Difficulty: Easy
Correct Answer: Correct
Explanation:
Introduction / Context:
An RC “integrator” is simply an RC network driven by a pulse or step where the output of interest is usually taken across the capacitor. Whether it truly integrates or instead reaches near steady states during each pulse depends on the relationship between the time constant τ = R * C and the pulse width (PW) and duty cycle. This question asks if a capacitor can effectively charge and discharge to its final values within each half-cycle when τ is less than one-fifth of the pulse width in a 50% duty train.
Given Data / Assumptions:
Concept / Approach:
First-order exponentials settle to about 63% in 1τ, 86% in 2τ, 95% in 3τ, 98% in 4τ, and 99.3% in 5τ. Thus, when a pulse high (or low) interval exceeds roughly 5τ, the capacitor will reach within about 1% of its final value for that interval. With a 50% duty waveform, there is a charging interval and a discharging interval of the same length; the same 5τ rule applies to both halves.
Step-by-Step Solution:
Verification / Alternative check:
Use v_C(t) = V_final + (V_initial − V_final) * exp(−t/τ). For t = 5τ, |v_C − V_final| ≈ 0.7% of the step. With τ < PW/5, t ≥ 5τ within each interval, yielding ≈99% settling.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing “integrator behavior” (which prefers τ ≫ PW) with “full charging” (which needs τ ≪ PW). The same RC can show either behavior depending on τ vs. PW.
Final Answer:
Correct — with τ < PW/5, the capacitor essentially reaches its final value each half-cycle.
Discussion & Comments