To get the truth table shown, from the follow | vedclass

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(B) The circuit consists of an $OR$ gate with inputs $A$ and $C$, where $C$ is the output of gate $G$ with inputs $A$ and $B$. Thus, the output is $Y

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$AND$

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(B) The circuit consists of an $OR$ gate with inputs $A$ and $C$, where $C$ is the output of gate $G$ with inputs $A$ and $B$. Thus, the output is $Y = A + C = A + (A 	ext{ gate } B)$.Let's test the options for gate $G$:$1$. If $G$ is $AND$, then $C = A \cdot B$. The output $Y = A + (A \cdot B)$.For $(A, B) = (0, 0)$, $Y = 0 + (0 \cdot 0) = 0$.For $(A, B) = (0, 1)$, $Y = 0 + (0 \cdot 1) = 0$.For $(A, B) = (1, 0)$, $Y = 1 + (1 \cdot 0) = 1$.For $(A, B) = (1, 1)$, $Y = 1 + (1 \cdot 1) = 1$.This matches the given truth table perfectly. Therefore, $G$ must be an $AND$ gate.

To get the truth table shown, from the following logic circuit, the Gate $G$ should be
$A$ $B$ $Y$
$0$ $0$ $0$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $1$

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To get the truth table shown, from the following logic circuit, the Gate $G$ should be$A$$B$$Y$$0$$0$$0$$0$$1$$0$$1$$0$$1$$1$$1$$1$