Physics & Mechanics

Capacitors in Series Calculator

Calculate the total equivalent capacitance of capacitors connected in series. Uses the reciprocal addition formula for accurate electrical engineering results.

F
F
F
F
F
Equivalent Capacitance (C_eq)
6.6667 × 10⁻⁵

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Capacitors in Series

Connecting capacitors in series is less common than in parallel, but it is a vital technique for specific engineering needs. In a series configuration, the total equivalent capacitance ($C_{eq}$) is always lower than the smallest individual capacitor in the string.

Why use Series Capacitance?

  1. Increased Voltage Rating: This is the primary reason to use series capacitors. While the total capacitance drops, the total voltage the string can handle is the sum of the individual voltage ratings. Two $100 , \text{V}$ capacitors in series can safely handle $200 , \text{V}$.
  2. Specific Tuning: Used in radio frequency (RF) circuits to achieve very precise, small capacitance values that aren't available as standard components.

The Formula

1Ceq=1C1+1C2+1C3+...\begin{aligned} \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... \end{aligned}

Where:
CeqC_{eq}=
Equivalent Capacitance
C1,C2,...C_1, C_2, ...=
Individual Capacitances

Example Calculation

You connect a $10 , \mu\text{F}$ and a $20 , \mu\text{F}$ capacitor in series.

  1. Calculate Reciprocals: $1/10 = 0.1$ and $1/20 = 0.05$.
  2. Sum them: $0.1 + 0.05 = 0.15$.
  3. Take the Reciprocal: $1 / 0.15 = 6.67 , \mu\text{F}$.

Frequently Asked Questions

Placing capacitors in series is like increasing the distance between the outermost plates. Since capacitance is inversely proportional to the distance between plates, the total ability to store charge decreases.

In a series circuit, every capacitor stores the exact same amount of charge ($Q$), regardless of its capacitance value. The total voltage is split among them, with the smallest capacitor taking the largest share of the voltage.

Technically yes, but it is risky. Differences in 'leakage resistance' can cause the voltage to distribute unevenly, potentially causing one capacitor to exceed its voltage rating and fail.