5.454545
µF
5,454.545
nF
65.454545
µC
392.727273
µJ
6.545455
V
3.272727
V
2.181818
V
5.454545
µF
5,454.545
nF
65.454545
µC
392.727273
µJ
6.545455
V
3.272727
V
2.181818
V
The Capacitors in Series Calculator determines the total equivalent capacitance when two or three capacitors are connected end-to-end in a series configuration. In a series circuit, each capacitor stores the same charge but may have different voltage drops across it, and the total capacitance is always less than the smallest individual capacitor.
Capacitors are fundamental passive components in electrical and electronic circuits. They store energy in an electric field between two conductive plates separated by a dielectric material. When engineers connect capacitors in series, they do so to achieve a lower total capacitance, to increase the overall voltage rating of the combination, or to create specific frequency-response characteristics in filter circuits.
The governing equation for capacitors in series uses the reciprocal sum formula: the reciprocal of the total capacitance equals the sum of the reciprocals of each individual capacitance. Mathematically, this is expressed as:
$$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$
This relationship arises because in a series configuration, the same charge \(Q\) appears on each capacitor, but the total voltage is the sum of the individual voltage drops: \(V = V_1 + V_2 + V_3\). Since \(V_i = Q / C_i\) for each capacitor, dividing through by \(Q\) yields the reciprocal formula.
An important consequence is that the total series capacitance is always smaller than any single capacitor in the chain. This is the opposite behavior from resistors in series, where the total resistance increases. For two identical capacitors of value \(C\) in series, the total is simply \(C/2\). For three identical capacitors, it is \(C/3\).
Series capacitor combinations are widely used in power electronics, radio-frequency (RF) tuning circuits, voltage dividers, and safety-rated capacitor configurations. In high-voltage applications, connecting capacitors in series allows the voltage to be shared among multiple components, preventing any single capacitor from exceeding its rated voltage. This calculator also computes the total stored charge and energy when an external voltage is applied, giving a complete picture of the series combination's electrical behavior.
Whether you are designing a filter network, balancing voltage ratings in a power supply, or studying circuit theory for an exam, this tool provides instant, accurate results for series capacitor calculations.
The calculator applies the reciprocal-sum rule for series-connected capacitors:
Two capacitors:
$$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} \quad \Rightarrow \quad C_t = \frac{C_1 \cdot C_2}{C_1 + C_2}$$
Three capacitors:
$$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$
If only two capacitors are needed, set \(C_3 = 0\) and the calculator ignores the third term.
Stored charge on each capacitor (identical in series):
$$Q = C_t \cdot V$$
where \(V\) is the total applied voltage and \(C_t\) is in µF, giving \(Q\) in µC.
Total energy stored:
$$E = \frac{1}{2} C_t V^2$$
Here \(C_t\) must be converted to farads (multiply by \(10^{-6}\)) to obtain energy in joules.
The total capacitance result is always less than the smallest individual capacitor in the series chain. The stored charge value represents the charge on each capacitor plate (identical for all series capacitors). The energy output shows the total electrostatic energy stored in the combination. For context, a typical ceramic capacitor is 0.1–100 µF, an electrolytic capacitor ranges from 1 µF to several thousand µF, and supercapacitors can reach farads.
Inputs
Results
Three capacitors of 10, 20, and 30 µF in series yield a total capacitance of about 5.45 µF. With 12 V applied, each capacitor stores 65.45 µC of charge.
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Results
Two 100 µF capacitors in series produce 50 µF total—exactly half of one capacitor. The voltage rating doubles, making this useful in high-voltage applications.
In a series connection, the same charge must pass through each capacitor, but the voltage divides among them. Since \(C = Q/V\) and the total voltage is the sum of individual voltages, the effective capacitance decreases. Physically, adding more capacitors in series increases the effective plate separation, which reduces capacitance.
The total capacitance is dominated by the smallest capacitor. For example, if \(C_1 = 1\) µF and \(C_2 = 1000\) µF in series, the total is approximately 0.999 µF—nearly equal to the smaller capacitor. The larger capacitor contributes almost no additional impedance.
The primary reasons are: (1) to increase the voltage rating of the combination—each capacitor shares a portion of the total voltage; (2) to reduce capacitance to a precise value not available from a single component; (3) in safety-critical applications where redundancy ensures the circuit remains safe even if one capacitor fails short.
Yes. In a series circuit, the same current flows through every component, so the same charge accumulates on each capacitor. However, the voltage across each capacitor differs depending on its capacitance: \(V_i = Q / C_i\). Smaller capacitors have larger voltage drops.
This calculator supports up to three capacitors. For more, apply the same reciprocal formula iteratively: compute the series combination of the first group, then combine that result in series with additional capacitors using the same formula.
For resistors in series, the total resistance is the simple sum: \(R_t = R_1 + R_2 + R_3\). For capacitors, it is the reciprocal sum. This is because resistors impede current (which is the same in series), while capacitors store charge relative to voltage (which divides in series). Interestingly, capacitors in parallel add directly, just as resistors in series do.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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