A 100 V DC source charges a capacitor. What is the capacitor voltage after one time constant τ from the start of charging?
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A37.8 V
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B38 V
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C63 V
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D101 V
Answer
Correct Answer: 63 V
Explanation
Introduction:First-order RC charging follows a well-known exponential. This question checks your ability to apply the standard time-constant result to find the capacitor voltage after one τ when charging from a step DC source.
Given Data / Assumptions:
- Step input Vs = 100 V applied at t = 0.
- Capacitor initially uncharged.
- Time of interest t = τ = R * C.
Concept / Approach:The capacitor voltage during charging is Vc(t) = Vs * (1 − e^(−t/τ)). Substituting t = τ yields Vc(τ) = Vs * (1 − e^(−1)). The numerical factor 1 − e^(−1) ≈ 0.632 applies to any RC with a step input.
Step-by-Step Solution:
Write Vc(t) = Vs * (1 − e^(−t/τ)).Set t = τ → Vc(τ) = Vs * (1 − e^(−1)).Compute 1 − e^(−1) ≈ 1 − 0.3679 ≈ 0.6321.With Vs = 100 V → Vc(τ) ≈ 100 * 0.6321 ≈ 63.21 V.Closest given option is 63 V.Verification / Alternative check:The complementary current at t = τ is i(τ) = (Vs/R) * e^(−1) ≈ 0.368 * (Vs/R); this pairs with Vc(τ) ≈ 0.632 * Vs, confirming standard RC behavior.
Why Other Options Are Wrong:
- 37.8 V or 38 V: These correspond to the remaining fraction e^(−1) of the initial step on the resistor, not the capacitor voltage at τ.
- 101 V: Exceeds the source; impossible without an inductor or overshoot mechanism.
Common Pitfalls:Confusing the capacitor's voltage with the resistor's drop at t = τ, or rounding errors that swap 0.632 and 0.368 fractions.
Final Answer:63 V.