The problem involves an RL circuit where the inductance $L = 584 \, \text{mH} = 0.584 \, \text{H}$, the resistance $R = 21 \, \Omega$, and the applied voltage $V = 120 \, \text{V}$. We are asked to calculate the current in the circuit after 1 time constant.

Formula for Time Constant ($\tau$):

The time constant for an RL circuit is given by:

$\tau = \frac{L}{R}$

Substituting the given values:

$\tau = \frac{0.584}{21} = 0.0278 \, \text{seconds}$

Formula for Current at Time $t$:

The current in an RL circuit after a time $t$ is given by:

$I(t) = I_{\text{max}} \left( 1 - e^{-\frac{t}{\tau}} \right)$

Where $I_{\text{max}}$ is the maximum current (when the circuit has reached steady state), and $e$ is the base of the natural logarithm.

The maximum current $I_{\text{max}}$ is:

$I_{\text{max}} = \frac{V}{R} = \frac{120}{21} = 5.714 \, \text{A}$

At $t = \tau$, which is 1 time constant, the current $I(\tau)$ is:

$I(\tau) = 5.714 \left( 1 - e^{-1} \right)$

We know that $e^{-1} \approx 0.3679$, so:

$I(\tau) = 5.714 \left( 1 - 0.3679 \right)$ $I(\tau) = 5.714 \times 0.6321 = 3.615 \, \text{A}$

Thus, the current after 1 time constant is approximately 3.615 A.

---

Would you like further clarification or have any questions? Here are some related questions to expand on this topic:

1. How does the current behave at $t = 5\tau$?
2. What happens to the time constant if the resistance is doubled?
3. How does the maximum current change if the applied voltage is increased to 240V?
4. What would the current be if the inductor were replaced with a 1 H inductor?
5. How can we calculate the power dissipated in this circuit after 1 time constant?

Tip: The current in an RL circuit asymptotically approaches its maximum value as time progresses, and the time constant determines how quickly this approach happens.