The truth table for the given logic circuit is
The truth table for the given logic circuit is


- P
- Q
- S
- R
Solution
From the logic circuit,
$\mathrm{Y}=\overline{(\overline{\mathrm{A}+\mathrm{B}}) \cdot(\mathrm{A} \cdot \mathrm{~B})}$
Using De-Morgan's law, $\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}$
$\begin{aligned}
& \mathrm{Y}=\overline{(\overline{\mathrm{A}} \cdot \overline{\mathrm{~B}}) \cdot(\mathrm{A} \cdot \mathrm{~B})} \\
& \mathrm{Y}=\overline{(\overline{\mathrm{A}} \cdot \mathrm{~A}) \cdot(\overline{\mathrm{B}} \cdot \mathrm{~B})} \\
& \mathrm{Y}=\overline{0.0}=1
\end{aligned}$
$\therefore \quad$ Output is always 1.
~
Asked in: MHT CET 2024 (04 May Shift 2)
Practice more Semiconductors questions on Aicharya