Step 1: Understanding the Question:
This question asks for the relationship between the time constant ($\tau$) of a first-order system and the time it takes for its step response to reach $63.2\%$ of its ultimate steady-state value.
Step 2: Key Formula or Approach:
The transfer function of a first-order system with gain $K_p$ and time constant $\tau$ is:
\[ G(s) = \frac{Y(s)}{X(s)} = \frac{K_p}{\tau s + 1} \]
For a unit step input, $X(s) = 1/s$. The response in the Laplace domain is:
\[ Y(s) = \frac{K_p}{s(\tau s + 1)} \]
Taking the inverse Laplace transform gives the time-domain response:
\[ y(t) = K_p \left(1 - e^{-t/\tau}\right) \]
Step 3: Detailed Explanation:
The final (steady-state) value of the response as $t \to \infty$ is:
\[ y(\infty) = \lim_{t \to \infty} K_p \left(1 - e^{-t/\tau}\right) = K_p \]
Let us evaluate the response at time $t = \tau$ (one time constant):
\[ y(\tau) = K_p \left(1 - e^{-\tau/\tau}\right) = K_p \left(1 - e^{-1}\right) \]
The value of the mathematical constant $e$ is approximately $2.71828$.
Calculating $e^{-1}$:
\[ e^{-1} \approx \frac{1}{2.71828} \approx 0.3678 \]
Substituting this back into the equation:
\[ y(\tau) = K_p (1 - 0.3678) = 0.6322 K_p \]
Thus, at $t = \tau$, the response reaches exactly $63.2\%$ of its ultimate value $K_p$.
This is a fundamental definition of the time constant $\tau$ for any first-order system, representing the speed of the system's response.
Step 4: Final Answer
Therefore, the system reaches $63.2\%$ of its final value at $t = \tau$ (one time constant), corresponding to option (B).