In a digital circuit for three input signals | vedclass

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(D) To find the correct circuit, we analyze the given truth table conditions where $Y = 0$:$1$. For $(A, B, C) = (0, 0, 0)$, $Y = 0$.$2$. For $(A, B,

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(D) To find the correct circuit, we analyze the given truth table conditions where $Y = 0$:$1$. For $(A, B, C) = (0, 0, 0)$, $Y = 0$.$2$. For $(A, B, C) = (0, 0, 1)$, $Y = 0$.$3$. For $(A, B, C) = (0, 1, 0)$, $Y = 0$.Let's evaluate the circuit in option $D$ (image $814-$d281), which consists of an $AND$ gate with inputs $B$ and $C$, followed by an $OR$ gate with inputs $A$ and the output of the $AND$ gate. The Boolean expression is $Y = A + (B \cdot C)$.Testing the inputs:- For $(0, 0, 0)$: $Y = 0 + (0 \cdot 0) = 0 + 0 = 0$.- For $(0, 0, 1)$: $Y = 0 + (0 \cdot 1) = 0 + 0 = 0$.- For $(0, 1, 0)$: $Y = 0 + (1 \cdot 0) = 0 + 0 = 0$.- For $(0, 1, 1)$: $Y = 0 + (1 \cdot 1) = 0 + 1 = 1$.- For $(1, 0, 0)$: $Y = 1 + (0 \cdot 0) = 1 + 0 = 1$.- For $(1, 0, 1)$: $Y = 1 + (0 \cdot 1) = 1 + 0 = 1$.- For $(1, 1, 0)$: $Y = 1 + (1 \cdot 0) = 1 + 0 = 1$.- For $(1, 1, 1)$: $Y = 1 + (1 \cdot 1) = 1 + 1 = 1$.This matches the required output conditions perfectly. Therefore, the correct circuit is $D$.

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