A logic gate whose output will be in logic 0 state only when all inputs are in logic 1 state is called

Question

A logic gate whose output will be in logic 0 state only when all inputs are in logic 1 state is called (A) AND (B) OR (C) NOR (D) NAND

Tardigrade Admin · Accepted Answer

For AND gate, Y = A.B = 1.1 = 1 For OR gate, Y = A + B = 1 + 1 = 1 For NOR gate, Y = overline A + B = overline1 + 1 = overline1 = 0 Also Y = overline1 + 0 = overline1 = 0 For NAND gate, Y= overlineA·B = overline1 .1 = overline1 =0

Q. A logic gate whose output will be in logic 0 state only when all inputs are in logic 1 state is called

AND

OR

NOR

NAND

For $A N D$ gate, $Y = A . B = 1.1 = 1$
For $OR$ gate, $Y = A + B = 1 + 1 = 1$
For $NOR$ gate, $Y = \overline{A + B} = \overline{1 + 1} = \overline{1} = 0$
Also $Y = \overline{1 + 0} = \overline{1} = 0$
For $N A N D$ gate, $Y = \overline{A \cdot B} = \overline{1.1} = \overline{1} = 0$

Q. A logic gate whose output will be in logic 0 state only when all inputs are in logic 1 state is called

AND

OR

NOR

NAND

For AND gate,Y=A.B=1.1=1 For OR gate, Y=A+B=1+1=1 For NOR gate, Y=A+B​=1+1​=1=0 Also Y=1+0​=1=0 For NAND gate, Y=A⋅B=1.1=1=0

For $A N D$ gate, $Y = A . B = 1.1 = 1$
For $OR$ gate, $Y = A + B = 1 + 1 = 1$
For $NOR$ gate, $Y = \overline{A + B} = \overline{1 + 1} = \overline{1} = 0$
Also $Y = \overline{1 + 0} = \overline{1} = 0$
For $N A N D$ gate, $Y = \overline{A \cdot B} = \overline{1.1} = \overline{1} = 0$