For the following logic circuit, the truth table is:

For the following logic circuit, the truth table is:

$A \quad B \quad Y$ \\ $0 \quad 0 \quad 0$ \\ $0 \quad 1 \quad 0$ \\ $1 \quad 0 \quad 0$ \\ $1 \quad 1 \quad 1$
$A \quad B \quad Y$ \\ $0 \quad 0 \quad 1$ \\ $0 \quad 1 \quad 1$ \\ $1 \quad 0 \quad 1$ \\ $1 \quad 1 \quad 0$
$A \quad B \quad Y$ \\ $0 \quad 0 \quad 0$ \\ $0 \quad 1 \quad 1$ \\ $1 \quad 0 \quad 1$ \\ $1 \quad 1 \quad 1$
$\begin{aligned}A & B & Y \\0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 1 \\1 & 1 & 0 \\\end{aligned}$

Solution

The output of the given circuit can be depicted schematically as follows:

Hence. from the above diagram, the output can be written as

$\begin{array}{rcl} Y & = & A \cdot B \\ = & A + B \\ = & A + B \end{array}$

Thus, the given circuit represents an $OR$ logic, the truth table for which is given below:

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

Asked in: NEET 2023 (All India)

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