RC time constant behavior: for a first-order RC response to a step, does the capacitor voltage (or charge) move about 63% of the remaining difference toward its final value over each successive interval of one time constant?

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
The first-order RC time constant (tau = R * C) governs how quickly a capacitor charges or discharges toward a new steady state after a step. Understanding the “63% rule” helps engineers estimate settling time without a calculator.


Given Data / Assumptions:

  • Single-pole RC circuit driven by a step.
  • Ideal linear components.
  • We consider changes over equal intervals of length tau.


Concept / Approach:
For a step from V_initial to V_final, the solution is V(t) = V_final + (V_initial - V_final) * exp(-t/tau). Over any interval of duration tau starting at t0, the change toward the final value equals (1 - e^-1) ≈ 0.632 of the remaining difference at t0. This is true for charge and discharge; only the sign of the change differs.


Step-by-Step Solution:

1) Write V(t0 + tau) - V_final = (V(t0) - V_final) * e^-1. 2) Therefore, reduction in the remaining gap over that tau is (1 - e^-1) ≈ 0.632 (≈ 63%). 3) This holds for each successive tau, not just the first.


Verification / Alternative check:
At t = tau from the initial step, the state is ~63% closer to final than at t = 0. Reapplying the same interval later yields the same fractional progress on the then-remaining difference.


Why Other Options Are Wrong:
“Incorrect” contradicts the exponential law; discharge vs charge and temperature qualifiers do not alter the first-order mathematics.


Common Pitfalls:
Confusing “63% of total change” with “63% of remaining change” when looking at later intervals; mixing series vs parallel RC forms.


Final Answer:
Correct

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