Kirchhoff's Rules

When series and parallel are not enough

Some circuits cannot be reduced to one resistor. An example is a circuit with two batteries in different branches. For these, we use Kirchhoff's two rules. Each one is a law of conservation.

The junction rule: charge is conserved

A junction is a point where three or more wires meet. At a junction, the current that flows in is equal to the current that flows out.

Charge cannot pile up at a point in a wire. So what comes in each second must go out each second. It is like water in a pipe junction.

An example. At a junction, $3\ \text{A}$ and $2\ \text{A}$ flow in. One wire takes $1\ \text{A}$ out. The other wire takes $3\ \text{A} + 2\ \text{A} - 1\ \text{A} = 4\ \text{A}$ out.

The loop rule: energy is conserved

Go round any closed loop of a circuit. The changes in potential add up to zero.

A charge that goes once round a loop comes back to the same point, with the same potential. The energy that it gains in the batteries is the energy that it gives to the resistors.

Signs

Choose a direction to go round the loop. Then follow these two rules.

• Going across a battery from $-$ to $+$, the potential rises. Write $+E$. From $+$ to $-$, write $-E$.
• Going across a resistor in the same direction as the current, the potential falls. Write $-IR$. In the opposite direction, write $+IR$.

You do not have to guess the direction of every current correctly. Choose any direction for each branch. If an answer is negative, the current simply flows the other way. That is not a mistake. It is an answer.

You should find that the middle current always equals the sum of the other two, and that the right battery's current turns negative when the left battery is much stronger.

A real battery has resistance inside

A real battery has a small resistance $r$ inside it. We call it the internal resistance. The loop rule for a battery of emf $E$ and a resistor $R$ gives

The voltage at the terminals is what is left after the loss inside the battery.

An example. A cell has $E = 6\ \text{V}$ and $r = 1\ \Omega$. It is connected to a $5\ \Omega$ resistor. The current is $I = \dfrac{6\ \text{V}}{5\ \Omega + 1\ \Omega} = 1\ \text{A}$. The terminal voltage is $6\ \text{V} - 1\ \text{A} \times 1\ \Omega = 5\ \text{V}$.

A plan for every circuit

1. Draw an arrow for the current in every branch.
2. Write the junction rule at every junction except one.
3. Write the loop rule for enough loops, so that you have as many equations as unknown currents.
4. Solve the equations.
5. Check: the power from the batteries must be equal to the heat in the resistors, $\sum I^2R$.

A worked example

The left branch has an $8\ \text{V}$ battery and a $2\ \Omega$ resistor. The right branch has a $2\ \text{V}$ battery and a $1\ \Omega$ resistor. A $2\ \Omega$ resistor sits in the middle. The positive terminals of both batteries are at the top wire.

Let $I_1$ and $I_2$ go up through the batteries. Let $I_3$ go down through the middle resistor.

We write each loop in volts, with the currents in amperes and the resistances in ohms.

The junction rule: $I_1 + I_2 = I_3$.

The left loop: $8 - 2I_1 - 2I_3 = 0$.

The right loop: $2 - 1 \times I_2 - 2I_3 = 0$.

From the loop equations, $I_1 = 4 - I_3$ and $I_2 = 2 - 2I_3$. We put them in the junction rule.

So $I_1 = 2.5\ \text{A}$ and $I_2 = -1\ \text{A}$. The minus sign means that the $2\ \text{V}$ battery has a current going down through it. It is being charged by the other battery.

Practise with new numbers

Try it in Python

Each rule gives one linear equation. So the three equations become one matrix equation. This program solves them with numpy. Then it checks the energy.

Remember these

• Junction rule: current in = current out. It comes from conservation of charge.
• Loop rule: the potential changes around a loop add up to zero. It comes from conservation of energy.
• A negative current only means that it flows the other way.
• For a battery with internal resistance, $I = E/(R + r)$ and $V_{\text{terminal}} = E - Ir$.

Check yourself

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

1. At a junction, $6\ \text{A}$ flows in through one wire, and two wires take current out. One carries $2.5\ \text{A}$. What does the other carry?

2. A cell of emf $1.5\ \text{V}$ and internal resistance $0.5\ \Omega$ drives a $2.5\ \Omega$ resistor. Find the current and the terminal voltage.

3. You go round a loop against the current in a resistor. Is the $IR$ term positive or negative?

4. In the worked example, which battery is being charged, and how do you know?