Combining capacitors and stored energy
Series and parallel
We often join several capacitors together. The two simplest ways are in series, one after another, and in parallel, side by side.
In series, C₁ and C₂ are joined end to end across a battery. The battery puts +Q on the first plate. That pulls −Q onto the facing plate, which leaves +Q on the next one. Every capacitor carries the same charge Q.
The voltages add: V = V₁ + V₂ = Q/C₁ + Q/C₂. Divide by Q: the combination acts like one capacitor with 1/C = 1/C₁ + 1/C₂.
So in series, the total is less than the smallest capacitor. For 2 μF and 3 μF in series, 1/C = 1/2 + 1/3 = 5/6, so C is only 1.2 μF.
In parallel, both capacitors are connected across the same battery, so each has the same voltage V. Their charges are Q₁ = C₁V and Q₂ = C₂V.
The total charge is Q = Q₁ + Q₂ = (C₁ + C₂)V. So in parallel, C = C₁ + C₂: capacitances simply add. 2 μF and 3 μF in parallel make 5 μF.
Worked example: four 10 μF capacitors on a 500 V supply. C₁, C₂ and C₃ are in series, and C₄ is connected in parallel with that chain.
The chain gives 1/C = 3/10, so C = 10/3 μF, about 3.3 μF. Adding C₄ in parallel gives 13.3 μF in all.
C₄ has the full 500 V across it, so its charge is 10 μF × 500 V = 5 × 10⁻³ C. The chain carries (10/3 μF) × 500 V, about 1.7 × 10⁻³ C, on each of C₁, C₂ and C₃, and each has about 167 V across it.
Energy stored in a capacitor
Charging a capacitor takes work. Imagine moving charge from one plate to the other, a tiny amount δQ at a time.
When the plates already hold Q′, the potential difference is Q′/C. Moving δQ across takes work (Q′/C) δQ. It is cheap at first and gets harder as the charge builds up.
On a graph of V against Q this is a straight line. The total work is the area under it: a triangle, half of Q times V.
So the energy stored is U = Q²/2C = ½CV² = ½QV. All three say the same thing.
Where is this energy? In the field between the plates. For a parallel plate capacitor, put C = ε₀A/d and V = Ed into ½CV². It gives ½ε₀E² × Ad.
Ad is the volume between the plates. So the energy per unit volume, the energy density, is u = ½ε₀E². This is true for any electric field.
Example: our 44 pF capacitor at 100 V stores ½ × 4.4 × 10⁻¹¹ × 100², about 2.2 × 10⁻⁷ J. Its field is 5 × 10⁴ V/m, so u is about 0.011 J/m³, spread through 2 × 10⁻⁵ m³.
Now charge a 4 μF capacitor to 200 V. It holds 800 μC and stores ½ × 4 × 10⁻⁶ × 200², which is 0.08 J. Disconnect it, and join it across an uncharged 2 μF capacitor.
Charge flows until both are at the same voltage. The 800 μC is shared over 6 μF, so V = 133 V. The energy is now ½ × 6 × 10⁻⁶ × 133², about 0.053 J.
About 0.027 J has gone, as heat in the wires and as electromagnetic radiation. When capacitors share charge, the charge is conserved, but the stored energy is not.
Try it: capacitor networks
- This is the worked example: C₁, C₂, C₃ in series, with C₄ across them, on 500 V. C_eq = 13.3 μF.
- Change C₄: only its own charge changes. Change C₁: the charge on the whole chain changes.
- Pick a series and a parallel combination from the list and check 1/C = Σ1/Cᵢ and C = ΣCᵢ.
- The energies ½CV² of the separate capacitors add up to ½C_eqV².
Try it: sharing charge
- C₁ = 4 μF is charged to 200 V. Predict the common voltage, then press “Flip switch”.
- Compare the total charge before and after: it is the same. Now compare the energy.
- Set C₂ equal to C₁: exactly half the energy is lost, whatever the voltage.
Three in series
Capacitors of 2 μF, 3 μF and 6 μF are joined in series across a 12 V battery. What are the equivalent capacitance and the charge on each capacitor?
- 1 μF and 12 μC
- 11 μF and 132 μC
- 1 μF and 4 μC
- 11 μF and 12 μC
Answer
A. 1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1, so C = 1 μF. In series, every capacitor carries the same charge, Q = CV = 1 μF × 12 V = 12 μC. Their voltages are 6 V, 4 V and 2 V, adding up to 12 V. (11 μF is the parallel formula; 4 μC splits Q three ways.)
Where did the energy go?
A 4 μF capacitor charged to 200 V is disconnected and joined to an uncharged 2 μF capacitor. How much energy is lost?
- None, because charge is conserved
- About 0.027 J
- About 0.053 J
- 0.08 J, all of it
Answer
B. Before: U = ½ × 4 × 10⁻⁶ × 200² = 0.08 J, Q = 800 μC. After: V = 800 μC / 6 μF = 133 V, U = ½ × 6 × 10⁻⁶ × 133² ≈ 0.053 J. Lost: 0.08 − 0.053 ≈ 0.027 J, as heat and electromagnetic radiation. Charge is conserved; energy stored in the capacitors is not.
Summary
- Series:
1/C = 1/C₁ + 1/C₂ + … (same Q) - Parallel:
C = C₁ + C₂ + … (same V) - Energy stored:
U = Q²/2C = ½CV² = ½QV - Energy density:
u = ½ε₀E²
- In series every capacitor has the same charge and the voltages add; C is less than the smallest one.
- In parallel every capacitor has the same voltage and the charges add.
- The energy of a charged capacitor is stored in its electric field, ½ε₀E² per unit volume.
- When a charged capacitor shares charge with an uncharged one, charge is conserved but some energy is lost as heat and radiation.