[amp_mcq option1=”0.151 mA” option2=”157 raA” option3=”0.995 mA” option4=”2.02 A E. None of the above” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{A. }0.151\\text{ mA}}$.<\/p>\n
The current through a capacitor is given by the following equation:<\/p>\n
$$I = \\dfrac{V}{X_C}$$<\/p>\n
where $V$ is the voltage across the capacitor, $X_C$ is the capacitive reactance, and $I$ is the current through the capacitor.<\/p>\n
The capacitive reactance is given by the following equation:<\/p>\n
$$X_C = \\dfrac{1}{2\\pi fC}$$<\/p>\n
where $f$ is the frequency of the AC source, and $C$ is the capacitance of the capacitor.<\/p>\n
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:<\/p>\n
$$X_C = \\dfrac{1}{2\\pi (100\\text{ Hz})(0.02\\mu\\text{F})} = 159\\text{ ohms}$$<\/p>\n
Substituting this value into the equation for current, we get:<\/p>\n
$$I = \\dfrac{12\\text{ V}}{159\\text{ ohms}} = 0.151\\text{ mA}$$<\/p>\n
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.<\/p>\n
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.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0.151 mA” option2=”157 raA” option3=”0.995 mA” option4=”2.02 A E. None of the above” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[682],"tags":[],"class_list":["post-19087","post","type-post","status-publish","format-standard","hentry","category-electronic-principles","no-featured-image-padding"],"yoast_head":"\n