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A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0 and goes back LOW at t = 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t = 3 ms. The output pulse.
- a)comes back high at t = 1 ms
- b)comes back high at t = 3 ms
- c)goes low at t = 0.8 ms
- d)goes low at t = 0
Correct answer is option 'A,C'. Can you explain this answer?
Verified Answer
A pulse is applied to each input of a 2 input NAND gate, one pulse goe...
A NAND gate is given by 
Truth table is.
It is low when both input signals are high otherwise it goes high.
The correct answers are: goes low at t = 0.8 ms, comes back high at t = 1 ms
Truth table is.It is low when both input signals are high otherwise it goes high.
The correct answers are: goes low at t = 0.8 ms, comes back high at t = 1 ms
Most Upvoted Answer
A pulse is applied to each input of a 2 input NAND gate, one pulse goe...
Explanation:
To understand the behavior of the NAND gate, let's analyze the inputs and outputs at different time intervals.
Time interval: t = 0 to t = 1 ms
- One input pulse goes high at t = 0 and goes back LOW at t = 1 ms.
- The other input pulse is not yet high at this time interval.
Time interval: t = 0.8 ms to t = 3 ms
- The second input pulse goes high at t = 0.8 ms and goes back LOW at t = 3 ms.
- The first input pulse is still LOW during this time interval.
Output behavior:
- The output of a NAND gate is HIGH when any of its inputs are LOW.
- The output of a NAND gate is LOW only when both of its inputs are HIGH.
Analysis:
1. At t = 0, the first input pulse goes HIGH, making the first input of the NAND gate LOW. Since the other input is still LOW, the output of the NAND gate is HIGH.
2. At t = 0.8 ms, the second input pulse goes HIGH, making the second input of the NAND gate LOW. The first input is still LOW, so the output of the NAND gate remains HIGH.
3. At t = 1 ms, the first input pulse goes back LOW, making the first input of the NAND gate HIGH. The second input is still LOW, so the output of the NAND gate goes LOW.
4. At t = 3 ms, the second input pulse goes back LOW, making the second input of the NAND gate HIGH. Both inputs are now HIGH, so the output of the NAND gate goes back HIGH.
Conclusion:
Based on the analysis, we can conclude that:
- The output of the NAND gate comes back HIGH at t = 1 ms because both inputs are LOW at that time.
- The output of the NAND gate goes LOW at t = 0.8 ms because one of the inputs is HIGH at that time.
Therefore, the correct answers are option 'A' (comes back high at t = 1 ms) and option 'C' (goes low at t = 0.8 ms).
To understand the behavior of the NAND gate, let's analyze the inputs and outputs at different time intervals.
Time interval: t = 0 to t = 1 ms
- One input pulse goes high at t = 0 and goes back LOW at t = 1 ms.
- The other input pulse is not yet high at this time interval.
Time interval: t = 0.8 ms to t = 3 ms
- The second input pulse goes high at t = 0.8 ms and goes back LOW at t = 3 ms.
- The first input pulse is still LOW during this time interval.
Output behavior:
- The output of a NAND gate is HIGH when any of its inputs are LOW.
- The output of a NAND gate is LOW only when both of its inputs are HIGH.
Analysis:
1. At t = 0, the first input pulse goes HIGH, making the first input of the NAND gate LOW. Since the other input is still LOW, the output of the NAND gate is HIGH.
2. At t = 0.8 ms, the second input pulse goes HIGH, making the second input of the NAND gate LOW. The first input is still LOW, so the output of the NAND gate remains HIGH.
3. At t = 1 ms, the first input pulse goes back LOW, making the first input of the NAND gate HIGH. The second input is still LOW, so the output of the NAND gate goes LOW.
4. At t = 3 ms, the second input pulse goes back LOW, making the second input of the NAND gate HIGH. Both inputs are now HIGH, so the output of the NAND gate goes back HIGH.
Conclusion:
Based on the analysis, we can conclude that:
- The output of the NAND gate comes back HIGH at t = 1 ms because both inputs are LOW at that time.
- The output of the NAND gate goes LOW at t = 0.8 ms because one of the inputs is HIGH at that time.
Therefore, the correct answers are option 'A' (comes back high at t = 1 ms) and option 'C' (goes low at t = 0.8 ms).
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A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer?
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A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer?.
A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer?.
Solutions for A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer?, a detailed solution for A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? has been provided alongside types of A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A pulse is applied to each input of a 2 input NAND gate, one pulse goes high at t = 0and goes back LOW at t= 1 ms. The other pulse goes high at f = 0.8 ms and goes Mark 0.00 out of 2.00 back LOW at t= 3 ms. The output pulse.a)comes back high at t= 1 msb)comes back high at t = 3 msc)goes low at t = 0.8 msd)goes low at t = 0Correct answer is option 'A,C'. Can you explain this answer? tests, examples and also practice Physics tests.
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