JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 12)

The output from NAND gate having inputs A and B given below will be,

JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 44 English

JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 44 English Option 1
JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 44 English Option 2
JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 44 English Option 3
JEE Main 2023 (Online) 13th April Evening Shift Physics - Semiconductor Question 44 English Option 4

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

$\mathrm{A}$ $\mathrm{B}$ $\mathrm{Y}$
1 1 0
0 0 1
0 1 1
1 0 1
1 1 0
0 0 1
0 1 1

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