Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:First-order RC charging follows an exponential approach to the final value. Because exponentials are asymptotic, the capacitor voltage gets closer and closer to the source voltage but mathematically reaches it only as time approaches infinity. Practical rules of thumb use percentages at specific multiples of the time constant τ = R * C.
Given Data / Assumptions:
Concept / Approach:At t = 1τ, v_C ≈ 63%; 2τ ≈ 86%; 3τ ≈ 95%; 4τ ≈ 98%; 5τ ≈ 99.3%. Many practitioners call 5τ effectively full for practical purposes. Therefore, the claim that the capacitor is fully charged in 4τ is inaccurate; at 4τ it is close (about 98%), but not fully, and 5τ is the more common benchmark for 'essentially charged.'
Step-by-Step Solution:
Compute v_C(4τ) = V * (1 − e^(−4)) ≈ 98% of V.Compute v_C(5τ) = V * (1 − e^(−5)) ≈ 99.3% of V.Note asymptotic approach: exact equality requires infinite time.Conclude the statement is incorrect as written.Verification / Alternative check:Oscilloscope measurements of step response match the exponential curve and common 5τ guideline used in design texts.
Why Other Options Are Wrong:
Correct: contradicts the asymptotic nature of the response.Other qualifiers (sinusoidal, R = 0) do not fix the misstatement; the 4τ figure is still not 'fully charged.'Common Pitfalls:Equating 'nearly full' with 'fully' without specifying tolerance; ignoring that many specifications require explicit percentage thresholds.
Final Answer:Incorrect
Discussion & Comments