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.

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:

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:

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