Question E5B01:
What is the term for the time required for the capacitor in an RC circuit to be charged to 63.2% of the applied voltage or to discharge to 36.8% of its initial voltage?  

Choices:  
A: An exponential rate of one  
B: One time constant  
C: One exponential period  
D: A time factor of one    

Correct Answer is  B: One time constant  

To calculate RC timing behavior, remember the time constant:
$$\tau = RC$$

Why B (One time constant) is correct:
- In an RC circuit, the capacitor voltage during charging is:
  $$V_C(t) = V_{\text{final}}\left(1 - e^{-t/\tau}\right)$$
  At $t = \tau$, 
  $$V_C(\tau) = V_{\text{final}}\left(1 - e^{-1}\right) \approx 0.632\,V_{\text{final}}$$
- During discharge:
  $$V_C(t) = V_0\,e^{-t/\tau}$$
  At $t = \tau$, 
  $$V_C(\tau) = V_0\,e^{-1} \approx 0.368\,V_0$$
- Therefore, the time to reach 63.2% of the final voltage (or decay to 36.8%) is exactly one time constant.

Why the other choices are incorrect:
- A: “An exponential rate of one” — Not a defined term in RC timing; “rate of one” is meaningless here.
- C: “One exponential period” — Exponentials are not periodic; “period” applies to sinusoidal signals, not exponential growth/decay.
- D: “A time factor of one” — Not a standard term used in RC timing.

Useful formulas and rules of thumb:
- Time constant:
  $$\tau = RC \quad\text{(seconds)}$$
- Fraction of final value after $n$ time constants (charging): 
  $$1 - e^{-n}$$
- Fraction remaining after $n$ time constants (discharging): 
  $$e^{-n}$$
- Quick reference:
  - 1τ ≈ 63.2%
  - 2τ ≈ 86.5%
  - 3τ ≈ 95.0%
  - 4τ ≈ 98.2%
  - 5τ ≈ 99.3%

Note: For RL circuits, the time constant is $\tau = L/R$, with the same percentage behavior.