JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
For the given logic gate circuit, which of the following is the correct truth table?
- A
n m z 0 0 1 0 1 0 1 1 0 1 0 0 - B
n m z 0 0 0 0 1 1 1 1 0 1 0 1 - C
n m z 0 0 1 0 1 0 1 1 1 1 0 0 - D
n m z 0 0 1 0 1 1 1 1 0 1 0 0
Answer & Solution
Correct Answer
(D) n m z 0 0 1 0 1 1 1 1 0 1 0 0
Step-by-step Solution
Detailed explanation
Gate 1 : At bottom there is an OR gate with inputs n & m out put \(=n+m\) Gate 2 : A NAND gate, its Input are direct n & the output of OR gate ( \(n+m\) ) out put \(z =\overline{ n \cdot( n + m )}\) since \(n .( n + m )=( n . n )+( n . m )\) \(= n + n \cdot m = n (1+ m )= n\) ∴…
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