Kirchhoff's laws in circuits for A-Level Physics
Kirchhoff's two circuit laws are direct consequences of conservation of charge and conservation of energy. The first law (KCL) says the total current flowing into any junction equals the total current flowing out. The second law (KVL) says the sum of the EMFs around a closed loop equals the sum of the potential differences across the components in that loop.
This guide covers both laws, the conservation principles that justify them, the sign convention you need for AQA A-Level Physics, and the question types where Kirchhoff's laws come up most. Expect to see them in series and parallel circuit calculations, in potential divider analysis, and in any multi-loop network where Ohm's law alone is not enough.
First law: Charge is conserved
At any junction, current in equals current out. No charge piles up or disappears at a node.
Second law: Energy is conserved
Around any closed loop, the sum of EMFs equals the sum of the voltage drops across resistors and other components.
Sign convention matters
Choose a loop direction first, then assign + or – consistently to each EMF and voltage drop. Getting the signs wrong is the most common slip.
Kirchhoff's first law: The junction rule
Kirchhoff's first law states that the algebraic sum of currents at any junction is zero. Equivalently, the total current flowing into a junction equals the total current flowing out. This follows directly from conservation of charge: Charge cannot accumulate at a single point in a wire, so whatever comes in must go out.
For AQA A-Level Physics, the law is usually written as ΣI(in) = ΣI(out). Currents flowing in are taken as positive, currents flowing out as negative, and they must add to zero. The law applies at any node in a circuit, no matter how many wires meet there.
Worked example for the first law Three wires meet at a junction. Wire 1 carries 3 A in. Wire 2 carries 2 A in. Wire 3 carries current I out. By the first law, 3 + 2 = I, so I = 5 A. The 5 A flowing out balances the 5 A flowing in.
Kirchhoff's second law: The loop rule
Kirchhoff's second law states that the sum of the EMFs around any closed loop equals the sum of the potential differences across the components in that loop. Mathematically, ΣEMF = ΣIR for a loop containing resistors. This is a direct consequence of conservation of energy: Energy supplied by EMF sources must equal energy dissipated by resistors in any complete loop.
The law applies to any closed loop in a circuit, not just the outer loop. In a multi-loop network you can apply the second law to each independent loop, generating one equation per loop. Combined with the first law at each junction, you get enough equations to solve for every unknown current and voltage.
| Quantity | Symbol | How it appears in KVL |
|---|---|---|
| EMF of a source | ε (or E) | Adds energy to the loop, positive in the direction of travel |
| Potential difference across a resistor | V = IR | Drops energy, negative in the direction of current flow |
| Internal resistance of a cell | r | Treated as a resistor in series with the EMF source |
How the two laws connect to GCSE circuit rules
If you remember the GCSE rules for series and parallel circuits, you already know Kirchhoff's laws in disguise. "Current is the same everywhere in a series circuit" is the first law applied at every point. "Current is shared between branches of a parallel circuit" is the first law applied at the branching junctions.
"Voltages around a series loop add up to the EMF" is the second law. "Each parallel branch has the same voltage as the source" is the second law applied to a loop containing one branch and the source. At A-Level, you just need the general form because you may have multiple sources or asymmetric branches.
From GCSE intuition to A-Level practice At GCSE you learn that voltages add in series. At A-Level you apply the same principle in a multi-loop network where the algebra gets longer but the physics stays identical. Conservation of charge and energy are the only ideas you need.
Sign convention for KVL
Choosing signs is where most students lose marks. Pick a direction (clockwise or anticlockwise) for each loop and stick with it. When you travel through an EMF source from – to +, the EMF is positive in your equation. When you travel from + to –, it is negative. When you cross a resistor in the same direction as the current flow, the potential difference IR is negative (you are losing potential). When you cross against the current, IR is positive.
This sounds fiddly. The trick is to draw the chosen loop direction as a curved arrow on your diagram before writing any equation, then walk through the loop in that direction and write down each term as you meet it.
Worked example: Two cells and two resistors
A 12 V cell with internal resistance 1 Ω is connected to a 4 Ω resistor in series with a second cell of EMF 4 V (assume negligible internal resistance) and a 5 Ω resistor. The two cells are connected so that they drive current in the same direction. Find the current in the circuit.
Apply Kirchhoff's second law around the single loop. Total EMF = 12 + 4 = 16 V. Total resistance = 1 + 4 + 5 = 10 Ω. Therefore I = 16 / 10 = 1.6 A. Now check with Ohm's law on each component: V across 4 Ω resistor = 1.6 × 4 = 6.4 V. V across 5 Ω resistor = 1.6 × 5 = 8.0 V. V across internal resistance = 1.6 × 1 = 1.6 V. Sum = 16 V, which matches the total EMF, confirming KVL holds.
Worked example: A parallel network
A 6 V cell drives current through a 2 Ω resistor in series with a parallel combination of a 4 Ω and a 6 Ω resistor. Find the current in each resistor.
Step 1: Parallel combination. 1/R = 1/4 + 1/6 = 5/12, so R(parallel) = 2.4 Ω. Total resistance = 2 + 2.4 = 4.4 Ω. Total current from cell = 6 / 4.4 = 1.36 A.
Step 2: Apply Kirchhoff's first law at the junction. The 1.36 A splits between the 4 Ω and 6 Ω branches. By KVL, both branches have the same voltage across them. That voltage = 1.36 × 2.4 = 3.27 V. Current through 4 Ω = 3.27 / 4 = 0.82 A. Current through 6 Ω = 3.27 / 6 = 0.54 A. Check: 0.82 + 0.54 = 1.36 A, satisfying the first law at the junction.
Where Kirchhoff's laws appear on AQA papers
AQA A-Level Physics examines Kirchhoff's laws in Paper 1 (3.5 Electricity) and sometimes in synoptic questions in Paper 3. They appear most often in: Multi-loop circuit analysis, potential divider problems where you need to compare branches, internal resistance questions where the loop sums EMF against terminal voltage and ir drop, and questions that combine current and voltage measurements from a meter to deduce the rest of the circuit.
In the practical endorsement, you will use both laws implicitly when measuring EMF and internal resistance using the gradient and intercept of a V-against-I graph.
Internal resistance: Where KVL really earns its keep For a cell of EMF ε and internal resistance r connected to a load R, KVL around the single loop gives ε = IR + Ir. Rearranging: V = ε – Ir, where V is the terminal voltage across the load. This is the equation you plot from V-I data to find ε (y-intercept) and r (negative gradient).
Where students lose marks
AQA examiner reports flag the same set of issues every year on Kirchhoff's law questions. Most are about signs and bookkeeping rather than missing physics.
Common mistakes that cost easy marks Mixing up signs when traversing a resistor against the current. Forgetting to include internal resistance in the KVL sum. Stating only one law in an answer when the question asks for both. Confusing junction with loop. Writing ΣV = 0 without showing which terms are EMFs and which are IR drops. Not labelling the loop direction on a circuit diagram before applying KVL.
Key facts to memorise for the exam
- First law: Sum of currents into a junction = sum of currents out (conservation of charge)
- Second law: Sum of EMFs around a closed loop = sum of IR drops (conservation of energy)
- Choose a loop direction (clockwise or anticlockwise) before writing KVL
- EMF positive when traversed from – to +, negative when traversed + to –
- Voltage drop IR negative when crossing a resistor in the direction of current
- Internal resistance r is always included as a series resistor inside the cell
- Terminal voltage V = ε – Ir for a cell driving current I through internal resistance r
- GCSE series-parallel rules are special cases of Kirchhoff's two laws
