RC integrator design: A 15 V input pulse having a width equal to two time constants is applied to an RC integrator. To what voltage will the capacitor charge by the end of the pulse?

Introduction / Context:This problem checks time-domain charging behavior of a first-order RC integrator when it is driven by a finite-width rectangular pulse. Knowing how far a capacitor charges during a pulse is essential for pulse shaping, timing, and analog signal conditioning design.

Given Data / Assumptions:

• Input pulse amplitude, Vin = 15 V.
• Pulse width equals two time constants, t = 2 * tau.
• Ideal RC integrator, no leakage or series parasitics beyond R and C.
• Capacitor initially uncharged just before the pulse edge (worst-case rise computation).

Concept / Approach:For a step of amplitude Vin applied across an RC, the capacitor voltage follows vC(t) = Vin * (1 - e^(−t/tau)). For a finite-width pulse of duration t, the capacitor reaches vC(t) at the end of the pulse, provided the RC remains in its charging phase throughout that interval.

Step-by-Step Solution:Given t = 2 * tauvC(t) = Vin * (1 - e^(−t/tau))vC(2 * tau) = 15 * (1 - e^(−2))e^(−2) ≈ 0.1353 ⇒ 1 − e^(−2) ≈ 0.8647vC ≈ 15 * 0.8647 ≈ 12.97 V ≈ 12.9 V

Verification / Alternative check:As a rule of thumb, after 1 tau a capacitor reaches ~63.2% of its final value; after 2 tau it reaches ~86.5%. For 15 V, 0.865 * 15 ≈ 12.97 V, confirming the calculation.

Why Other Options Are Wrong:

• 19.45 V and 15 V: Both exceed or equal the source; a passive RC cannot charge above Vin in this simple case.
• 8.6 V: Would correspond to approximately 1 tau, not 2 tau.

Common Pitfalls:Confusing one time constant with two, or forgetting to apply the exponential factor e^(−t/tau). Always convert the pulse width to multiples of tau before applying the formula.

Final Answer:12.9 V