) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y_1 = A + B$.This output $Y_1$ is connected to both inputs of a $N

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$NOR$ gate

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(C) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y_1 = A + B$.This output $Y_1$ is connected to both inputs of a $NAND$ gate. Let the inputs of the $NAND$ gate be $X_1$ and $X_2$,where $X_1 = X_2 = Y_1 = A + B$.The output of the $NAND$ gate is $Y = \overline{X_1 \cdot X_2} = \overline{(A + B) \cdot (A + B)} = \overline{A + B}$.The expression $\overline{A + B}$ represents the Boolean operation of a $NOR$ gate.Therefore,the combination is equivalent to a $NOR$ gate.

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

The output of an $OR$ gate is connected to both inputs of a $NAND$ gate. This combination is equivalent to a:

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$A$ $B$ $Y$

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