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NEET · Physics · STD 12 - 14. Semicondutor electronics

દર્શાવેલ લોજીક પરિપથ માટે સત્યાર્થ સારણી .......... છે 

  1. A \(\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{array}\)
  2. B \(\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}\)
  3. C \(\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}\)
  4. D \(\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}\)

Step-by-step Solution

Detailed explanation

\(Y=\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}=A \cdot B=\) \(AND\) gate \(\begin{array}{lll}\text {A} & \text {B} & \text {Y} \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}\)
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