[amp_mcq option1=”5 uF” option2=”30 uF” option3=”45 uF” option4=”50 uF” correct=”option2″]<\/p>\n
The correct answer is $\\boxed{\\text{A. }5 \\mu F}$.<\/p>\n
When capacitors are connected in series, the equivalent capacitance is less than any of the individual capacitances. This is because the charge on each capacitor must be the same, but the voltage across each capacitor is different. The formula for the equivalent capacitance of capacitors in series is:<\/p>\n
$$C_e = \\frac{1}{C_1 + C_2 + C_3 + …}$$<\/p>\n
In this case, we have three capacitors with capacitance $C_1 = C_2 = C_3 = 15 \\mu F$. Substituting these values into the formula, we get:<\/p>\n
$$C_e = \\frac{1}{15 \\mu F + 15 \\mu F + 15 \\mu F} = \\frac{1}{45 \\mu F} = 5 \\mu F$$<\/p>\n
Therefore, the net capacitance of three 15 uF capacitors connected in series is 5 uF.<\/p>\n
Option B is incorrect because it is the sum of the individual capacitances. Option C is incorrect because it is the product of the individual capacitances. Option D is incorrect because it is the square of the individual capacitances.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”5 uF” option2=”30 uF” option3=”45 uF” option4=”50 uF” correct=”option2″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[971],"tags":[],"class_list":["post-44273","post","type-post","status-publish","format-standard","hentry","category-electrostatics","no-featured-image-padding"],"yoast_head":"\n