The following configuration of gates is equiv | vedclass

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(C) The output of the $OR$ gate is $(A + B)$.The output of the $NAND$ gate is $(\overline{A \cdot B})$.These two outputs are fed into an $AND$ gate.Th

Vedclass · Accepted Answer

$	ext{XOR}$

Vedclass · Answer

(C) The output of the $OR$ gate is $(A + B)$.The output of the $NAND$ gate is $(\overline{A \cdot B})$.These two outputs are fed into an $AND$ gate.Therefore,the final output $Y$ is given by:$Y = (A + B) \cdot (\overline{A \cdot B})$Using De Morgan's theorem,$\overline{A \cdot B} = \overline{A} + \overline{B}$.$Y = (A + B) \cdot (\overline{A} + \overline{B})$$Y = A \cdot \overline{A} + A \cdot \overline{B} + B \cdot \overline{A} + B \cdot \overline{B}$Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$,we get:$Y = A \cdot \overline{B} + \overline{A} \cdot B$This expression represents the Boolean function for an $	ext{XOR}$ gate.

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

The following configuration of gates is equivalent to $:-$

Similar Questions

The logic circuit is shown below:
$A$ $B$ $Y$
$0$ $0$ $1$
$0$ $1$ $0$
$1$ $0$ $0$
$1$ $1$ $1$

To obtain the given truth table,which logic gate should be placed at $G$?

The output of the following combination is the same as that of:

If the output of $Y = A \cdot \bar B$ is $Y = 1$,then the inputs are:

The symbolic representations of four logic gates are given below. The logic symbols for $OR$,$NOT$,and $NAND$ gates are respectively:

Identify the correct truth table of the given logical circuit.

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