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The following configuration of gate is equivalent to A. `NAND` gateB. `XOR` gateC. `OR` gateD. `NOR` gate

Answer» Correct Answer - B
`Y=(A+B).(overline(A.B))=(A+B).(overline(A)+overline(B))`
`=A. overline(A) + A. overline(A) + overline(A).B+ B. overline(B)`
`Y=(A.overline(B))+(overline(A).B)`. It is `XOR` gate.


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