RC charging curve: After one time constant (t = τ = R * C), what percentage of the initial charging current still flows through the series resistor in a first-order RC charging circuit?

Introduction:Charging a capacitor through a resistor is a classic first-order transient. The current starts at a maximum at t = 0 and decays exponentially as the capacitor voltage builds. Understanding the percentage remaining at specific multiples of τ (the time constant) is fundamental for timing and filter design.

Given Data / Assumptions:

• Ideal RC circuit with step input
• Time t = τ = R * C
• Initial current I0 = V / R at t = 0+
• Charging process (not discharging)

Concept / Approach:

For charging, current i(t) = I0 * e^(-t / (R * C)). At t = τ, i(τ) = I0 * e^(-1) ≈ 0.3679 * I0, which is 36.79% of the initial current, typically rounded to 36.8%.

Step-by-Step Solution:

Verification / Alternative check:

Complementary capacitor voltage at t = τ is 63.2% of final value. Since current is proportional to the rate of change of voltage, it is the remaining 36.8% of its initial value, matching the exponential relationship.

Why Other Options Are Wrong:

• 63.2%: This is the capacitor voltage fraction at t = τ, not current remaining.
• 13.5% and 5.0%: These correspond to approximately 2τ and 3τ current levels, not 1τ.
• 0%: Current never becomes exactly zero at finite time in an ideal RC.

Common Pitfalls:

• Confusing voltage rise (63.2%) with current decay (36.8%) at t = τ.
• Assuming current is zero at t = τ, which is incorrect.

Final Answer:

36.8%