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A NOR gate and a NAND gate are connected...

A NOR gate and a NAND gate are connected shown. Two different sets of inputs are given to this set up. In the first case, the inputs to the gates are A = 0, B = 0, C = 0. In the second case, the inputs are A = 1, B = 0, C = 1. The output D in the first case and second case respectively are

A

0 and 0

B

0 and 1

C

1 and 0

D

1 and 1

Text Solution

Verified by Experts

The correct Answer is:
D

From given combination of gates, the output is
`Y=bar(bar(A+B).C)`
`Y=bar(bar(A+B))+bar(C)"(By De-Morgan.s theorem)"`
`Y=(A+B+bar(C))`
Now, if A = 0, B = 0 and C = 0,
then output, `Y=(0+0+bar(0))=1.(becausebar(0)=1)`
Now if A = 1, B = 0 and C = 1,
then output, `Y=(1+0+bar(1))=1.(becausebar(1)=0)`
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