The following logic gate combination is equiv | vedclass

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(C) The given circuit consists of an $OR$ gate and a $NAND$ gate whose outputs are fed into an $AND$ gate.Let the inputs be $A$ and $B$.The output of

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$XOR$ gate

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(C) The given circuit consists of an $OR$ gate and a $NAND$ gate whose outputs are fed into an $AND$ gate.Let the inputs be $A$ and $B$.The output of the $OR$ gate is $(A + B)$.The output of the $NAND$ gate is $(\overline{A \cdot B})$.These are the inputs to the final $AND$ gate.Therefore,the final output $Y$ is given by:$Y = (A + B) \cdot (\overline{A \cdot B})$Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$.$Y = (A + B) \cdot (\bar{A} + \bar{B})$$Y = A \cdot \bar{A} + A \cdot \bar{B} + B \cdot \bar{A} + B \cdot \bar{B}$Since $A \cdot \bar{A} = 0$ and $B \cdot \bar{B} = 0$,we have:$Y = 0 + A \cdot \bar{B} + B \cdot \bar{A} + 0$$Y = A \cdot \bar{B} + \bar{A} \cdot B$This is the Boolean expression for an $XOR$ gate.

The following logic gate combination is equivalent to

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