# Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 41)

41.

An OR gate with schematic "bubbles" on its inputs performs the same functions as a(n)________ gate.

Discussion:

4 comments Page 1 of 1.
PRITHA said:
7 years ago

According to Demorgan's 1st theorem,

(A.B)'=A'+B'.

Here, the input of the OR gate becomes,

A',B' as it is passed through the bubble before applied to the gate . Thus the output becomes

A'+B' which gives (A+B)' by De morgan's theorem which is the output of NAND gate.

(A.B)'=A'+B'.

Here, the input of the OR gate becomes,

A',B' as it is passed through the bubble before applied to the gate . Thus the output becomes

A'+B' which gives (A+B)' by De morgan's theorem which is the output of NAND gate.

Umesh chander said:
8 years ago

Question is saying "bubbles" on its inputs.

So inputs to the OR gate is now A' and B' and the on the addition of A'+B' will get the output as NAND gate i.e.

A B A' B' A'+B'

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

1 1 0 0 0

So inputs to the OR gate is now A' and B' and the on the addition of A'+B' will get the output as NAND gate i.e.

A B A' B' A'+B'

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

1 1 0 0 0

Saurav said:
9 years ago

The answer should be Option (A) NOR.

Steven said:
1 decade ago

why is it a NOR gate. Shouldn't a nor gate be (A+B)(bar) while the and gate has A(bar) + B(bar)?

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