The problem describes an RL circuit with a voltage source, a resistor, and an inductor. The given values are: Voltage $V = 12.0 \, \text{V}$ Resistance $R = 6.00 \, \Omega$ Inductance $L = 30.0 \, \text{mH} = 30.0 \times 10^{-3} \, \text{H}$ We need to find: a. The time constant of the RL circuit. b. The current in the circuit at $t = 2 \, \text{ms}$.

I will solve question II.

1. Problem Description

The problem describes an RL circuit with a voltage source, a resistor, and an inductor.

The given values are:

Voltage $V = 12.0 \, \text{V}$

Resistance $R = 6.00 \, \Omega$

Inductance $L = 30.0 \, \text{mH} = 30.0 \times 10^{-3} \, \text{H}$

We need to find:

a. The time constant of the RL circuit.

b. The current in the circuit at $t = 2 \, \text{ms}$.

2. Solution Steps

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

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

Substituting the given values:

$\tau = \frac{30.0 \times 10^{-3} \, \text{H}}{6.00 \, \Omega} = 5.0 \times 10^{-3} \, \text{s} = 5.0 \, \text{ms}$

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:

$I(t) = \frac{12.0 \, \text{V}}{6.00 \, \Omega} (1 - e^{-t/5.0 \times 10^{-3}})$

$I(t) = 2.0 \, \text{A} (1 - e^{-t/5.0 \times 10^{-3}})$

We need to find the current at $t = 2 \, \text{ms} = 2.0 \times 10^{-3} \, \text{s}$:

$I(2 \times 10^{-3}) = 2.0 (1 - e^{-2.0 \times 10^{-3}/5.0 \times 10^{-3}})$

$I(2 \times 10^{-3}) = 2.0 (1 - e^{-0.4})$

$I(2 \times 10^{-3}) = 2.0 (1 - 0.67032)$

$I(2 \times 10^{-3}) = 2.0 (0.32968)$

$I(2 \times 10^{-3}) = 0.65936 \, \text{A}$

$I(2 \times 10^{-3}) \approx 0.66 \, \text{A}$

3. Final Answer

a. The time constant of the RL circuit is $5.0 \, \text{ms}$.

b. The current in the circuit at $t = 2 \, \text{ms}$ is approximately $0.66 \, \text{A}$.