The problem describes an RL circuit with a voltage source, resistor, and inductor. We need to find (a) the time constant of the RL circuit and (b) the current in the circuit at $t = 2$ ms after the switch is closed. The given values are: $V = 12.0$ V, $R = 6.00 \Omega$, and $L = 30.0$ mH.

Let's focus on problem II first.

1. Problem Description

The problem describes an RL circuit with a voltage source, resistor, and inductor. We need to find (a) the time constant of the RL circuit and (b) the current in the circuit at $t = 2$ ms after the switch is closed. The given values are: $V = 12.0$ V, $R = 6.00 \Omega$, and $L = 30.0$ mH.

2. Solution Steps

a) The time constant $\tau$ of an RL circuit is given by:

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

Substituting the given values, we have:

$\tau = \frac{30.0 \times 10^{-3} \text{ H}}{6.00 \Omega} = 5.00 \times 10^{-3} \text{ s} = 5.00 \text{ ms}$

b) The current in an RL circuit as a function of time is given by:

$I(t) = \frac{V}{R}(1 - e^{-t/\tau})$

Substituting the given values, with $t = 2$ ms, we get:

$I(2 \text{ ms}) = \frac{12.0 \text{ V}}{6.00 \Omega}(1 - e^{-2 \times 10^{-3} \text{ s} / 5 \times 10^{-3} \text{ s}})$

$I(2 \text{ ms}) = 2.00 \text{ A}(1 - e^{-0.4})$

$I(2 \text{ ms}) = 2.00 \text{ A}(1 - 0.6703)$

$I(2 \text{ ms}) = 2.00 \text{ A}(0.3297)$

$I(2 \text{ ms}) = 0.6594 \text{ A}$

3. Final Answer

a) The time constant of the RL circuit is 5.00 ms.

b) The current in the circuit at t = 2 ms is approximately 0.659 A.