A $5.0 - \mu \mathrm { F }$, a $14 - \mu \mathrm { F }$, and a $21 - \mu \mathrm { F }$ capacitor are connected in series. How much capacitance would a single capacitor need to have to replace the three capacitors?
A) $3.6 \mu \mathrm { F }$
B) $3.1 \mu \mathrm { F }$
C) $40 \mu \mathrm { F }$
D) $2.0 \mu \mathrm { F }$

Question

Quizplus · Accepted Answer

The equivalent capacitance $C_{\text{eq}}$ for capacitors in series is given by $ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} $. Calculating this gives $C_{\text{eq}} = 3.1 \mu \mathrm{F}$.

A 5.0−μF5.0 - \mu \mathrm { F }5.0−μF , A 14−μF14 - \mu \mathrm { F }14−μF

A $5.0 - \mu \mathrm { F }$ , A $14 - \mu \mathrm { F }$