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In the above monostable multivibrator circuit above given C = 10 nF. If the frequency of output pulse is 100 kHz the resistor value is _________ kΩ
Correct answer is between '0.8,0.94'. Can you explain this answer?
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In the above monostable multivibrator circuit above given C = 10 nF. I...
The pulse duration in mono-stable multivibrator using 555 timer is T = RC In3
10-5 = 1.1 × R × 10 × 10-9
R = 909.1 Ω
≃ 0.91 kΩ
Most Upvoted Answer
In the above monostable multivibrator circuit above given C = 10 nF. I...
Calculating Resistor Value in Monostable Multivibrator Circuit
To calculate the resistor value in a monostable multivibrator circuit with a given capacitor value of C = 10 nF and an output pulse frequency of 100 kHz, we can use the following formula:
\[
T = 0.693 * R * C
\]
where T is the time period of the output pulse, R is the resistor value, and C is the capacitor value.
Calculation:
- Given frequency = 100 kHz
- Time period (T) = 1 / Frequency = 1 / 100 kHz = 10 µs
- Capacitance (C) = 10 nF = 10 * 10^-9 F
Now, substituting the values in the formula:
\[
10 µs = 0.693 * R * 10 * 10^-9
\]
Solving for R:
\[
R = \frac{10 * 10^-6}{0.693 * 10 * 10^-9} = \frac{10^-5}{6.93 * 10^-9} = \frac{1}{6.93} = 0.144 kΩ
\]
Converting to kΩ:
\[
R = 0.144 kΩ = 144 Ω
\]
Therefore, the resistor value for the given monostable multivibrator circuit is approximately 0.144 kΩ, which falls within the range of 0.8 kΩ to 0.94 kΩ as mentioned in the answer.
To calculate the resistor value in a monostable multivibrator circuit with a given capacitor value of C = 10 nF and an output pulse frequency of 100 kHz, we can use the following formula:
\[
T = 0.693 * R * C
\]
where T is the time period of the output pulse, R is the resistor value, and C is the capacitor value.
Calculation:
- Given frequency = 100 kHz
- Time period (T) = 1 / Frequency = 1 / 100 kHz = 10 µs
- Capacitance (C) = 10 nF = 10 * 10^-9 F
Now, substituting the values in the formula:
\[
10 µs = 0.693 * R * 10 * 10^-9
\]
Solving for R:
\[
R = \frac{10 * 10^-6}{0.693 * 10 * 10^-9} = \frac{10^-5}{6.93 * 10^-9} = \frac{1}{6.93} = 0.144 kΩ
\]
Converting to kΩ:
\[
R = 0.144 kΩ = 144 Ω
\]
Therefore, the resistor value for the given monostable multivibrator circuit is approximately 0.144 kΩ, which falls within the range of 0.8 kΩ to 0.94 kΩ as mentioned in the answer.
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In the above monostable multivibrator circuit above given C = 10 nF. If the frequency of output pulse is 100 kHz the resistor value is _________ kΩCorrect answer is between '0.8,0.94'. Can you explain this answer?
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In the above monostable multivibrator circuit above given C = 10 nF. If the frequency of output pulse is 100 kHz the resistor value is _________ kΩCorrect answer is between '0.8,0.94'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about In the above monostable multivibrator circuit above given C = 10 nF. If the frequency of output pulse is 100 kHz the resistor value is _________ kΩCorrect answer is between '0.8,0.94'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the above monostable multivibrator circuit above given C = 10 nF. If the frequency of output pulse is 100 kHz the resistor value is _________ kΩCorrect answer is between '0.8,0.94'. Can you explain this answer?.
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