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

Question

Quizplus · Accepted Answer

The equivalent capacitance of the two capacitors in series is:$$C_{eq} = \frac{C_1C_2}{C_1+C_2} = \frac{(5.0\mu F)(12.0\mu F)}{5.0\mu F+12.0\mu F} = 3.75\mu F$$The equivalent capacitance of the three capacitors is then:$$C_{eq} = C_{eq} + C_3 = 3.75\mu F + 29.0\mu F = 32.75\mu F$$Therefore, a single capacitor with a capacitance of 32.75 μF would be needed to replace this combination of three capacitors.

A 50-μF and a 12 29.0−μF29.0 - \mu \mathrm { F }29.0−μF

A 50-μF and a 12 $29.0 - \mu \mathrm { F }$