The following configuration of gates is equiv | vedclass

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(C) The output of the $OR$ gate is $(A + B)$.The output of the $NAND$ gate is $(\overline{A \cdot B})$.These two outputs are fed into an $AND$ gate.Th

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$	ext{XOR}$

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(C) The output of the $OR$ gate is $(A + B)$.The output of the $NAND$ gate is $(\overline{A \cdot B})$.These two outputs are fed into an $AND$ gate.Therefore,the final output $Y$ is given by:$Y = (A + B) \cdot (\overline{A \cdot B})$Using De Morgan's theorem,$\overline{A \cdot B} = \overline{A} + \overline{B}$.$Y = (A + B) \cdot (\overline{A} + \overline{B})$$Y = A \cdot \overline{A} + A \cdot \overline{B} + B \cdot \overline{A} + B \cdot \overline{B}$Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$,we get:$Y = A \cdot \overline{B} + \overline{A} \cdot B$This expression represents the Boolean function for an $	ext{XOR}$ gate.

The following configuration of gates is equivalent to $:-$

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