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).
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The core idea
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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.
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.
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
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
Problem
Study the worked solution
Sort currents by direction
Method
I₁ enters; I₂ and I₃ leave.Reason
Charge does not accumulate at the steady-state junction.Working
∑ Iᵢₙ = ∑ IₒᵤₜWrite the junction equation
Method
I₁ = I₂ + I₃.Reason
The total current entering equals the total current leaving.Working
2.5 = 1.1 + I₃Calculate
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
Problem
Try this before viewing the solution
Hints
Hint 1: keep the assumed arrow
View solution step by step
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 + I₄Solve 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 AInterpret 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
Learner claim
Try this before viewing the solution
View solution step by step
Write conservation before choosing an operation
Method
I₁ = I₂ + I₃.Reason
One incoming branch supplies both outgoing branches.Working
3.2 = 0.60 + I₃Isolate 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)
Examination question
Try this before viewing the solution
View solution step by step
Write the junction equation
1 markMethod
I₁ + I₂ = I₃ + I₄ + I₅.Reason
Total current entering equals total current leaving.Working
1.5 + 0.80 = 0.40 + 0.70 + I₅Rearrange
1 markMethod
Subtract the two known outgoing currents.Reason
This isolates the remaining branch current.Working
I₅ = 1.5 + 0.80-0.40-0.70State the result
1 markMethod
I₅ = 1.20 A leaving.Reason
The positive value agrees with the stated direction.Working
I₅ = 1.20 A
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the junction equation, rearrangement and stated current.
Challenge 5
Checking a stated current direction
Independent transfer
Try this before viewing the solution
Hints
Hint 1: separate sign and magnitude
View solution step by step
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”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.
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Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027