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14. Semicondutor electronics — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics14. Semicondutor electronics1 Mark
MCQ
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}}$
  • ✓
    $ {\begin{array}{*{20}{c}}
    A&B&Y\\
    0&0&1\\
    0&1&1\\
    1&0&0\\
    1&1&0
    \end{array}} $
✓

Answer

Correct option: D.
$ {\begin{array}{*{20}{c}}
A&B&Y\\
0&0&1\\
0&1&1\\
1&0&0\\
1&1&0
\end{array}} $
d
$C=A+B$

and  $y=\overline{A . C}$

$\begin{array}{l|l|l|l|l|}\hline A & {\mathrm{B}} & {\mathrm{C}=(\mathrm{A}+\mathrm{B})} & {\mathrm{A} \mathrm{C}} & {\mathrm{y}=\overline{\mathrm{A} . \mathrm{C}}} \\ \hline 0 & {0} & {0} & {0} & {1} \\ {0} & {1} & {1} & {0} & {1} \\ {1} & {0} & {1} & {1} & {0} \\ {1} & {1} & {1} & {1} & {0} \\ \hline\end{array}$

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