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Enter the capacitance of each capacitor in series. Add more capacitors with the button below. The total capacitance is calculated from all filled fields.
When capacitors are connected in series, the total capacitance is always less than the smallest individual capacitor. Each capacitor adds an additional gap for the electric field to cross, which reduces the overall ability to store charge. The formula for \( n \) capacitors in series is: \[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n} \] Or equivalently: \[ C_{\text{total}} = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots + \dfrac{1}{C_n}} \]
This is the same form as resistors in parallel, which is a useful analogy if you already have that formula memorized. The reason for the reciprocal relationship is that in series, all capacitors carry the same charge \( Q \), so the total voltage is the sum of the individual voltages: \[ V_{\text{total}} = V_1 + V_2 + \cdots + V_n = \frac{Q}{C_1} + \frac{Q}{C_2} + \cdots + \frac{Q}{C_n} \] Dividing both sides by \( Q \) gives the series formula directly.
For exactly two capacitors in series the formula simplifies to the product-over-sum form, which is often faster to compute by hand: \[ C_{\text{total}} = \frac{C_1 \cdot C_2}{C_1 + C_2} \]
When all \( n \) capacitors have the same value \( C \), the total simplifies further: \[ C_{\text{total}} = \frac{C}{n} \] For example, three 30 µF capacitors in series give a total of 10 µF.
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