For the logic circuit shown, the truth table is : - Vidyadip

MCQ

For the logic circuit shown, the truth table is :

A
$\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{array}$
✓
$\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}$
C
$\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}$
D
$\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}$

✓

Answer

Correct option: B.

$\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}$

b
$Y=\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}=A \cdot B=$ $AND$ gate

$\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}$

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