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The following configuration of gates is equivalent toA. `HAND`B. `XOR`C. `OR`D. None of these

Answer» Correct Answer - A
Output of `G_(1)+A+B`
Output of `G_(1)=bar(A.B)`
Output of `G_(3)`
`=(A+B).bar(A.B)=(A+B).(bar(A)+bar(B))`
If `A=1` and `B=1`, then
`A+B=1`
`A+B=0`
So, output is zero,
If `A=0` and `B=0`, then
`A+B=0`
`bar(A)+bar(B)=1+1=1`
So, output is zero.
Clearly, the given combination is `XOR` gate.


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