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Question
Truth table for the given circuit (Figure) is ______.
Options
A B E 0 0 1 0 1 0 1 0 1 1 1 0 A B E 0 0 1 0 1 0 1 0 0 1 1 1 A B E 0 0 1 0 1 1 1 0 0 1 1 1 A B E 0 0 0 0 1 1 1 0 1 1 1 0
Solution
| A | B | E |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Explanation:
In this problem, the input C of the OR gate is an output of the AND gate. So, “C equals A AND B” or C = A · B and “D equals Not A AND B” or D = Ā · B and “E equals C AND D" or E = C + D = (A · B) + (A · B)
Now we can generate the truth table of this arrangement of gates can be given by
| A | B | Ā | C = A · B | d = Ā · B | E = (C + D) |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 | 0 | 1 |
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Useful Constants and Relations:
| 1. | Charge of a proton | (e) | =1.6 × 10-19C |
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| 3. | Mass of an electron | (m) | = 9·1× 10-31 kg |
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