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(B) The truth table shows that the output $Y$ is $1$ only when $A=1$. Specifically,when $A=0$,$Y=0$ regardless of $B$. When $A=1$,$Y=1$ regardless of

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(B) The truth table shows that the output $Y$ is $1$ only when $A=1$. Specifically,when $A=0$,$Y=0$ regardless of $B$. When $A=1$,$Y=1$ regardless of $B$. This corresponds to the Boolean expression $Y = A$. Let us analyze the circuits:Option $A$: $Y = (A+B) \cdot B$. If $A=0, B=1$,$Y = (0+1) \cdot 1 = 1$. This does not match the truth table.Option $B$: $Y = (A+B) \cdot A$. If $A=0, B=0$,$Y = (0+0) \cdot 0 = 0$. If $A=0, B=1$,$Y = (0+1) \cdot 0 = 0$. If $A=1, B=0$,$Y = (1+0) \cdot 1 = 1$. If $A=1, B=1$,$Y = (1+1) \cdot 1 = 1$. This matches the given truth table perfectly.

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

Similar Questions

In the Boolean algebra,$(\overline {\bar A \cdot \bar B} ) \cdot A$ is equal to:

The Boolean expression of the circuit given in the figure is

Sketch the output $Y$ from a $NAND$ gate having inputs $A$ and $B$ as shown in the figure.

Which option represents the correct truth table for the logic circuit shown below?

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

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