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theBEAST
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Homework Statement
The Attempt at a Solution
Is my solution correct or can I simplify it even further?
Boolean Algebra Question (Digital Logic Circuits)
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The discussion revolves around a Boolean algebra problem related to digital logic circuits, focusing on the correctness and simplification of a solution. Participants question whether the final gate should be OR or NOR and discuss the application of DeMorgan's theorem. One contributor realizes they can apply DeMorgan's theorem to three terms, which could simplify their calculations significantly. There is a consensus that further simplification is possible and beneficial. Overall, the conversation emphasizes the importance of understanding and applying Boolean algebra concepts effectively.
Is my solution correct or can I simplify it even further?
- #2
NascentOxygen
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Is your final gate OR or NOR?
- #3
theBEAST
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NascentOxygen said:
Oh, that should be a NOR. Other than that mistake is all my work correct?
- #4
FOIWATER
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I see you applied de morgan's theorem to the AB term in the beginning, why did you not opt to apply it on the NOR function?
- #5
theBEAST
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FOIWATER said:
Oh right, when I did this I didn't know you could use DeMorgan's on three terms, I just found this out today. I will simplify it further now with this new knowledge.
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FOIWATER
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That should eliminate the need for so many calculations, although it is important to learn them as well.
I am not certain but I think applying de morgan's theorem to the NOR function might get you to a final solution in three steps
I am not certain but I think applying de morgan's theorem to the NOR function might get you to a final solution in three steps
The answer is (B) but I don't really understand why.
Based on formula of Young Modulus:
$$x=\frac{FL}{AE}$$
The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A##
I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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