For Capacitors Wired in Series, the Equivalent Capacitance Is Calculated $\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}$

Short Answer

For capacitors wired in series, the equivalent capacitance is calculated using $\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}$ . (See Eq. A72.) What is $C$ if $C_{1}=24$ microfarads $(\mu \mathrm{F})$ and $C_{2}=15 \mu \mathrm{F}$ ?