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(C) The given circuit consists of two $NAND$ gates acting as $NOT$ gates (inverters) followed by a third $NAND$ gate.Let the inputs be $A$ and $B$.The

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$OR$

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(C) The given circuit consists of two $NAND$ gates acting as $NOT$ gates (inverters) followed by a third $NAND$ gate.Let the inputs be $A$ and $B$.The first $NAND$ gate with both inputs tied to $A$ gives an output of $\bar{A}$.The second $NAND$ gate with both inputs tied to $B$ gives an output of $\bar{B}$.These outputs $\bar{A}$ and $\bar{B}$ are fed into the third $NAND$ gate.The final output $Y$ is given by:$Y = \overline{\bar{A} \cdot \bar{B}}$Using De Morgan's theorem, $\overline{x \cdot y} = \bar{x} + \bar{y}$, we get:$Y = \overline{\bar{A}} + \overline{\bar{B}}$Since $\overline{\bar{A}} = A$ and $\overline{\bar{B}} = B$, we have:$Y = A + B$This is the Boolean expression for an $OR$ gate.Therefore, the equivalent logic gate is an $OR$ gate.

Identify the equivalent logic gate for the following circuit.

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