In a certain 2-input logic gate,when inputs A | vedclass

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(C) The truth table for the given conditions is:$A=0, B=0 \implies C=1$$A=0, B=1 \implies C=1$Let us check the truth tables for the given options:$1$.

Vedclass · Accepted Answer

$NAND$

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(C) The truth table for the given conditions is:$A=0, B=0 \implies C=1$$A=0, B=1 \implies C=1$Let us check the truth tables for the given options:$1$. $OR$ gate: $0+0=0, 0+1=1, 1+0=1, 1+1=1$. (Does not match)$2$. $AND$ gate: $0 \cdot 0=0, 0 \cdot 1=0, 1 \cdot 0=0, 1 \cdot 1=1$. (Does not match)$3$. $NAND$ gate: $\overline{0 \cdot 0}=1, \overline{0 \cdot 1}=1, \overline{1 \cdot 0}=1, \overline{1 \cdot 1}=0$. (Matches the given conditions)$4$. $NOR$ gate: $\overline{0+0}=1, \overline{0+1}=0, \overline{1+0}=0, \overline{1+1}=0$. (Does not match)Thus,the logic gate is a $NAND$ gate.

In a certain $2$-input logic gate,when inputs $A=0$ and $B=0$,then output $C=1$. And also when inputs $A=0, B=1$,then again output $C=1$. The gate must be

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