Single 30 V, 6 ms pulse into an RC integrator (R = 3.3 kΩ, C = 2 µF): To what voltage does the capacitor charge by the end of the pulse?

Introduction / Context:
This computation illustrates exponential charging during a finite pulse. Designers frequently need to predict the capacitor's voltage after a given pulse width to set thresholds, timing margins, and subsequent stage behavior.

Given Data / Assumptions:

• R = 3.3 kΩ, C = 2 µF ⇒ tau = R * C.
• Input pulse amplitude Vin = 30 V.
• Pulse width t = 6 ms.
• Assume vC(0) = 0 V (uncharged) for a conservative estimate.

Concept / Approach:
Capacitor charging under a step input follows vC(t) = Vin * (1 − e^(−t/tau)). The pulse ends at t = 6 ms; the capacitor voltage at that instant is the charged value reached during the pulse.

Step-by-Step Solution:
tau = R * C = 3.3e3 * 2e−6 = 6.6e−3 s = 6.6 mst/tau = 6 ms / 6.6 ms ≈ 0.9091vC = 30 * (1 − e^(−0.9091))e^(−0.9091) ≈ 0.402vC ≈ 30 * (1 − 0.402) = 30 * 0.598 ≈ 17.94 V ≈ 17.91 V

Verification / Alternative check:
Since 6 ms is slightly less than one time constant (6.6 ms), a value a bit under 63.2% of 30 V (i.e., 18.96 V) is expected. Our result, ~17.9 V, is consistent because the pulse ends before one full tau.

Why Other Options Are Wrong:

• 30 V: Requires infinite time; not possible with a 6 ms pulse.
• 12.09 V or 10.3 V: Too low; they correspond to shorter effective charge times.

Common Pitfalls:
Forgetting to compute tau, or using linear instead of exponential charging. Always use vC(t) = Vin * (1 − e^(−t/tau)).

Final Answer:
17.91 V