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How to add capacitors in series?

Published in Circuitry Fundamentals 3 mins read

Adding capacitors in series involves connecting them end-to-end in a continuous line, which decreases the total capacitance of the circuit.

Understanding Series Connection

To connect capacitors in series, you arrange them one after another, forming a single path for the current. Imagine them lined up like beads on a string. In this configuration, the positive plate of one capacitor is connected to the negative plate of the next, and so on.

A key characteristic of capacitors wired in series is that the charge stored on each individual capacitor is the same throughout the entire series combination. However, the total voltage supplied by the source is distributed among the capacitors, meaning the individual voltages across all capacitors add up to the total voltage of the voltage source. This arrangement is often used to achieve a lower overall capacitance or to increase the total voltage rating that the combination can safely handle.

Calculating Equivalent Capacitance in Series

When capacitors are connected in series, their combined effect, known as the equivalent capacitance ($C_{eq}$), is less than the capacitance of any single capacitor in the series. This is because connecting them in series effectively increases the distance between the plates and decreases the effective plate area.

The formula for calculating the equivalent capacitance ($C_{eq}$) of capacitors in series is:

$$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n} $$

Where:

  • $C_{eq}$ is the equivalent capacitance of the series combination.
  • $C_1, C_2, C_3, ..., C_n$ are the capacitances of the individual capacitors.

For the special case of just two capacitors ($C_1$ and $C_2$) in series, the formula can be simplified to:

$$ C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2} $$

Important Note: The equivalent capacitance ($C_{eq}$) for capacitors in series will always be smaller than the smallest individual capacitance in the series. For a more detailed explanation of series and parallel capacitor arrangements, you can refer to resources like this guide on Capacitors in Series.

Practical Example

Let's say you have three capacitors with the following values:

  • $C_1 = 10 \mu F$ (microfarads)
  • $C_2 = 15 \mu F$
  • $C_3 = 30 \mu F$

To find the equivalent capacitance when they are connected in series:

  1. Apply the formula:
    $$ \frac{1}{C_{eq}} = \frac{1}{10 \mu F} + \frac{1}{15 \mu F} + \frac{1}{30 \mu F} $$

  2. Find a common denominator (which is 30 in this case):
    $$ \frac{1}{C_{eq}} = \frac{3}{30 \mu F} + \frac{2}{30 \mu F} + \frac{1}{30 \mu F} $$

  3. Add the fractions:
    $$ \frac{1}{C_{eq}} = \frac{3 + 2 + 1}{30 \mu F} = \frac{6}{30 \mu F} $$

  4. Solve for $C_{eq}$:
    $$ C_{eq} = \frac{30 \mu F}{6} = 5 \mu F $$

As you can see, the equivalent capacitance ($5 \mu F$) is indeed smaller than any of the individual capacitances ($10 \mu F$, $15 \mu F$, or $30 \mu F$).

Key Characteristics Summarized

  • Connection Method: Capacitors are connected end-to-end in a single line.
  • Charge: The charge ($Q$) stored on each capacitor is the same.
  • Voltage: The total voltage across the series combination is the sum of the individual voltages across each capacitor ($V_{total} = V_1 + V_2 + ... + V_n$).
  • Capacitance: The equivalent capacitance ($C_{eq}$) is always less than the smallest individual capacitance. This configuration reduces the overall capacitance.