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 $C_{2}=15 \mu \mathrm{F}$ ?

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The Answer of 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 $C_{2}=15 \mu \mathrm{F}$ ?

For Capacitors Wired in Series, the Equivalent Capacitance Is Calculated 1C=1C1+1C2\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}C1​=C1​1​+C2​1​

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