Kirchhoff's First Law

Key idea: Apply Kirchhoff’s first law (sum of currents at a junction is zero) to solve circuit junction current problems (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Recall circuit symbols and draw or interpret circuit diagrams.
  • Draw circuit diagrams containing sources, switches, resistors, meters, lamps, thermistors, light-dependent resistors and diodes.
  • Apply resistance and resistivity, interpret I–V characteristics and explain temperature effects.
  • Analyse e.m.f., terminal potential difference and internal resistance in real sources.
  • Analyse series, parallel and potential-divider resistor networks.
  • Combine capacitors in series and parallel.
  • Analyse charging and discharging in RC circuits using the time constant.
Syllabus note (9478)

Kirchhoff’s laws are not explicitly listed in the 9478 syllabus, but they are a useful (and widely used) method for analysing more complex d.c. circuits.

1. Definitions (Must Know)

A. Kirchhoff’s first law (junction rule / current law)

Kirchhoff’s first law states that at any junction:

  • total current entering = total current leaving

Equivalently, the algebraic sum of currents at a junction is zero:

∑ I = 0

2. Key Ideas (What Earns Marks)

  • It is a statement of conservation of charge: charge does not build up at an ideal junction in steady state.
  • You must choose a sign convention at the junction:
    • currents entering are positive, leaving negative (or vice versa), then apply ∑ I = 0.
  • Current is the same everywhere in a series branch, but it can split between parallel branches.
Fast workflow

Circle the junction → write currents in and out → write one equation using ∑ I = 0 → solve.

3. Detailed Explanations

A. Why it is true (charge conservation)

Current is rate of flow of charge: I = Δ Q/Δ t.

At a junction in steady state, charge cannot accumulate (otherwise the junction would become increasingly charged).

So the rate charge enters must equal the rate charge leaves:

∑ Iᵢₙ = ∑ Iₒᵤₜ

B. Reading a junction diagram

Four currents at a junctionCurrents I1 and I4 point into one junction, while currents I2 and I3 point away. Charge conservation gives I1 plus I4 equals I2 plus I3.I₁I₂I₃I₄I₁ and I₄ enter; I₂ and I₃ leavesum entering = sum leaving
At a junction: sum of currents in equals sum of currents out.

For example, if I₁ and I₄ enter the junction while I₂ and I₃ leave, then:

I₁ + I₄ = I₂ + I₃

4. Common Mistakes

  • Mixing up “in” and “out” currents at the same junction.
  • Writing multiple junction equations that are not independent (wastes time).
  • Confusing conventional current direction with electron drift direction (Kirchhoff’s laws use conventional current).

5. Exam Tips

  • If the question gives you current directions, use them. If not, choose directions; a negative answer means the real direction is opposite.
  • Use Kirchhoff’s first law together with:
    • Kirchhoff’s second law (loops) and
    • potential divider ideas for multi-step circuit questions.

6. Worked Examples

Modelled example 1

Simple junction

Core

Problem

At a junction, I₁ = 2.5 A enters and I₂ = 1.1 A leaves along one branch. Find the current I₃ leaving along the other branch.
Study the worked solution
  1. Sort currents by direction

    Method

    I₁ enters; I₂ and I₃ leave.

    Reason

    Charge does not accumulate at the steady-state junction.

    Working

    ∑ Iᵢₙ = ∑ Iₒᵤₜ
  2. Write the junction equation

    Method

    I₁ = I₂ + I₃.

    Reason

    The total current entering equals the total current leaving.

    Working

    2.5 = 1.1 + I₃
  3. Calculate

    Method

    I₃ = 1.4 A leaving the junction.

    Reason

    The positive result agrees with the stated direction.

    Working

    I₃ = 2.5-1.1 = 1.4 A

Guided practice 2

Unknown direction

About 4 min

Problem

At a junction, 0.80 A and 0.50 A enter, while 1.50 A leaves. A fourth current I₄ is assumed to leave. Calculate I₄ and determine its actual direction.

Try this before viewing the solution

Unit: A
Actual direction of I₄

Hints

Hint 1: keep the assumed arrow
Start with 0.80 + 0.50 = 1.50 + I₄ and let the sign test the direction.
View solution step by step
  1. Use the assumed direction

    Method

    Place I₄ with the leaving currents.

    Reason

    An arbitrary current direction is valid if its sign is interpreted afterward.

    Working

    0.80 + 0.50 = 1.50 + I₄
  2. Solve algebraically

    Method

    I₄ = -0.20 A relative to the assumed arrow.

    Reason

    The known outgoing current exceeds the two known incoming currents.

    Working

    I₄ = 0.80 + 0.50-1.50 = -0.20 A
  3. Interpret the sign

    Method

    I₄ actually enters with magnitude 0.20 A.

    Reason

    The negative sign reverses the assumed leaving direction.

    Working

    I₄ = 0.20 A entering

Common misconception 3

Find two unknown branch currents

Find and correct the mistake

Learner claim

At a junction, I₁ = 3.2 A enters while I₂ = 0.60 A and I₃ leave. A learner calculates I₃ = 3.2 + 0.60 = 3.8 A. Diagnose the operation and find I₃.

Try this before viewing the solution

Unit: A

View solution step by step
  1. Write conservation before choosing an operation

    Method

    I₁ = I₂ + I₃.

    Reason

    One incoming branch supplies both outgoing branches.

    Working

    3.2 = 0.60 + I₃
  2. Isolate the unknown branch

    Method

    I₃ = 2.6 A.

    Reason

    Subtract the known outgoing current from the total entering current.

    Working

    I₃ = 3.2-0.60 = 2.6 A

Examiner practice 4

Multiple in and out (algebraic sum)

3 marks

Examination question

At a junction, currents I₁ = 1.5 A and I₂ = 0.80 A enter. Currents I₃ = 0.40 A, I₄ = 0.70 A and I₅ leave. Find I₅. [3 marks]

Try this before viewing the solution

Unit: A

View solution step by step
  1. Write the junction equation

    1 mark

    Method

    I₁ + I₂ = I₃ + I₄ + I₅.

    Reason

    Total current entering equals total current leaving.

    Working

    1.5 + 0.80 = 0.40 + 0.70 + I₅
  2. Rearrange

    1 mark

    Method

    Subtract the two known outgoing currents.

    Reason

    This isolates the remaining branch current.

    Working

    I₅ = 1.5 + 0.80-0.40-0.70
  3. State the result

    1 mark

    Method

    I₅ = 1.20 A leaving.

    Reason

    The positive value agrees with the stated direction.

    Working

    I₅ = 1.20 A

Challenge 5

Checking a stated current direction

Minimal support

Independent transfer

At a junction, you assume I₁ enters and I₂ leaves. Your algebra gives I₂ = -0.30 A. State the magnitude and actual direction of I₂.

Try this before viewing the solution

Unit: A
Actual direction

Hints

Hint 1: separate sign and magnitude
The sign refers to the chosen reference arrow; the magnitude is |I₂|.
View solution step by step
  1. Read the algebraic sign

    Method

    The negative sign reverses the assumed arrow.

    Reason

    The variable was defined positive for a current leaving the junction.

    Working

    I₂ = -0.30 A relative to “leaving”
  2. State the physical current

    Method

    I₂ has magnitude 0.30 A and enters the junction.

    Reason

    Physical magnitude is non-negative; direction carries the sign information.

    Working

    |I₂| = 0.30 A entering

7. Mind Stretchers

Mind stretcher 1: Interpret the sign of an unknown currentExtension

At a junction, I₁ and I₂ enter and I₃, I₄, I₅ leave. Given I₁ = 3.0 A, I₂ = 0.80 A, I₃ = 1.2 A, and I₄ = 1.9 A, find I₅ and interpret the sign.

Show Answer

Apply the junction rule: I₁ + I₂ = I₃ + I₄ + I₅ I₅ = 3.0 + 0.80-1.2-1.9 = 0.70 A

I₅ is positive, so it really leaves the junction as assumed.

Mind stretcher 2: Why only one junction equation is often enoughExtension

In a circuit with several junctions, explain why many junction equations are not independent.

Show Answer

In a complete circuit, junction currents are linked by the branch currents. Writing junction equations at multiple junctions often repeats the same information in a different form.

Usually, only a minimal set of junction equations (often just one) is independent; the rest can be obtained by adding/subtracting others.

Mind stretcher 3: Optional (Enrichment)Extension

A. Video walkthrough

B. When the junction rule can break down (beyond this model)

Kirchhoff’s first law assumes no net charge accumulation at a junction (steady-state / quasi-static circuits).

In very rapid transients or high-frequency situations, charge can temporarily accumulate (e.g. due to stray capacitance), so the simple “current in = current out instantly” picture can fail without a more complete electromagnetic treatment.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027