Kirchhoff's first law
Apply Kirchhoff’s first law (sum of currents at a junction is zero) to solve circuit junction current problems.
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ContinueWhere a circuit branches, the current divides; where branches meet, the currents combine. Kirchhoff’s first law states how the currents at a junction are related. It follows directly from conservation of charge.
Kirchhoff’s laws are not explicitly listed in the H2 syllabus, but they are a useful (and widely used) method for analysing more complex D.C. circuits.
The junction rule
Kirchhoff’s first law states that at any junction, the total current entering equals the total current leaving:
Current is the rate of flow of charge, I = Δ Q/Δ t. In a steady state, charge cannot build up at a junction; otherwise the junction would become more and more charged. So charge must leave at the same rate as it arrives.
Worked example 1
Simple junction
Problem
Show full solution
Sort currents by direction
Method
I₁ enters; I₂ and I₃ leave.Reason
Charge does not accumulate at the steady-state junction.Working
Answer
∑ Iᵢₙ = ∑ IₒᵤₜWrite the junction equation
Method
I₁ = I₂ + I₃.Reason
The total current entering equals the total current leaving.Working
Answer
2.5 = 1.1 + I₃Calculate
Method
I₃ = 1.4 A leaving the junction.Reason
The positive result agrees with the stated direction.Working
Answer
I₃ = 2.5-1.1 = 1.4 A
Quick check
Write the equation from a diagram
If I₁ and I₄ enter the junction while I₂ and I₃ leave, then
Equivalently, count currents entering as positive and currents leaving as negative. The algebraic sum of the currents at the junction is then zero:
The current is the same everywhere along a series branch, but it can split between parallel branches. Kirchhoff’s laws use conventional current, not the direction of electron drift.
Check your understanding 1 Read arrows before arithmetic
On paper sketch a junction with I₁ and I₄ entering and I₂ and I₃ leaving. Write its equation in both “in equals out” and signed-sum forms. Explain why the rule follows from charge conservation, and why current is not used up in the junction.
Show answer
Compare your arrows and equations with the displayed junction: I₁ + I₄ = I₂ + I₃ and I₁ + I₄-I₂-I₃ = 0. Each current is a rate of charge flow. With negligible charge accumulation at the node, equal charge must enter and leave per second.
Spot the mistake 2
Find a missing outgoing branch current
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.
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Show solution step by step
Write conservation before choosing an operation
Method
I₁ = I₂ + I₃.Reason
One incoming branch supplies both outgoing branches.
Working
Answer
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
Answer
I₃ = 3.2-0.60 = 2.6 A
Common mistake
Adding 0.60 A would make the outgoing total exceed the current entering the junction.
Exam-style question 1
Several currents in and out
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]
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Show solution step by step
Write the junction equation
1 markMethod
I₁ + I₂ = I₃ + I₄ + I₅.Reason
Total current entering equals total current leaving.
Working
Answer
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
Answer
I₅ = 1.5 + 0.80-0.40-0.70State the result
1 markMethod
I₅ = 1.2 A leaving.Reason
The positive value agrees with the stated direction.
Working
Answer
I₅ = 1.2 A
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Self-mark the junction equation, rearrangement and stated current.
Unknown current directions
If a question shows the current directions, use them. If not, choose a direction for each unknown current and write the junction equation. A negative answer means the current flows opposite to your arrow. Report its size as a positive magnitude with the actual direction.
Guided practice 3
Unknown direction
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 the magnitude of I₄ and determine its actual direction. Keep the signed equation in your working.
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Hints
Hint 1
keep the assumed arrow
Which currents enter and which leave when you keep the assumed arrow?
Show 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
Answer
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
Answer
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
Answer
I₄ = 0.20 A entering
Quick check
Incoming current is then 0.80 + 0.50 + 0.20 = 1.50 A, equal to the outgoing current.
Try it yourself 4
Checking a stated current direction
Problem
At a junction, 1.60 A enters while 0.70 A and 1.40 A leave. Assume the fourth current I₂ leaves. Calculate its magnitude and actual direction.
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Hints
Hint 1
separate sign and magnitude
The sign refers to the chosen reference arrow; the magnitude is |I₂|.
Show 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
Answer
1.60 = 0.70 + 1.40 + I₂, so I₂ = -0.50 A relative to “leaving”.
State the physical current
Method
I₂ has magnitude 0.50 A and enters the junction.
Reason
Physical magnitude is non-negative; direction carries the sign information.
Working
Answer
|I₂| = 0.50 A entering; 1.60 + 0.50 = 0.70 + 1.40.
Quick check
Do not report a “negative magnitude”; report the positive magnitude with the reversed direction.
Circuits with several junctions
A larger circuit has a junction equation at every junction, but these equations are not all independent. Kirchhoff’s second law supplies the other equations, from the potential differences around each loop.
Check your understanding 2 Why only one junction equation is often enough
A two-node circuit has three branches. At node A, I₁ enters and I₂ and I₃ leave. Write the equations at A and B. Does the second equation provide another independent relation?
Show answer
At A, I₁ = I₂ + I₃. Each branch connects the same two nodes, so the directions reverse at B: I₂ + I₃ = I₁. This is the same equation and adds no independent relation. Loop equations are needed to determine the separate branch currents.
Mind stretcher 1 When charge builds up at a nodeExtension
Charge is always conserved. Why can the simple junction equation still fail for a very rapidly changing current?
Show answer
The full statement of charge conservation at a node is
The usual junction equation assumes that the net charge stored at the node is negligible. For very rapid changes, charge can build up on stray capacitance, so that capacitance must be included in the circuit model. This changes which branch currents you count; it does not make charge conservation fail.
Common mistakes
- Mixing up the currents entering and leaving the same junction.
- Adding a known outgoing current to the incoming current instead of subtracting it.
- Reporting a negative magnitude. Give the positive size and reverse the direction.
- Writing junction equations that are not independent, which wastes time and gives no new information.
- Using the electron drift direction instead of conventional current.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30