Capacitor networks and symmetry

A board that is neither series nor parallel

Open the back of an old radio and you may find five capacitors in a diamond, with one more across the middle. Try to find two of them in series. You cannot, because charge can leave the joint through the middle capacitor. Try to find two in parallel. You cannot either, because no pair shares both ends.

The rules of series and parallel are not enough for such a board. We need two rules that always work.

Two rules that always work

The rule of the loop. Go round any closed loop of the network. The rises and falls of potential add to zero.

The rule of the node. A node is a set of plates joined by wires. If no battery touches the node, charge cannot enter or leave it. So the charges on its plates add to the charge the node held at the start. For a node that was never charged, they add to zero.

One habit is needed. The charge on a plate is $q = C\,(V_{\text{plate}} - V_{\text{other plate}})$. The plate at the higher potential carries positive charge.

Here is the method.

1. Give each battery terminal its potential, such as $V$ and $0$.
2. Name the potential of every other node.
3. At each of those nodes, add $C\,(V_{\text{node}} - V_{\text{neighbour}})$ over the capacitors joined there. Put the sum equal to zero. If the node carried a charge $q_0$ at the start, put the sum equal to $q_0$.
4. Solve. Then find each charge from $q = C\,\Delta V$.

A first use: three capacitors at one node

Capacitors of $3.0\ \mu\text{F}$, $6.0\ \mu\text{F}$ and $3.0\ \mu\text{F}$ meet at a point O, and nothing else touches O. Their other plates are held at $100\ \text{V}$, $40\ \text{V}$ and $0\ \text{V}$. What is the potential of O?

The three plates at O add to zero:

That gives $12\ \mu\text{F} \times V_O = 540\ \mu\text{C}$, so $V_O = 45\ \text{V}$. The plates at O carry $3.0 \times (45 - 100) = -165\ \mu\text{C}$, then $6.0 \times (45 - 40) = +30\ \mu\text{C}$, then $3.0 \times 45 = +135\ \mu\text{C}$. They add to zero, as they must.

Look at the result. The potential of O is an average of the three potentials, and the capacitances are the weights. This is the same average as in charge sharing: a big capacitor pulls the node close to its own potential.

The bridge

Now take the five capacitors of the diamond. $C_1$ and $C_2$ are in the upper wire, $C_3$ and $C_4$ are in the lower wire, and $C_5$ joins the two middle corners M and N. A battery of voltage $V$ is joined to the corners A and B. So A is at $V$ and B is at $0$. The potentials of M and N are the unknowns.

A worked example. Take $C_1 = 2.0\ \mu\text{F}$, $C_2 = 3.0\ \mu\text{F}$, $C_3 = 4.0\ \mu\text{F}$, $C_4 = 14\ \mu\text{F}$, $C_5 = 10\ \mu\text{F}$ and $V = 40\ \text{V}$. The rule of the node at M, with each term in $\mu\text{C}$, is

and at N it is

1. Solve. From the first equation, $V_M = \dfrac{80 + 10V_N}{15}$. Put it into the second: $320\,V_N = 3200$. So $V_N = 10\ \text{V}$ and $V_M = 12\ \text{V}$.
2. The charges. $C_1$ holds $2.0\ \mu\text{F} \times (40 - 12)\ \text{V} = 56\ \mu\text{C}$. $C_2$ holds $3.0 \times 12 = 36\ \mu\text{C}$. $C_3$ holds $4.0 \times (40 - 10) = 120\ \mu\text{C}$. $C_4$ holds $14 \times 10 = 140\ \mu\text{C}$. $C_5$ holds $10 \times (12 - 10) = 20\ \mu\text{C}$, positive on the side of M.
3. Check at M. $56\ \mu\text{C}$ comes in. $36\ \mu\text{C}$ goes to $C_2$ and $20\ \mu\text{C}$ goes to $C_5$. At N, $120 + 20 = 140\ \mu\text{C}$ goes to $C_4$.
4. The whole network. The battery sends out $56 + 120 = 176\ \mu\text{C}$. So $C = \dfrac{176\ \mu\text{C}}{40\ \text{V}} = 4.4\ \mu\text{F}$.

When does the middle capacitor hold no charge? When M and N are at the same potential. The pair $C_1$ and $C_2$ is a chain in series, so M is at $V\,\dfrac{C_1}{C_1 + C_2}$. The pair $C_3$ and $C_4$ gives N at $V\,\dfrac{C_3}{C_3 + C_4}$. They agree when

This is the balance of the bridge. Then $C_5$ may be taken away, or may have any value. The charge on $C_5$ turns out to be proportional to $C_1C_4 - C_2C_3$, so it changes sign as the bridge goes through balance.

With $C_4 = 6.0\ \mu\text{F}$ the example above is balanced, since $2.0 \times 6.0 = 3.0 \times 4.0$. M and N are both at $16\ \text{V}$. The network is then two chains in parallel: $\dfrac{2.0 \times 3.0}{5.0} + \dfrac{4.0 \times 6.0}{10} = 1.2 + 2.4 = 3.6\ \mu\text{F}$.

Move the red point along the curve. At what value of $C_4$ does the charge become zero? Now change $C_5$. Does the zero move? Then look at the sign: which side of $C_5$ is positive when $C_4$ is larger than the balance value?

Equal potentials from symmetry

Sometimes the shape of the network shows that two points are at the same potential. If the battery cannot tell the two points apart, they are at the same potential. Then two things are allowed.

• Join the two points by a wire. Nothing changes, because no charge needs to move between them.
• A capacitor joined between the two points has no voltage, so it holds no charge. It can be taken out.

A cube of capacitors. Twelve equal capacitors $C$ are the edges of a cube. A battery is joined to the two ends of a long diagonal, A and G. The three corners next to A are alike. The battery cannot tell them apart, so they are at one potential. The same is true of the three corners next to G.

Join the three equal corners together. Between A and the first group there are 3 edges in parallel, so $3C$. Between the two groups there are 6 edges, so $6C$. Between the second group and G there are 3 edges, so $3C$. Now it is a chain in series:

With $C = 5.0\ \mu\text{F}$, the whole cube is worth $6.0\ \mu\text{F}$. This is the same cube as the cube of resistors, with the capacitances in the place of the conductances.

Take care. Joining points is allowed only when they really are at the same potential. In a bridge that is not balanced, M and N are not at the same potential, and a wire between them changes the answer.

An endless ladder

A ladder has a series capacitor $C_1$ and a shunt capacitor $C_2$ in each section, and it goes on for ever. What is the equivalent capacitance between its input terminals?

Call it $x$. Cut off the first section. What is left is the same ladder, so it also has the value $x$. The first section then has $C_1$ in series with ($C_2$ in parallel with $x$):

We take the positive root. For $C_1 = 4.0\ \mu\text{F}$ and $C_2 = 2.0\ \mu\text{F}$, the equation is $x^2 + 2x - 8 = 0$, so $x = 2.0\ \mu\text{F}$. Check: $4.0$ in series with $(2.0 + 2.0)$ gives $\dfrac{4.0 \times 4.0}{8.0} = 2.0\ \mu\text{F}$.

Put $48\ \text{V}$ across the input. The first series capacitor and the rest of the ladder $(2.0 + 2.0 = 4.0\ \mu\text{F})$ are in series with equal capacitances, so each has $24\ \text{V}$. The first rung has $24\ \text{V}$. The next rung has half of it, $12\ \text{V}$, and so on. The charge from the battery is $2.0\ \mu\text{F} \times 48\ \text{V} = 96\ \mu\text{C}$.

Practise with new numbers

Try it in Python

The program solves a network with the node method. It puts $1\ \text{V}$ on one corner and $0\ \text{V}$ on another, writes the rule of the node for every other node, and solves all the equations at once. Then it finds the equivalent capacitance of the cube for three different pairs of corners, and the bridge of this lesson.

Common mistakes

• Calling a bridge series and parallel. If charge can leave through the middle capacitor, no pair is in series. Use the rule of the node.
• Joining points that are not at the same potential. Symmetry or balance must show that they are.
• Using the average of the potentials. The potential of a node is weighted by the capacitances.
• Forgetting that a capacitor holds charge of both signs. A capacitor has $+q$ on one plate and $-q$ on the other. The rule of the node adds only the plates at that node.
• Writing the rule of the node at a node that touches a battery. The battery can give or take charge there. Use the rule only at nodes that touch no battery.

Remember these

• Loop: the voltages round a loop add to zero.
• Node: the charges on the plates at an isolated node add to its starting charge, which is zero if it was never charged.
• Plate charge: $q = C\,(V_{\text{plate}} - V_{\text{other plate}})$.
• Bridge: balanced when $\dfrac{C_1}{C_2} = \dfrac{C_3}{C_4}$. Then the middle capacitor holds no charge.
• Symmetry: points that the battery cannot tell apart are at one potential. They can be joined.
• Cube between opposite corners: $\dfrac{6C}{5}$.
• Endless ladder: $x^2 + C_2x - C_1C_2 = 0$.

Check yourself

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

1. Capacitors of $6.0\ \mu\text{F}$ and $12\ \mu\text{F}$ meet at a point O that touches nothing else. Their other plates are held at $30\ \text{V}$ and $0\ \text{V}$. What is the potential of O?

2. A student says: "In a balanced bridge the middle capacitor holds no charge, so no charge comes from the battery." Spot the mistake.

3. In a bridge, $C_1 = C_2 = C_3 = C_4 = 5.0\ \mu\text{F}$ and the middle capacitor is $9.0\ \mu\text{F}$. What is the equivalent capacitance between A and B?

4. Why is the equivalent capacitance of an endless ladder always smaller than the series capacitor $C_1$?

5. In the cube, why are the three corners next to A at the same potential?