MHT CET · Physics · Semiconductors
The truth table for the given logic circuit is
- A P
- B Q
- C S
- D R
Answer & Solution
Correct Answer
(A) P
Step-by-step Solution
Detailed explanation
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.
\(\mathrm{Y}=\overline{(\overline{\mathrm{A}+\mathrm{B}}) \cdot(\mathrm{A} \cdot \mathrm{~B})}\)
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.
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