Question

For the logic circuit, the truth table is :

• A. A B Y
  $0$ $0$ $1$
  $0$ $1$ $1$
  $1$ $0$ $1$
  $1$ $1$ $0$
• B. A B Y
  $0$ $0$ $1$
  $0$ $1$ $0$
  $1$ $0$ $0$
  $1$ $1$ $0$
• C. A B Y
  $0$ $0$ $0$
  $0$ $1$ $0$
  $1$ $0$ $0$
  $1$ $1$ $1$
• D. A B Y
  $0$ $0$ $0$
  $0$ $1$ $1$
  $1$ $0$ $1$
  $1$ $1$ $1$

Answer 1

The correct option is C A B Y

$0$ $0$ $0$
$0$ $1$ $0$
$1$ $0$ $0$
$1$ $1$ $1$
Here,
$Y=\overline{\overline{A}+\overline{B}}=\overline{\overline{A}}+\overline{\overline{B}}=A.B$
(using De Morgan's Rule)

If $A=0$ and $B=0$ then $Y=0$

Similarly, for other inputs.

Hence, option $c$ is correct.