In RC charging theory for first-order circuits, engineers use the time constant τ = R * C to estimate charging progress. After how many time constants can a capacitor be considered 'practically' fully charged (about 99% of the supply voltage) in typical design practice?

Introduction:
Time constant analysis provides a quick way to estimate how fast a capacitor charges in a first-order RC circuit. This question targets the rule of thumb engineers use to declare a capacitor essentially fully charged without waiting for true mathematical completion.

Given Data / Assumptions:

• Series RC charging from a DC source.
• Time constant τ defined as R * C.
• Fully charged in practice means near the final value.

Concept / Approach:
Capacitor voltage during charging is Vc(t) = Vs * (1 - exp(-t / τ)). Exact final value is approached asymptotically, but design convention treats about 99% as practically full.

Step-by-Step Solution:
1) At t = 1τ, Vc ≈ 0.632 * Vs.2) At t = 3τ, Vc ≈ 0.950 * Vs.3) At t = 5τ, Vc ≈ 0.993 * Vs, commonly accepted as practically full charge.4) Beyond 5τ the change becomes negligible for most applications.

Verification / Alternative check:
Compute exp(-5) ≈ 0.0067, so Vc(5τ) ≈ (1 - 0.0067) * Vs ≈ 0.993 * Vs, which is near 99.3% of final value.

Why Other Options Are Wrong:
First: only ~63% charged.Third: about 95%, not typically called fully charged.Seventh: unnecessary waiting; 5τ already meets practical criteria.

Common Pitfalls:
Assuming 100% occurs at a finite time; in first-order systems, the exponential never truly reaches the final value, so engineers use 5τ as a practical benchmark.

Final Answer:
fifth