JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 12)
The output from NAND gate having inputs A and B given below will be,
_13th_April_Evening_Shift_en_12_2.png)
_13th_April_Evening_Shift_en_12_3.png)
_13th_April_Evening_Shift_en_12_4.png)
_13th_April_Evening_Shift_en_12_5.png)
Explanation
$$
\begin{aligned}
&\text {Truth table for NAND gate is }\\\\
&\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}} \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
\end{aligned}
$$
On the basis of given input $\mathrm{A}$ and $\mathrm{B}$ the truth table is
On the basis of given input $\mathrm{A}$ and $\mathrm{B}$ the truth table is
$\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{Y}$ |
---|---|---|
1 | 1 | 0 |
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
0 | 0 | 1 |
0 | 1 | 1 |
Comments (0)
