Kirchhoff's second law

Kirchhoff's second law says that around any closed loop in a circuit, the sum of the emfs equals the sum of the potential differences: Σε = ΣIR. It is conservation of energy for electric charge: the energy each coulomb gains from the cells in a loop equals the energy it transfers in the components. In a loop with a 12 V cell and two resistors, the potential differences across the resistors must add up to 12 V. It is also called Kirchhoff's voltage law, or the loop rule.

Kirchhoff's second law

AQA's A-level specification covers the relationships between currents, voltages and resistances in series and parallel circuits, and conservation of charge and conservation of energy in dc circuits. Kirchhoff's second law is that conservation of energy written as a rule for loops, and Kirchhoff's first law is the conservation of charge.

Why it works

Potential difference is energy transferred per unit charge. A coulomb of charge that goes all the way round a loop ends up back where it started, at the same potential. So whatever energy it gains in the cells, it must transfer in the components. Per coulomb, energy gained is the emf and energy transferred is the potential difference.

Kirchhoff's first law

The first law, the junction rule, says that the total current into a junction equals the total current out of it. Charge is not created or destroyed, so it cannot pile up at a junction. You usually need both laws together for circuits with more than one loop.

Using Kirchhoff's second law in a series circuit

Example 1: one loop, two resistors

A 12 V cell with negligible internal resistance is connected to resistors of 4.0 Ω and 2.0 Ω in series. Find the current and the potential difference across each resistor.

Round the loop, the emf equals the sum of the potential differences:

The potential differences are 2.0 × 4.0 = 8.0 V and 2.0 × 2.0 = 4.0 V. Check: 8.0 + 4.0 = 12 V, the emf.

Example 2: internal resistance

A cell of emf 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Find the current and the terminal potential difference.

The internal resistance is in the loop too, so Kirchhoff's second law gives ε = I(R + r), the equation in AQA's data booklet:

The terminal pd is IR = 2.0 × 4.0 = 8.0 V. The other 1.0 V, the "lost volts", is across the internal resistance.

Example 3: two cells in a loop

A 6.0 V cell and a 1.5 V cell are connected in the same loop so that they oppose each other, with a total resistance of 9.0 Ω. Find the current.

When cells push in opposite directions round the loop, their emfs subtract:

The current flows in the direction the bigger cell pushes. The 1.5 V cell is being charged, rather than supplying energy.

Using both laws: a circuit with two loops

Example 4: a resistor in series with a parallel pair

A 12 V supply is connected to a 2.0 Ω resistor in series with a parallel pair of 6.0 Ω and 3.0 Ω resistors. Find the current in each resistor.

The parallel pair has a combined resistance of (6.0 × 3.0) ÷ (6.0 + 3.0) = 2.0 Ω, so the total is 4.0 Ω and the supply current is 12 ÷ 4.0 = 3.0 A.

• Loop through the supply, the 2.0 Ω resistor and the 6.0 Ω resistor: 12 = 3.0 × 2.0 + V, so V = 6.0 V across the parallel pair.
• Current in the 6.0 Ω resistor: 6.0 ÷ 6.0 = 1.0 A. Current in the 3.0 Ω resistor: 6.0 ÷ 3.0 = 2.0 A.
• Kirchhoff's first law check: 1.0 + 2.0 = 3.0 A, the current into the junction.

The loop that goes round just the two parallel resistors has no cell in it, so the sum of the potential differences round it is zero: the 6.0 V across one equals the 6.0 V across the other. That is why components in parallel always have the same potential difference.

Here is a quick check.

The Circuit Builder lets you build any DC circuit with meters and see the live readings, so you can check that the potential differences round each loop add up to the emf. Our guides to resistors in series and parallel and potential difference cover the basics.

Try it

Circuit Builder

Build any DC circuit with meters, lamps, LEDs and sensors, and see live readings

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How to apply Kirchhoff's second law to any loop

1. Choose a closed loop and a direction to go round it.
2. Add up the emfs, counting a cell as positive if it pushes current in your direction and negative if it pushes against it.
3. Add up the potential differences IR across each component, using the current through that component.
4. Set the two totals equal and solve.
5. If a current comes out negative, it flows the opposite way to the one you assumed. The size is still right.

Common mistakes

• Forgetting the internal resistance. It is part of the loop, with its own pd, Ir.
• Adding opposing emfs. Cells facing opposite ways round a loop subtract.
• Using the wrong current. In a multi-loop circuit, each resistor has its own current; use the current through that resistor.
• Confusing the two laws. The first law is about currents at a junction; the second is about potential differences round a loop.
• Not checking. Add up the potential differences at the end: they should equal the emf in every loop.

Practise electricity in a short timed paper

A-level Physics: Year 1 gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so circuits turn up regularly but not in every paper. The Electricity course covers current, potential difference and resistance from the start.

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