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Question:
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Truth table for the given circuit is
Options:
- A
- B
- C
- D
Solution:
2223 Upvotes
Verified Answer
The correct answer is:
As the given figure the output of $\mathrm{C}, \mathrm{D}$ and $\mathrm{E}$ are :
$$
\begin{aligned}
&\mathrm{C}=\mathrm{A} \cdot \mathrm{B} \text { and } \mathrm{D}=\overline{\mathrm{A}} \cdot \mathrm{B} \\
&\mathrm{E}=\mathrm{C}+\mathrm{D}=(\mathrm{A} \cdot \mathrm{B})+(\overline{\mathrm{A}} \cdot \mathrm{B})
\end{aligned}
$$
So, the truth table of given arrangement of gates can be written as :
$$
\begin{array}{c|c|c|c|c|c}
\hline \mathrm{A} & \mathrm{B} & \overline{\mathrm{A}} & \mathrm{C}=\mathrm{A} \cdot \mathrm{B} & \mathrm{D}=\overline{\mathrm{A}} \cdot \mathrm{B} & \mathrm{E}=(\mathrm{C}+\mathrm{D}) \\
\hline 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 1 & 0 & 1 \\
\hline
\end{array}
$$
$$
\begin{aligned}
&\mathrm{C}=\mathrm{A} \cdot \mathrm{B} \text { and } \mathrm{D}=\overline{\mathrm{A}} \cdot \mathrm{B} \\
&\mathrm{E}=\mathrm{C}+\mathrm{D}=(\mathrm{A} \cdot \mathrm{B})+(\overline{\mathrm{A}} \cdot \mathrm{B})
\end{aligned}
$$
So, the truth table of given arrangement of gates can be written as :
$$
\begin{array}{c|c|c|c|c|c}
\hline \mathrm{A} & \mathrm{B} & \overline{\mathrm{A}} & \mathrm{C}=\mathrm{A} \cdot \mathrm{B} & \mathrm{D}=\overline{\mathrm{A}} \cdot \mathrm{B} & \mathrm{E}=(\mathrm{C}+\mathrm{D}) \\
\hline 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 1 & 0 & 1 \\
\hline
\end{array}
$$
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