Combining Capacitors and Stored Energy: Quick Revision

A brief note on Combining Capacitors and Stored Energy. Read it before a test. Try each question at the end before you open its answer.

Series and parallel

• Parallel: the same $V$. The charges add, $Q = Q_1 + Q_2 + \dots$, and $C = C_1 + C_2 + \dots$
• Series: the same $Q$. The voltages add, and $\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots$
• Two in series: $C = \dfrac{C_1C_2}{C_1 + C_2}$ and $V_1 = V\,\dfrac{C_2}{C_1 + C_2}$. The smaller capacitor has the larger voltage.
• Identical capacitors: $n$ in series give $\dfrac{C}{n}$. $n$ in parallel give $nC$.
• A network: replace plain series or parallel groups one at a time, then work back for the charge and voltage of each part.
• A rating: in a series chain, find the charge at which each capacitor reaches its limit. The smallest sets the limit.

Energy

• Stored: $U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}$.
• A battery: it does work $QV$. Half is stored, and half is lost as heat in the resistance, whatever its value.
• In parallel: $U \propto C$. In series: $U \propto \dfrac{1}{C}$.
• Density of a field: $u = \tfrac{1}{2}\varepsilon_0E^2$.
• Force between the plates: $F = \dfrac{Q^2}{2\varepsilon_0A} = \tfrac{1}{2}QE$.

Joining and changing

• Common voltage: $V' = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2}$. Use a minus sign for a capacitor joined the wrong way round.
• Energy lost: $\dfrac{C_1C_2}{2(C_1 + C_2)}(V_1 - V_2)^2$. It is zero only when $V_1 = V_2$.
• If $C \to nC$: with the battery on, $Q \to nQ$ and $U \to nU$. With the battery off, $V \to \dfrac{V}{n}$ and $U \to \dfrac{U}{n}$.

A worked example

A $7.0\ \mu\text{F}$ capacitor is charged to $50\ \text{V}$ and joined to an uncharged $3.0\ \mu\text{F}$ capacitor. Find the common voltage and the fraction of the energy that is left.

1. The voltage. $V' = \dfrac{7.0\ \mu\text{F} \times 50\ \text{V}}{7.0\ \mu\text{F} + 3.0\ \mu\text{F}} = 35\ \text{V}$.
2. The energy. The fraction left is $\dfrac{C_1}{C_1 + C_2} = \dfrac{7.0}{10} = 0.70$.

Practice with new numbers

A program to try

The program finds the energy in a capacitor in three ways.

Mistakes to avoid

• Mixing up the rules. Parallel capacitors share the voltage. Series capacitors share the charge.
• Writing $U = QV$. The energy stored is $\tfrac{1}{2}QV$.
• Averaging the voltages when capacitors are joined. Use the total charge over the total capacitance.
• Saying that energy is conserved on joining. It is lost unless the voltages are equal.

Check yourself

1. Capacitors of $1.5\ \mu\text{F}$ and $4.5\ \mu\text{F}$ are in series. What is the equivalent capacitance?

2. A $10\ \mu\text{F}$ capacitor stores $20\ \text{mJ}$. What is its voltage?

3. Two identical capacitors with equal and opposite charges are joined, plate to plate. How much energy remains?

4. A battery charges a capacitor and does $0.80\ \text{J}$ of work. How much heat is made?