An RC integrator is driven by a 24 V input pulse whose width equals five time constants (t = 5RC). To what voltage does the capacitor charge by the end of the pulse?

Introduction / Context:First-order charging behavior dictates how close an RC node gets to its final value within a given number of time constants. This question uses the canonical result at five time constants for a DC step/pulse of sufficient width.

Given Data / Assumptions:

• RC integrator receiving a 24 V pulse.
• Pulse width equals 5 * tau where tau = R * C.
• Initial capacitor voltage assumed 0 V (uncharged) at the pulse start.

Concept / Approach:For a step of amplitude V applied to an RC from rest, the capacitor voltage during charging is v_C(t) = V * (1 - exp(-t / tau)). At t = 5 * tau, the exponential term exp(-5) ≈ 0.006737, so the node is within about 0.67% of its final value V.

Step-by-Step Solution:

Verification / Alternative check:A common engineering rule: after 5 tau, a first-order system reaches ~99.3% of its final value. For 24 V, that is ≈ 23.84 V, which rounds to 24 V in typical multiple-choice contexts.

Why Other Options Are Wrong:

• 15.12 V: Corresponds to ~63% of 24 V (one time constant), not five.
• 20.64 V: ~86% (two time constants), still far from 99.3%.
• 12 V: Exactly half the target; unrelated to standard time-constant milestones.

Common Pitfalls:

• Using 1 - 1/5 instead of the exponential relation.
• Confusing 5 tau with 5% error rather than 0.67% error.

Final Answer:24 V