[amp_mcq option1=”0.151 mA” option2=”157 raA” option3=”0.995 mA” option4=”2.02 A E. None of the above” correct=”option3″]
The correct answer is $\boxed{\text{A. }0.151\text{ mA}}$.
The current through a capacitor is given by the following equation:
$$I = \dfrac{V}{X_C}$$
where $V$ is the voltage across the capacitor, $X_C$ is the capacitive reactance, and $I$ is the current through the capacitor.
The capacitive reactance is given by the following equation:
$$X_C = \dfrac{1}{2\pi fC}$$
where $f$ is the frequency of the AC source, and $C$ is the capacitance of the capacitor.
In this case, we are given that $V = 12\text{ V}$, $f = 100\text{ Hz}$, and $C = 0.02\mu\text{F}$. Substituting these values into the equation for capacitive reactance, we get:
$$X_C = \dfrac{1}{2\pi (100\text{ Hz})(0.02\mu\text{F})} = 159\text{ ohms}$$
Substituting this value into the equation for current, we get:
$$I = \dfrac{12\text{ V}}{159\text{ ohms}} = 0.151\text{ mA}$$
Therefore, the current that flows through a 0.02 micro F capacitor that is operating from a 12-V ac, 100-Hz source is 0.151 mA.
Option B is incorrect because it is too small. Option C is incorrect because it is too large. Option D is incorrect because it is much larger than the actual current. Option E is incorrect because it is not one of the possible answers.