RC integrator charging: A 12 V input pulse with a width equal to one time constant is applied to an RC integrator. What voltage does the capacitor reach at the end of the pulse?

Introduction / Context:This question targets the exponential charging law of capacitors in first-order RC networks, a foundational concept in electronics used for timing, filtering, and pulse shaping. It specifically examines the capacitor voltage after exactly one time constant of charging.

Given Data / Assumptions:

• Input pulse amplitude, Vin = 12 V.
• Pulse width t = tau (one time constant).
• Ideal RC network, initially vC(0) = 0 V (uncharged).

Concept / Approach:For a step input of amplitude Vin applied to an RC, the capacitor charges as vC(t) = Vin * (1 - e^(−t/tau)). At t = tau, the result is vC = Vin * (1 − e^(−1)) ≈ Vin * 0.6321.

Step-by-Step Solution:vC(tau) = 12 * (1 − e^(−1))e^(−1) ≈ 0.3679 ⇒ 1 − e^(−1) ≈ 0.6321vC ≈ 12 * 0.6321 ≈ 7.585 V ≈ 7.56 V (rounded to match option)

Verification / Alternative check:Rule of thumb: one tau corresponds to ~63.2% of the final value. 0.632 * 12 ≈ 7.58 V, consistent with our calculation.

Why Other Options Are Wrong:

• 12 V: Would require infinite time (full charge), not one tau.
• 6.3 V: A rough 0.525 * 12 V number would be too low; the correct 63.2% gives ~7.58 V.
• 0 V: Only at t = 0; not after charging for tau.

Common Pitfalls:Using linear intuition instead of the exponential model, or misremembering 63.2% as 50%. Always apply the exact exponential term for accuracy.

Final Answer:7.56 V