Modern Physics 6 Question 3
3. The truth table for the circuit given in the figure is
(Main 2019, 12 April I)
(a) $\left|\begin{array}{lll}A & B & Y \ 0 & 0 & 1 \ 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 1\end{array}\right|$
(c) $\left|\begin{array}{lll}A & B & Y \ 0 & 0 & 1 \ 0 & 1 & 1 \ 1 & 0 & 0 \ 1 & 1 & 0\end{array}\right|$
(b) $\left|\begin{array}{lll}A & B & Y \ 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \ 1 & 1 & 0\end{array}\right|$
(d) $\left|\begin{array}{lll}A & B & Y \ 0 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & 1 \ 1 & 1 & 1\end{array}\right|$
Show Answer
Solution:
- Given circuit is
Let the intermediate state $X$ of OR gate is shown in figure.
Clearly, $Y=\overline{A X}$
Here, $X=A+B$
$\therefore \quad Y=\overline{A(A+B)}=\overline{A A+A B}$
$$ \begin{aligned} & =\overline{A+A B} \\ & =\overline{A(1+B)} \end{aligned} $$
$$ (\because A A=A) $$
$\Rightarrow \quad Y=\bar{A}$
$(\because 1+B=1)$
So, truth table shown in option (c) is correct.
Alternate Solution We can solve it using truth table
| $\boldsymbol{A}$ | $\boldsymbol{B}$ | $\boldsymbol{X}=\boldsymbol{A}+\boldsymbol{B}$ | $\boldsymbol{Y}=\overline{\boldsymbol{A} \boldsymbol{X}}$ |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 |
