JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
The truth table for the circuit given in the fig is

- A \( {\begin{array}{*{20}{c}}
A&B&Y\\
0&0&1\\
0&1&0\\
1&0&0\\
1&1&0
\end{array}} \) - B \( {\begin{array}{*{20}{c}}
A&B&Y\\
0&0&0\\
0&1&0\\
1&0&1\\
1&1&1
\end{array}}\) - C \( {\begin{array}{*{20}{c}}
A&B&Y\\
0&0&1\\
0&1&1\\
1&0&1\\
1&1&1
\end{array}}\) - D \( {\begin{array}{*{20}{c}}
A&B&Y\\
0&0&1\\
0&1&1\\
1&0&0\\
1&1&0
\end{array}} \)
Answer & Solution
Correct Answer
(D) \( {\begin{array}{*{20}{c}}
A&B&Y\\
0&0&1\\
0&1&1\\
1&0&0\\
1&1&0
\end{array}} \)
Step-by-step Solution
Detailed explanation
\(C=A+B\) and \(y=\overline{A . C}\)…
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