Kirchhoff's Laws

When circuits become more complex than simple series or parallel arrangements, you need a systematic method to analyse them. Kirchhoff's two laws provide exactly that. They are based on two fundamental conservation principles and together can solve any circuit problem.

Kirchhoff's First Law (Current Law)

At any junction (node) in a circuit, the sum of the currents entering the junction equals the sum of the currents leaving the junction.

Mathematically: ΣI_in = ΣI_out

Or equivalently, if we assign positive signs to currents entering and negative signs to currents leaving: ΣI = 0

Physical Basis

Kirchhoff's first law is a statement of the conservation of charge. Charge cannot accumulate at a junction — every electron that arrives must leave. If this were not true, charge would build up at junctions, which does not happen in steady-state circuits.

Worked Example 1

At a junction, three wires meet. Wire A carries 3.0 A into the junction. Wire B carries 1.0 A into the junction. What current flows in wire C, and in which direction?

Solution:

Currents in = 3.0 + 1.0 = 4.0 A
By Kirchhoff's first law, currents out must also equal 4.0 A
Wire C carries 4.0 A out of the junction

Worked Example 2: Four-Wire Junction

At a junction, wire P carries 5.0 A in, wire Q carries 2.0 A out, wire R carries 1.5 A in, and wire S carries current I_S. Find I_S and its direction.

Solution:

ΣI_in = ΣI_out
5.0 + 1.5 = 2.0 + I_S
I_S = 6.5 - 2.0 = 4.5 A out of the junction

Kirchhoff's Second Law (Voltage Law)

Around any closed loop in a circuit, the sum of the EMFs equals the sum of the potential differences.

Mathematically: Σε = ΣIR

Or equivalently: the algebraic sum of all potential changes around any closed loop is zero.

Physical Basis

Kirchhoff's second law is a statement of the conservation of energy. As a unit charge travels around any closed loop and returns to its starting point, it must end up with the same energy it started with. The energy gained from EMF sources must equal the energy lost in resistors.

How to Apply Kirchhoff's Second Law

flowchart TD
    A[Choose a closed loop] --> B[Choose a direction: CW or ACW]
    B --> C[Assign current directions to each branch]
    C --> D[Write the loop equation]
    D --> E{Traversing a battery from − to +?}
    E -->|Yes| F[+ε on the left side]
    E -->|No| G[−ε on the left side]
    D --> H{Traversing a resistor in the direction of assumed I?}
    H -->|Yes| I[+IR on the right side]
    H -->|No| J[−IR on the right side]
    F --> K[Solve the equation]
    G --> K
    I --> K
    J --> K
    K --> L{Negative current?}
    L -->|Yes| M[Current actually flows opposite to assumed direction]
    L -->|No| N[Assumed direction was correct]

Worked Example 3

A single loop contains a 12 V battery with internal resistance 1.0 Ω and two external resistors of 3.0 Ω and 8.0 Ω in series. Find the current.

Solution: Applying Kirchhoff's second law around the loop:

Σε = ΣIR

12 = I(1.0) + I(3.0) + I(8.0)

12 = I(12)

I = 1.0 A

Applying Both Laws to Complex Circuits

For circuits with multiple loops and junctions, you typically need both laws together. The general approach is:

Label all unknown currents with symbols (I₁, I₂, I₃, etc.) and assign assumed directions.
Apply Kirchhoff's first law at each junction to relate the currents.
Apply Kirchhoff's second law around enough independent loops to create a system of equations.
Solve the simultaneous equations for the unknown currents.

How Many Equations Do You Need?

If the circuit has b branches with unknown currents, you need b independent equations. You get:

(n - 1) equations from Kirchhoff's first law, where n = number of nodes
The rest from Kirchhoff's second law applied to independent loops

Worked Example 4: Two-Battery Circuit

A circuit has two batteries: a 6.0 V battery with internal resistance 1.0 Ω, and a 4.0 V battery with internal resistance 0.50 Ω. They are connected in parallel, both driving current through a 5.0 Ω external resistor.

Let I₁ = current from the 6.0 V battery, I₂ = current from the 4.0 V battery, and I₃ = current through the 5.0 Ω resistor.

At the junction: I₃ = I₁ + I₂ (Kirchhoff's first law)

Loop 1 (6.0 V battery and external resistor): 6.0 = I₁(1.0) + I₃(5.0) = I₁ + 5.0(I₁ + I₂) = 6.0I₁ + 5.0I₂

Loop 2 (4.0 V battery and external resistor): 4.0 = I₂(0.50) + I₃(5.0) = 0.50I₂ + 5.0(I₁ + I₂) = 5.0I₁ + 5.5I₂

Solving these simultaneous equations: From equation 1: 6.0 = 6.0I₁ + 5.0I₂ → I₁ = (6.0 - 5.0I₂)/6.0 Substituting into equation 2: 4.0 = 5.0 × (6.0 - 5.0I₂)/6.0 + 5.5I₂ 4.0 = 5.0 - 4.167I₂ + 5.5I₂ 4.0 = 5.0 + 1.333I₂ I₂ = -0.75 A

The negative sign means I₂ actually flows in the opposite direction to our assumption — the 4.0 V battery is being charged by the 6.0 V battery.

I₁ = (6.0 - 5.0 × (-0.75))/6.0 = (6.0 + 3.75)/6.0 = 1.625 A I₃ = I₁ + I₂ = 1.625 + (-0.75) = 0.875 A

Verification

Always verify your solution by substituting back into both loop equations:

Loop 1: 6.0I₁ + 5.0I₂ = 6.0(1.625) + 5.0(-0.75) = 9.75 - 3.75 = 6.0 ✓
Loop 2: 5.0I₁ + 5.5I₂ = 5.0(1.625) + 5.5(-0.75) = 8.125 - 4.125 = 4.0 ✓

Worked Example 5: Wheatstone Bridge Analysis

A Wheatstone bridge circuit has a battery (ε = 6.0 V, negligible internal resistance) across two parallel branches. Branch 1 has 10 Ω and 30 Ω in series. Branch 2 has 20 Ω and 60 Ω in series. A galvanometer connects the midpoints. Is the bridge balanced? What is the galvanometer current?

Solution: Check the balance condition: R₁/R₂ = R₃/R₄

10/30 = 1/3 and 20/60 = 1/3

Since the ratios are equal, the bridge is balanced and the galvanometer current is zero.

When balanced, the p.d. across R₁ equals the p.d. across R₃, so no current flows through the galvanometer.

Sign Conventions: Getting the Signs Right

The most common source of errors with Kirchhoff's laws is getting signs wrong. Here is a reliable approach:

Pick a loop direction (clockwise or anticlockwise).
Travelling through a battery from − to +: this is an EMF gain → +ε
Travelling through a battery from + to −: this is an EMF loss → −ε
Travelling through a resistor in the same direction as the current: this is a voltage drop → +IR (on the right side of Σε = ΣIR)
Travelling through a resistor against the current: this is a voltage gain → −IR

If you are consistent with the signs, the algebra will give you the correct answer — including telling you (via negative values) if you guessed a current direction wrong.

When Do You Need Kirchhoff's Laws?

For simple series or parallel circuits, the series/parallel rules are quicker. You need Kirchhoff's laws when:

The circuit contains multiple sources of EMF in different branches.
Components are arranged in a way that is neither purely series nor purely parallel (e.g., a Wheatstone bridge).
You need to find the current through a specific branch in a complex network.

Common Exam Mistakes with Kirchhoff's Laws

Mistake	How to Avoid It
Inconsistent sign convention	Choose one loop direction and stick with it. Write the rules on your paper before starting.
Not enough equations	Count the unknowns. You need as many independent equations as unknowns.
Using the same loop twice	Each loop equation must trace a different path. Check that at least one branch is unique to each loop.
Panicking at a negative answer	Negative current simply means the actual direction is opposite to what you assumed. The magnitude is still correct.
Forgetting internal resistance	Internal resistance is inside the battery. The current through the battery also flows through r.
Not verifying the answer	Always substitute back into the original equations to check consistency.

A-Level Deep Dive: Kirchhoff's Laws

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic 3.6 (Kirchhoff's laws) establishes Kirchhoff's first law (the algebraic sum of currents at any junction is zero, $\Sigma I = 0$ , equivalently $\Sigma I_\text{in} = \Sigma I_\text{out}$ ) and Kirchhoff's second law (around any closed loop, the sum of EMFs equals the sum of $IR$ drops, $\Sigma \varepsilon = \Sigma IR$ ), together with their use in analysing networks containing more than one source of EMF (refer to the official Pearson Edexcel specification document for exact wording). This sub-topic sits at the analytical heart of Topic 3: it generalises the series and parallel rules of 3.4 to networks that are neither purely series nor purely parallel, and it is the formal statement of the conservation principles already implicit in 3.5 (where $\varepsilon = I(R + r)$ is itself a one-loop KVL equation). Questions on Kirchhoff's laws are examined on Paper 1 (9PH0-01) as multi-part calculations and on Paper 3 (9PH0-03) as analytical reasoning about practical circuits. The Edexcel formula booklet does not list Kirchhoff's laws as equations — they are conservation statements that must be quoted in words and then applied to write the appropriate algebraic system.

Worked example with full mark scheme

Question (8 marks):

A two-loop network contains two cells and three resistors. Cell 1 has EMF $\varepsilon_1 = 9.0\ \text{V}$ and internal resistance $r_1 = 1.0\ \Omega$ . Cell 2 has EMF $\varepsilon_2 = 6.0\ \text{V}$ and internal resistance $r_2 = 0.50\ \Omega$ . The two cells are connected in parallel (positive terminals together), and the parallel combination drives current through an external resistor $R = 4.0\ \Omega$ .

Let $I_1$ be the current from cell 1, $I_2$ the current from cell 2, and $I_3$ the current through $R$ , all assumed to flow toward the external resistor.

(a) State Kirchhoff's first and second laws. (2)

(b) Write the junction equation and the two independent loop equations for this circuit. (3)

Solution with mark scheme:

(a) B1 — Kirchhoff's first law: at any junction, the algebraic sum of currents is zero, equivalently $\Sigma I_\text{in} = \Sigma I_\text{out}$ (charge conservation).

B1 — Kirchhoff's second law: around any closed loop, the sum of EMFs equals the sum of $IR$ drops, $\Sigma \varepsilon = \Sigma IR$ (energy conservation).

(b) Step 1 — junction. At the node where the two cell branches meet the external resistor, currents $I_1$ and $I_2$ flow in and $I_3$ flows out:

$I_3 = I_1 + I_2$

M1 — correct junction equation.

Step 2 — loop 1 (cell 1, then through $R$ , returning through cell 1):

$\varepsilon_1 = I_1 r_1 + I_3 R \implies 9.0 = 1.0\, I_1 + 4.0\, (I_1 + I_2) = 5.0\, I_1 + 4.0\, I_2$

M1 — correct loop-1 equation with both $I_1 r_1$ and $I_3 R$ accounted for, and $I_3$ replaced using the junction relation.

Step 3 — loop 2 (cell 2, then through $R$ , returning through cell 2):

$\varepsilon_2 = I_2 r_2 + I_3 R \implies 6.0 = 0.50\, I_2 + 4.0\, (I_1 + I_2) = 4.0\, I_1 + 4.5\, I_2$

M1 — correct loop-2 equation, again using the junction substitution.

(c) Step 1 — eliminate $I_1$ . From loop 1: $I_1 = (9.0 - 4.0\, I_2)/5.0 = 1.80 - 0.80\, I_2$ . Substitute into loop 2:

$6.0 = 4.0\, (1.80 - 0.80\, I_2) + 4.5\, I_2 = 7.20 - 3.20\, I_2 + 4.5\, I_2 = 7.20 + 1.30\, I_2$

So $1.30\, I_2 = -1.20\$ , giving $I_2 = -0.92\ \text{A}$ (to 2 s.f.).

M1 — correct algebraic elimination and substitution.

Step 2 — back-substitute. $I_1 = 1.80 - 0.80 \times (-0.92) = 1.80 + 0.74 = 2.54\ \text{A}$ , and $I_3 = I_1 + I_2 = 2.54 - 0.92 = 1.62\ \text{A}$ .

A1 — $I_1 \approx 2.5\ \text{A}$ , $I_2 \approx -0.92\ \text{A}$ , $I_3 \approx 1.6\ \text{A}$ (2 s.f. throughout).

A1 — interpretation: the negative sign for $I_2$ means cell 2 is being charged by cell 1 — current actually flows into its positive terminal, opposite to the assumed direction. The two-cell mismatch (9.0 V driving against 6.0 V) creates a circulating current that uses cell 1's surplus EMF to push charge backwards through cell 2.

Total: 8 marks (B1 B1 + M1 M1 M1 + M1 A1 A1).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A circuit consists of a 12 V battery (negligible internal resistance) connected across two parallel branches. Branch X contains a 6.0 $\Omega$ resistor in series with a 3.0 $\Omega$ resistor. Branch Y contains a 4.0 $\Omega$ resistor in series with a 2.0 $\Omega$ resistor. A galvanometer (assume ideal, infinite resistance) connects the midpoint of branch X to the midpoint of branch Y.

(a) Use Kirchhoff's laws to find the current in each branch. (3)

(b) Calculate the potential difference across the galvanometer and state whether the bridge is balanced. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1b) — recognising that branches X and Y carry independent currents because the galvanometer is ideal: $I_X = 12/(6.0 + 3.0) = 1.33\ \text{A}$ .
M1 (AO1.1b) — $I_Y = 12/(4.0 + 2.0) = 2.0\ \text{A}$ .
A1 (AO1.1b) — both currents quoted to 2–3 s.f. with units.

(b)

M1 (AO2.1) — p.d. across the 3.0 $\Omega$ resistor in branch X: $V_X = I_X \times 3.0 = 1.33 \times 3.0 = 4.0\ \text{V}$ (measured from the midpoint to the lower rail).
M1 (AO2.1) — p.d. across the 2.0 $\Omega$ resistor in branch Y: $V_Y = I_Y \times 2.0 = 2.0 \times 2.0 = 4.0\ \text{V}$ .
A1 (AO2.5) — galvanometer p.d. $= V_X - V_Y = 0\ \text{V}$ , so the bridge is balanced. Equivalent justification: the ratios $6.0/3.0 = 2.0$ and $4.0/2.0 = 2.0$ are equal, satisfying the bridge-balance condition.

Total: 6 marks split AO1 = 3, AO2 = 3. This is a classic Wheatstone-bridge application of Kirchhoff's laws. Edexcel uses the bridge to test whether candidates can decompose a network into two simpler series chains and recognise the balance condition as a consequence of equal potentials at the two midpoints.

Synoptic links

Connects to:

Topic 3 sub-topic 3.1 — Charge conservation: Kirchhoff's first law is the steady-state form of charge conservation. The continuity equation $\partial \rho / \partial t + \nabla \cdot \mathbf{J} = 0$ reduces, in DC, to $\nabla \cdot \mathbf{J} = 0$ — and integrating over the surface of a junction gives exactly $\Sigma I = 0$ . Every circuit problem you solve with KCL is invoking the same physics that conserves charge in beta decay and pair production.
Topic 3 sub-topic 3.5 — Energy conservation and EMF: Kirchhoff's second law is the line-integral statement $\oint \mathbf{E} \cdot d\mathbf{l} = 0$ (for electrostatic fields with EMF sources treated as discrete energy contributions). The familiar one-loop equation $\varepsilon = I(R + r)$ is the simplest case, with one EMF and two resistive elements. KVL generalises this to any number of sources and resistors.

Kirchhoff's First Law (Current Law)

Physical Basis

Worked Example 1

Worked Example 2: Four-Wire Junction

Kirchhoff's Second Law (Voltage Law)

Physical Basis

How to Apply Kirchhoff's Second Law

Worked Example 3

Applying Both Laws to Complex Circuits

How Many Equations Do You Need?

Worked Example 4: Two-Battery Circuit

Verification

Worked Example 5: Wheatstone Bridge Analysis

Sign Conventions: Getting the Signs Right

When Do You Need Kirchhoff's Laws?

Common Exam Mistakes with Kirchhoff's Laws

A-Level Deep Dive: Kirchhoff's Laws

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Synoptic links

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