This tool calculates the capacitance of capacitors in series.

### Total Capacitance =

### Understanding Capacitors in Series

When capacitors are connected in series, the effective capacitance decreases. This happens because the charge \( Q \) across each capacitor remains the same, but the voltage across each adds up, resulting in a lower overall capacitance compared to any single capacitor in the series.

#### How to Calculate

The formula to find the total capacitance \( C_{\text{total}} \) of capacitors in series is given by the reciprocal of the sum of the reciprocals of each individual capacitor's capacitance:

##### \( \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_n} \)

where:

\( C_1, C_2, C_3, \dots, C_n \) are the capacitances of each individual capacitor in the series.

#### Example

Suppose you have three capacitors in series with capacitances of 4 µF, 6 µF, and 12 µF. To find the total capacitance, use the formula:

##### \( \frac{1}{C_{\text{total}}} = \frac{1}{4\, \mu F} + \frac{1}{6\, \mu F} + \frac{1}{12\, \mu F} \)

Calculating this, we find:

##### \( \frac{1}{C_{\text{total}}} = 0.25\, \mu F^{-1} + 0.1667\, \mu F^{-1} + 0.0833\, \mu F^{-1} = 0.5\, \mu F^{-1} \)

So, the total capacitance \( C_{\text{total}} \) is:

##### \( C_{\text{total}} = \frac{1}{0.5\, \mu F^{-1}} = 2\, \mu F \)

#### Visual Representation

Visualizing this in a circuit:

Imagine each capacitor connected end-to-end (like a daisy chain). The first plate of the first capacitor connects to the second plate of the second capacitor, and so on, with the outer plates connecting to the circuit.

#### Key Points

**Total Capacitance Decreases:**In series, the total capacitance is always less than the smallest individual capacitor's capacitance in the group.**Voltage Adds Up:**The voltage across each capacitor when in series adds up, which is why the overall capacitance decreases.