Question Detail
Question
A 3.0 mF and a 13.0 mF capacitor are connected in series, and the series arrangement is connected in parallel to a 25.0 mF capacitor. How much capacitance would a single capacitor need to replace the three capacitors?
Correct Answer
Option A is the correct answer.
Detailed Explanation
First, calculate the equivalent capacitance of the two capacitors connected in series (C1 = 3.0 mF, C2 = 13.0 mF):1/Cseries = 1/C1 + 1/C21/Cseries = 1/(3.0 mF) + 1/(13.0 mF) = (13 + 3) / (3 * 13) = 16 / 39 mF-1Cseries = 39 / 16 mF ≈ 2.4375 mFNext, this series combination is connected in parallel with a third capacitor (C3 = 25.0 mF). For parallel capacitors, the equivalent capacitance is the sum:Cequivalent = Cseries + C3Cequivalent = 2.4375 mF + 25.0 mF = 27.4375 mFRounding to the nearest whole number, the equivalent capacitance is 27 mF.
Hint
Remember the formulas for equivalent capacitance: 1/Ceq = 1/C1 + 1/C2 for series, and Ceq = C1 + C2 for parallel.