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The circuit as shown in figure is equiva...

The circuit as shown in figure is equivalent to

A

OR gate

B

NOR gate

C

NAND gate

D

AND gate

Text Solution

Verified by Experts

The correct Answer is:
A

`Ybar(bar(A.B).(bar(A+B)))`
`=barbar((A+B))+barbar((A+B))" (From De Morgan.s theorem)"`
`=A.B+A+B=A(B+1)+B=A+B`
It is the Boolean expression of the OR gate.
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