Question 29: Semiconductors and Logic Gates

Question

The truth table for the circuit given below is:
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m6qjablt/e2a33071-ed44-4aaa-b90d-8e484b1c6077/81446b10-e2fe-11ef-9d65-2b094293796d/file-jaoe38c1m6qjablu.png?format=png)

Accepted Answer

$ \begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{\bar{A}} & \mathrm{\bar{B}} & \mathrm{A\cdot\bar{B}} & \mathrm{\bar{A} \cdot B} & \mathrm{Y}=\mathrm{A\cdot\bar{B}} + \mathrm{\bar{A} \cdot B}\ 
\hline 0 & 0 & 1 & 1 & 0 & 0 & 0 \ \hline 0 & 1 & 1 & 0 & 0 & 1 & 1 \ \hline 1 & 0 & 0 & 1 & 1 & 0 & 1 \ \hline 1 & 1 & 0 & 0 & 0 & 0 & 0 \ \hline \end{array} $ $\$
Hence, option 4 is correct.
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m6xpk2l1/0a90b1ec-9366-424b-8a2b-262b9ba18da2/83b5e450-e6f0-11ef-b104-c92cb842bf9b/file-jaoe38c1m6xpk2l2.png?format=png)

Answer

$ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y}\ 
\hline 0 & 0 & 0 \ \hline 1 & 1 & 1 \ \hline 1 & 0 & 1 \ \hline 0 & 1 & 1 \ \hline \end{array} $

Answer

$ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y}\ 
\hline 0 & 0 & 0 \ \hline 1 & 0 & 1 \ \hline 0 & 1 & 0 \ \hline 1 & 1 & 0 \ \hline \end{array} $

Answer

$ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y}\ 
\hline 0 & 0 & 0 \ \hline 1 & 0 & 0 \ \hline 1 & 1 & 0 \ \hline 0 & 1 & 1 \ \hline \end{array} $

Answer

$ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y}\ 
\hline 0 & 0 & 0 \ \hline 0 & 1 & 1 \ \hline 1 & 0 & 1 \ \hline 1 & 1 & 0 \ \hline \end{array} $

Semiconductors and Logic Gates

The truth table for the circuit given below is:

Choose the correct answer: