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The given circuit is equivalent to
Options:
- A
- B
- C
- D
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The correct answer is:
Method 2:
$\begin{aligned} & Y=\overline{\bar{A}}+\overline{\bar{B}} \\ & Y=\overline{\bar{A}} \cdot \overline{\bar{B}}=A B\end{aligned}$
AND gate
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