Astable timing sanity check — for an oscillator frequency of 54.86 kHz, is the period 1.82 s or something else? Use T = 1 / f to evaluate.
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AIncorrect — the period is approximately 18.2 µs, not 1.82 s
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BCorrect — the period is 1.82 s
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CCorrect only for very low duty cycles
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DCannot be determined without RA, RB, and C
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ECorrect only with a CMOS 555
Answer
Correct Answer: Incorrect — the period is approximately 18.2 µs, not 1.82 s
Explanation
Introduction / Context:Whether using a 555 astable or any periodic oscillator, the fundamental relationship between frequency and period is T = 1 / f. An assertion that a 54.86 kHz signal has a period of 1.82 s is off by many orders of magnitude. This question reinforces quick order-of-magnitude checks that prevent unit mistakes when reading instruments or solving design problems.
Given Data / Assumptions:
- Frequency f = 54.86 kHz = 54,860 Hz.
- Ideal periodic signal for which T = 1/f applies.
- We are checking numerical consistency only.
Concept / Approach:Use T = 1 / f with careful unit handling. Since kilohertz means 10^3 Hz, a period for tens of kilohertz must be in tens of microseconds, not seconds. This mental check helps catch common calculator and unit-entry mistakes.
Step-by-Step Solution:
1) Convert the frequency: f = 54,860 Hz.2) Compute T = 1 / 54,860 ≈ 0.00001823 s.3) Convert to microseconds: 0.00001823 s ≈ 18.23 µs.4) Compare to the claimed 1.82 s: the claim is wrong by a factor of ~100,000.Verification / Alternative check:Quick magnitude check: 50 kHz roughly corresponds to 20 µs (since 1/50,000 = 20 × 10^−6 s).
Why Other Options Are Wrong:
“Correct — 1.82 s”: inconsistent with T = 1/f.“Cannot be determined”: frequency alone is sufficient to compute period.Device type (CMOS vs bipolar) or duty cycle does not change T = 1/f for a given measured f.Common Pitfalls:Confusing kHz with Hz; misplacing decimal points; using minutes or milliseconds accidentally; relying on calculator without sanity checks.
Final Answer:Incorrect — the period is approximately 18.2 µs