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(B) The given circuit consists of two $AND$ gates,two $NOT$ gates,and one $OR$ gate.$1$. The upper $AND$ gate receives inputs $A$ and $\overline{B}$ (

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$A\cdot \overline {B} + \overline {A} \cdot B$

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(B) The given circuit consists of two $AND$ gates,two $NOT$ gates,and one $OR$ gate.$1$. The upper $AND$ gate receives inputs $A$ and $\overline{B}$ (since $B$ passes through a $NOT$ gate). Its output is $A \cdot \overline{B}$.$2$. The lower $AND$ gate receives inputs $\overline{A}$ (since $A$ passes through a $NOT$ gate) and $B$. Its output is $\overline{A} \cdot B$.$3$. These two outputs are fed into an $OR$ gate. Therefore,the final output $Y$ is the sum of these two expressions: $Y = A \cdot \overline{B} + \overline{A} \cdot B$.This is the Boolean expression for an $XOR$ (Exclusive-$OR$) gate.

In the combination of the following gates,the output $Y$ can be written in terms of inputs $A$ and $B$ as

Similar Questions

Identify the logic operation carried out by the given circuit.

Choose the correct logic circuit for the given truth table having inputs $A$ and $B$.
Inputs Inputs Outputs
$A$ $B$ $Y$
$0$ $0$ $0$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $1$

The circuit diagram shows a combination of logic gates. When inputs $P$ and $Q$ change to state $0$ with inputs $R$ and $S$ still at state $1$,the states of outputs $X$,$Y$ and $Z$ change to:

The following configuration of gates is equivalent to

$A$ gate in which all the inputs must be low to get a high output is called:

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Inputs	Inputs	Outputs
$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$