RC charging time constant: in a standard first-order RC charging circuit, the time to reach practical full charge is approximately which multiple of RC?
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A5 RC
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BRC
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C6 RC
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Dnone of the above
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E3 RC
Answer
Correct Answer: 5 RC
Explanation
Introduction / Context:The time constant τ = R * C sets the speed of exponential charging in RC networks. Engineers often use a practical benchmark for “full” charge to estimate delays, settling times, and sampling windows in analog systems.
Given Data / Assumptions:
- Standard RC charging from a DC source through a resistor.
- No significant leakage or series resistance beyond R.
- “Full charge” means about 99% of final value (common engineering convention).
Concept / Approach:Capacitor voltage during charging is Vc(t) = Vs * (1 - e^(-t/RC)). At t = 5RC, e^(-5) ≈ 0.0067, so Vc ≈ 99.33% of Vs. This is widely accepted as the time to essentially reach the final value for practical purposes.
Step-by-Step Solution:
Write charging law: Vc(t) = Vs * (1 - e^(-t/RC)).Choose t = 5RC → Vc/Vs = 1 - e^(-5).Compute e^(-5) ≈ 0.0067 → Vc ≈ 0.9933 * Vs (≈ 99.3%).Verification / Alternative check:At 4RC, Vc ≈ 98.2% Vs; at 6RC, ≈ 99.75% Vs. Many references standardize on 5RC as a convenient “practical full charge.”
Why Other Options Are Wrong:
- RC: Only ~63.2% charged.
- 6 RC: Also near full, but the conventional benchmark used in design notes is 5RC.
- None of the above: Incorrect because 5RC is the conventional answer.
Common Pitfalls:
- Confusing time constant (63.2%) with practical full charge (~99%).
- Assuming exact “100%” is achievable in finite time; exponential never truly reaches it.
Final Answer:5 RC