A 5.0-?F and a 12.0-?F capacitor are connected in series, and the series arrangement is connected in parallel to a $29.0 - \mu F$ capacitor. How much capacitance would a single capacitor need to replace this combination of three capacitors?
A)33 ?F
B)13 ?F
C)16 ?F
D)38 ?F

Question

Quizplus · Accepted Answer

The equivalent capacitance of the two capacitors in series is:$$C_{eq}=\frac{C_1C_2}{C_1+C_2}=\frac{(5.0\mu F)(12.0\mu F)}{5.0\mu F+12.0\mu F}=3.33\mu F$$The equivalent capacitance of the three capacitors is then:$$C_{eq}=C_{eq}+C_3=3.33\mu F+29.0\mu F=32.33\mu F$$Therefore, a single capacitor with a capacitance of 32.33 μF would be needed to replace this combination of three capacitors.

A 50-?F and a 12 29.0−μF29.0 - \mu F29.0−μF Capacitor How Much Capacitance Would a Single Capacitor Need to and the Series

A 50-?F and a 12 $29.0 - \mu F$ Capacitor How Much Capacitance Would a Single Capacitor Need to and the Series