Difficulty: Easy
Correct Answer: When the capacitor voltage is essentially equal to the DC source voltage (about after 5τ) and charging current is ~0
Explanation:
Introduction:In RC charging, a capacitor does not reach the final value instantaneously; instead, its voltage approaches the supply asymptotically. This question tests practical engineering judgment about when a capacitor is considered ‘‘fully charged’’ and the relationship between the capacitor voltage, the source voltage, and the current in a first-order RC circuit.
Given Data / Assumptions:
Concept / Approach:For a step input, capacitor voltage follows Vc(t) = Vs * (1 − e^(−t/τ)) and current follows i(t) = (Vs/R) * e^(−t/τ). As t increases, Vc tends to Vs while current decays toward zero. Engineers typically declare ‘‘fully charged’’ at about 5τ, where Vc ≈ 0.993 * Vs and i ≈ 0.007 * (Vs/R).
Step-by-Step Solution:
At t = 0: Vc = 0, i = Vs/R (maximum inrush current).At t = τ: Vc ≈ 0.632 * Vs; still far from Vs; cannot be called fully charged.At t = 3τ: Vc ≈ 0.95 * Vs; current is small but non-zero; sometimes acceptable in rough contexts.At t ≈ 5τ: Vc ≈ 0.993 * Vs; current is negligible for most practical purposes.Hence, practical ‘‘fully charged’’ means Vc ≈ Vs and i ≈ 0.Verification / Alternative check:The exponential decay of current i(t) confirms the approach to zero. Measuring Vc with a voltmeter near Vs while the series current falls to near the meter's noise floor supports the 5τ rule of thumb.
Why Other Options Are Wrong:
Common Pitfalls:Confusing time constant with full charge, mixing AC filter corner concepts with DC charging, and assuming the capacitor ever reaches Vs exactly (it approaches it asymptotically).
Final Answer:When the capacitor voltage is essentially equal to the DC source voltage (about after 5τ) and the charging current is approximately zero.
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