JEE MAIN - Physics (2023 - 11th April Evening Shift - No. 11)
The logic operations performed by the given digital circuit is equivalent to:
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NOR
AND
NAND
OR
Explanation
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$$ \begin{aligned} Z =\overline{(A+B) \cdot(A \cdot B)} \\\\ Y =\bar{Z}=(A+B) \cdot(A \cdot B) \\\\ \end{aligned} $$
$$ Y=A \cdot(A B)+B \cdot(A B) $$
$$ \begin{aligned} & =A B+A B \\\\ & =(A B) \end{aligned} $$
$\therefore$ It is an AND gate.
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