Question

Which of the following represents the correct truth table for the electrical network shown in figure given below?

• A. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\\ 1& 1& 0\end{array}$
• B. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\\ 1& 1& 0\end{array}$
• C. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 1& 0& 1\\ 0& 1& 1\\ 1& 1& 1\end{array}$
• D. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 1& 0& 1\\ 0& 1& 1\\ 1& 1& 0\end{array}$

Answer 1

The correct option is B $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\\ 1& 1& 0\end{array}$

The output from first logic gate is
$Y_1=\overline{A+B}$

The input to the second logic gate will be $Y_1$ and $Y_1$, thus output from second logic gate;

$Y_2=\overline{Y_1+Y_1}$

$\Rightarrow Y_2=\overline{Y_1}(\because A+A=A)$

$\Rightarrow Y_2=\overline{\overline{A+B}}=A+B$

Now the input to the third gate will be $Y_2$, here the final output is,

$Y=\overline{Y_2}$

$\Rightarrow Y=\overline{A+B}$

Constructing the truth table

$A=0,B=0;Y=\overline{0+0}=1$
$A=1,B=0;Y=\overline{1+0}=0$
$A=0,B=1;Y=\overline{0+1}=0$
$A=1,B=1;Y=\overline{1+1}=0$

$\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\\ 1& 1& 0\end{array}$

Hence, option $\left(B\right)$ is correct.

Why this question? Tip: When logic gates are connected in series, the output from the preceding gate will act as an input to the next gate.