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A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.
Which of the following conclusions is/are correct based on the given data?
- a)The minimum number of NAND gates required will be 2.
- b)The minimum number of NAND gates required will be 3.
- c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.
- d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.
Correct answer is option 'C,D'. Can you explain this answer?
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Most Upvoted Answer
A combinational digital circuit design is attempted using only NAND lo...
Given Data:
The circuit has four inputs A, B, C, D and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.
To find:
The conclusions that are correct based on the given data.
Solution:
a) The minimum number of NAND gates required will be 2:
To implement the given logic using NAND gates, we can use the following circuit:
A ---- NAND ----\
B ---- NAND ---- NAND ---- Z
/
C ---- NAND ----
D ---- NAND ----/
In this circuit, we have used 4 NAND gates. Hence, the conclusion "The minimum number of NAND gates required will be 2" is incorrect.
b) The minimum number of NAND gates required will be 3:
From the above circuit, we can see that we can implement the given logic using 3 NAND gates. Hence, the conclusion "The minimum number of NAND gates required will be 3" is correct.
c) If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns:
The delay of each NAND gate is 10 ns. Since we have used 3 NAND gates in the circuit, the total delay will be 3 times the delay of a single NAND gate. Therefore, the output will appear after 3 * 10 ns = 30 ns, not 10 ns. Hence, the conclusion "If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns" is incorrect.
d) If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns:
As mentioned earlier, the total delay will be 3 times the delay of a single NAND gate. Therefore, the output will appear after 3 * 10 ns = 30 ns, not 20 ns. Hence, the conclusion "If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns" is incorrect.
Conclusion:
Based on the given data, the correct conclusions are:
- The minimum number of NAND gates required will be 3.
- If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 30 ns.
The circuit has four inputs A, B, C, D and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.
To find:
The conclusions that are correct based on the given data.
Solution:
a) The minimum number of NAND gates required will be 2:
To implement the given logic using NAND gates, we can use the following circuit:
A ---- NAND ----\
B ---- NAND ---- NAND ---- Z
/
C ---- NAND ----
D ---- NAND ----/
In this circuit, we have used 4 NAND gates. Hence, the conclusion "The minimum number of NAND gates required will be 2" is incorrect.
b) The minimum number of NAND gates required will be 3:
From the above circuit, we can see that we can implement the given logic using 3 NAND gates. Hence, the conclusion "The minimum number of NAND gates required will be 3" is correct.
c) If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns:
The delay of each NAND gate is 10 ns. Since we have used 3 NAND gates in the circuit, the total delay will be 3 times the delay of a single NAND gate. Therefore, the output will appear after 3 * 10 ns = 30 ns, not 10 ns. Hence, the conclusion "If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns" is incorrect.
d) If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns:
As mentioned earlier, the total delay will be 3 times the delay of a single NAND gate. Therefore, the output will appear after 3 * 10 ns = 30 ns, not 20 ns. Hence, the conclusion "If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns" is incorrect.
Conclusion:
Based on the given data, the correct conclusions are:
- The minimum number of NAND gates required will be 3.
- If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 30 ns.
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Community Answer
A combinational digital circuit design is attempted using only NAND lo...
As per design demand, the truth table is prepared as shown below:
Two level NAND implementation has the delay = 10 + 10
= 20 ns.
Hence, the correct options are (C) and (D).
= 20 ns.
Hence, the correct options are (C) and (D).
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A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer?
Question Description
A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer?.
A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer?.
Solutions for A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer?, a detailed solution for A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? has been provided alongside types of A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.Which of the following conclusions is/are correct based on the given data?a)The minimum number of NAND gates required will be 2.b)The minimum number of NAND gates required will be 3.c)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 10 ns.d)If each NAND gate has a delay of 10 ns, then on applying inputs, the output will appear after 20 ns.Correct answer is option 'C,D'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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