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Series and parallel capacitors

Combine two to four capacitors and compare equivalent capacitance, voltage, charge and stored energy for each one.

Workbench / Ideal modelIn your browser
01 / Adjust the inputs

Sliders with a wide range use a logarithmic scale. When zero is allowed, it has its own position. Type a number to set an exact value.

From 0 to 48 V. All capacitors start discharged.

From 1 pF to 100,000 µF.

From 1 pF to 100,000 µF.

Try an example

+VsC1C2One path

Labels C1–C4 identify each capacitor without relying on color. The diagram changes with the connection and number of components.

Equivalent capacitance6.66667 µF
Charge supplied80 µC
Stored energy480 µJ
  • C1: 10 µF · 8 V · 80 µC · 320 µJ
  • C2: 20 µF · 4 V · 80 µC · 160 µJ

In series, both capacitors carry the same charge magnitude.

Ceq = 1 / (1/10 + 1/20) = 6.66667 µF. Q = Ceq × 12 = 80 µC. U = ½ Ceq × 12² = 480 µJ.

Use it in your classroom

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The same pair, two behaviors

Connect 10 µF and 20 µF in series across 12 V. Their equivalent capacitance is 6.66667 µF, less than either capacitor. The source moves Q = 6.66667 × 12 = 80 µC. Each capacitor carries that charge magnitude: V1 = 80/10 = 8 V and V2 = 80/20 = 4 V.

C1 stores ½ × 10 × 8² = 320 µJ; C2 stores ½ × 20 × 4² = 160 µJ. Together they store 480 µJ. This does not mean the source delivered 160 µC: adding charges of series components counts the same charge displacement twice.

In the parallel model, both capacitors receive 12 V: Ceq = 30 µF, Q = 360 µC and U = 2160 µJ. Changing the selector recalculates a different configuration from its initial state; it does not simulate rewiring charged capacitors.

Series: add reciprocals; parallel: add capacitances

Series: 1/Ceq = 1/C1 + … + 1/Cn
Parallel: Ceq = C1 + … + Cn
Qsource = Ceq Vs
Series: Qi = Qsource; Vi = Qi/Ci
Parallel: Vi = Vs; Qi = Ci Vs
Ui = ½ Ci Vi²; Utotal = ½ Ceq Vs²

Convert every capacitance to farads, solve for Ceq and calculate Q. Then distribute voltage or charge according to the connection. The interface uses µF, µC and µJ for comparison; changing units preserves the physical quantity.

Initial conditions you cannot omit

Two to four positive capacitors from 1 pF to 100,000 µF, with a supply from 0 to 48 V. Capacitors start discharged and floating series nodes have zero net charge. The connection transient is not modeled: this calculates the final electrostatic state.

Leakage, internal resistance, tolerance, temperature, polarity and voltage ratings are omitted. In real series circuits, leakage can change voltage distribution over time. Do not use this tool to select a voltage rating by adding component ratings or to directly connect charged capacitors together.

The displayed energy is stored energy. It excludes losses during charging in wires, resistors or the supply.

Common mistakes

  • Adding series capacitances as if they were resistances.
  • Giving unequal series capacitors the same voltage: the smaller one gets more.
  • Confusing the magnitude of charge on one plate with the capacitor’s total net charge.
  • Assuming “ideal” means a real voltage rating may be exceeded.

Quick guide: keep one variable fixed

  1. Look at the initial 10 µF and 20 µF series pair and predict which capacitor has the higher voltage.
  2. Double only C1 to 20 µF. Voltages are now equal: 6 V each.
  3. Try four equal 10 µF capacitors: Ceq is 2.5 µF and each receives 3 V.
  4. Switch to parallel without changing capacitances: Ceq increases to 40 µF.

Try it yourself

Two 100 µF capacitors in series across 10 V: what are charge and energy?

Ceq = 50 µF, Q = 500 µC, 5 V across each. Total energy = 2500 µJ = 2.5 mJ.

What happens if the supply falls to zero?

In this initially discharged model, every voltage, charge and energy is zero. Ceq stays the same because it depends on the components and connection.