The correct answer is $\boxed{\text{B) 0.955 A}}$.
The current flowing through a choke is given by the following formula:
$$I = \frac{V}{Z}$$
where $V$ is the voltage, $Z$ is the impedance, and $I$ is the current.
The impedance of a choke is given by the following formula:
$$Z = \omega L$$
where $\omega$ is the angular frequency and $L$ is the inductance.
The angular frequency is given by the following formula:
$$\omega = 2\pi f$$
where $f$ is the frequency.
In this case, we have:
- $V = 12\text{ V}$
- $f = 100\text{ Hz}$
- $L = 0.02\text{ H}$
Therefore, the impedance is:
$$Z = \omega L = 2\pi (100\text{ Hz}) (0.02\text{ H}) = 4\pi\text{ Ω}$$
The current is:
$$I = \frac{V}{Z} = \frac{12\text{ V}}{4\pi\text{ Ω}} = 0.955\text{ A}$$
Option A is incorrect because it is the inductance, not the current.
Option C is incorrect because it is a time of day, not an electrical value.
Option D is incorrect because it is twice the correct value.