NIELIT 2016 DEC Scientist B (CS) - Section B: 4

Question

The Circuit is equivalent to:

• A. $OR$ Gate
• B. $NOR$ Gate
• C. $AND$ Gate
• D. $EX-OR$ Gate

Answer 1

1.  $\overline{A\times A}=\overline{A}$    

2.  $\overline{B\times B}=\overline{B}$     

3.  $\overline{\overline{A}\times \overline{B}}=\overline{\overline{A+B}}=A+B$     (Using De-Morgan's law  $\overline{A}\times\overline{B}=\overline{A+B}$)

4.  $\overline{(A+B)+(A+B)}=\overline{A+B}$     $\Rightarrow$   NOR gate

Option B is correct.

Comments

@haralk10

Instead of $\times$ you can use $\cdot,$ because in digital logic $.$ is called multiplication.

Answer 2

NAND and NAND implementation is equivalent to And OR.

So A.A+B.B = A + B

Giving same input to NAND or NOR gates will give its complement.

So its (A+B)'

Hence B is correct.

Answer 3

Circuit is equivalent to NOR Gate.