Joining and recharging capacitors

What happens at the switch?

A charged capacitor is joined by wires to a second one. A spark may jump at the switch. Charge flows through the wires, and then it stops. What are the charges and the voltages at the end? Is any energy lost?

Two rules answer all of this.

1. The charge is conserved. No charge leaves the two capacitors. It only moves from one to the other.
2. The voltages become equal. The two capacitors are now joined in parallel. Charge flows until they have the same voltage.

Plates of the same sign joined

The top plate of each capacitor carries positive charge. Join the top plates together, and the bottom plates together. The total charge on the top plates is $Q_1 + Q_2 = C_1V_1 + C_2V_2$. After joining, the capacitors are in parallel, with capacitance $C_1 + C_2$. So

The common voltage is a weighted average of $V_1$ and $V_2$. The bigger capacitor counts for more.

A worked example. A $4.0\ \mu\text{F}$ capacitor is charged to $90\ \text{V}$. It is joined to an uncharged $2.0\ \mu\text{F}$ capacitor.

1. The charge. $Q = 4.0\ \mu\text{F} \times 90\ \text{V} = 360\ \mu\text{C}$.
2. The common voltage. $V' = \dfrac{360\ \mu\text{C}}{4.0\ \mu\text{F} + 2.0\ \mu\text{F}} = 60\ \text{V}$.
3. The energy before. $\tfrac{1}{2} \times 4.0\ \mu\text{F} \times (90\ \text{V})^2 = 16.2\ \text{mJ}$.
4. The energy after. $\tfrac{1}{2} \times 6.0\ \mu\text{F} \times (60\ \text{V})^2 = 10.8\ \text{mJ}$.
5. The loss. $16.2\ \text{mJ} - 10.8\ \text{mJ} = 5.4\ \text{mJ}$, which is one third of the energy. The fraction left is $\dfrac{C_1}{C_1 + C_2} = \dfrac{2}{3}$.

Where does the lost energy go?

The charge was conserved, but the energy was not. A short algebra step gives the loss for any two capacitors:

Three things to see in it.

• It is never negative. Joining capacitors cannot create energy.
• It is zero only when $V_1 = V_2$. If the voltages already match, no charge flows and nothing is lost.
• It does not depend on the resistance of the wires. A thick copper wire gives a short, strong spark. A thin wire gives a weak, slow current. The heat is the same. With no resistance at all, the energy leaves as radio waves.

A worked example, with both charged. A $5.0\ \mu\text{F}$ capacitor at $80\ \text{V}$ is joined to a $3.0\ \mu\text{F}$ capacitor at $40\ \text{V}$.

1. Same sign joined. $V' = \dfrac{5.0\ \mu\text{F} \times 80\ \text{V} + 3.0\ \mu\text{F} \times 40\ \text{V}}{8.0\ \mu\text{F}} = 65\ \text{V}$.
2. The loss. $\Delta U = \dfrac{5.0\ \mu\text{F} \times 3.0\ \mu\text{F}}{2 \times 8.0\ \mu\text{F}} \times (40\ \text{V})^2 = 1500\ \mu\text{J} = 1.5\ \text{mJ}$.
3. Opposite signs joined. Now the top plate of one is joined to the bottom plate of the other. The charges partly cancel. Treat $V_2$ as $-40\ \text{V}$: $V' = \dfrac{400\ \mu\text{C} - 120\ \mu\text{C}}{8.0\ \mu\text{F}} = 35\ \text{V}$.
4. The loss now. $\Delta U = \dfrac{5.0\ \mu\text{F} \times 3.0\ \mu\text{F}}{2 \times 8.0\ \mu\text{F}} \times (120\ \text{V})^2 = 13\,500\ \mu\text{J} = 13.5\ \text{mJ}$. Joining the wrong way round wastes far more.

Start with the defaults. Now set the second voltage to $100\ \text{V}$, the same as the first. What is the energy lost? Then slide it to $-100\ \text{V}$. What is the common voltage, and how much of the energy is lost?

A battery that stays or goes

A parallel-plate capacitor is charged by a battery. Then something changes its capacitance, such as moving the plates or inserting a slab. What happens to $Q$, $V$ and $U$? It depends on one thing: is the battery still connected?

Suppose the change makes the capacitance $n$ times as large, $C \to nC$.

• The battery stays connected. The voltage is held at $V$ by the battery. So $Q = CV$ becomes $nQ$. Charge flows in from the battery if $n > 1$, and back to it if $n < 1$. The energy $\tfrac{1}{2}CV^2$ becomes $nU$.
• The battery is taken away first. The plates are isolated, so $Q$ cannot change. The voltage $V = Q/C$ becomes $V/n$. The energy $\dfrac{Q^2}{2C}$ becomes $\dfrac{U}{n}$.

A worked example. A $10\ \mu\text{F}$ capacitor is charged to $60\ \text{V}$. Then the plates are moved to one third of their distance, so $C \to 30\ \mu\text{F}$ and $n = 3$. The energy at the start is $\tfrac{1}{2} \times 10\ \mu\text{F} \times (60\ \text{V})^2 = 18\ \text{mJ}$.

• Battery connected: $V$ stays $60\ \text{V}$. $Q = 30\ \mu\text{F} \times 60\ \text{V} = 1800\ \mu\text{C}$, three times the old value. $U = 54\ \text{mJ}$.
• Battery removed: $Q$ stays $600\ \mu\text{C}$. $V = \dfrac{600\ \mu\text{C}}{30\ \mu\text{F}} = 20\ \text{V}$. $U = 6.0\ \text{mJ}$.

Notice the second case. The energy fell when the plates came closer, because they attract. The plates did work on whoever was holding them back. Pulling the plates apart does the opposite: with the battery off, $n < 1$, and $U$ rises by the work you do against the attraction.

Charging through a resistor

A battery of voltage $V$ charges a capacitor $C$ through a resistor $R$. At every instant, the current is the voltage left over, divided by $R$:

The current starts at $V/R$ and falls to zero as the capacitor fills. The time scale is set by $RC$, the time constant. A bigger resistor or a bigger capacitor makes charging slower. But whatever $R$ is, the battery does work $QV = CV^2$, the capacitor keeps $\tfrac{1}{2}CV^2$, and the other $\tfrac{1}{2}CV^2$ becomes heat in $R$.

The program charges a capacitor in tiny time steps and adds up the heat as it goes.

Can the loss be made smaller? Yes. Charge the capacitor in $N$ small steps, with a supply that rises by $V/N$ each time. Each step loses only $\tfrac{1}{2}C(V/N)^2$, so the total loss is $\dfrac{1}{N}$ of $\tfrac{1}{2}CV^2$. Very slow charging wastes very little.

Practise with new numbers

Common mistakes

• Averaging the voltages. The common voltage is $\dfrac{Q_1 + Q_2}{C_1 + C_2}$. Use the plain average only if the capacitors are equal.
• Saying that energy is conserved on joining. The charge is conserved. The energy is lost, unless the voltages were equal.
• Adding the charges when the signs are opposite. Give one charge a minus sign. The charges partly cancel.
• Mixing up the two cases. With the battery on, $V$ stays the same. With the battery off, $Q$ stays the same.
• Thinking that the heat depends on the resistance. The energy lost in charging or joining does not depend on $R$.

Remember these

• Joining: $V' = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2}$. Use a negative sign for a plate joined the wrong way round.
• Energy lost: $\Delta U = \dfrac{C_1C_2}{2(C_1 + C_2)}(V_1 - V_2)^2$. It is zero when $V_1 = V_2$.
• An uncharged capacitor joined: the fraction of energy left is $\dfrac{C_1}{C_1 + C_2}$.
• If $C \to nC$: with the battery on, $Q \to nQ$ and $U \to nU$. With the battery off, $V \to V/n$ and $U \to U/n$.
• Charging through a resistor: the battery does $CV^2$, the capacitor keeps $\tfrac{1}{2}CV^2$, and the rest is heat.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $10\ \mu\text{F}$ capacitor at $40\ \text{V}$ is joined to an uncharged $30\ \mu\text{F}$ capacitor. What is the common voltage?

2. What fraction of the energy of the $10\ \mu\text{F}$ capacitor in question 1 is left after joining?

3. Two identical capacitors, each charged to $20\ \text{V}$, are joined with the plates the wrong way round. What is the common voltage, and how much of the energy is lost?

4. A charged capacitor is taken off its battery. Its plates are then pulled to twice their distance. What happens to $Q$, $V$ and $U$? Where does the extra energy come from?

5. A student says: "Charge is conserved when two capacitors are joined, so energy is conserved too." Spot the mistake.