Calculate missing currents at a circuit junction
Balance all entering and leaving currents with explicit arrow directions and consistent units.
Section 4.3.2 · Series and parallel circuits
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Read current arrows and milliampere units.
- Add and subtract signed values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 600 mA in amperes?
What you will learn
- Use the entering-equals-leaving current balance.
- Interpret a negative inferred reference current.
Account for every connection
At a steady circuit junction, the sum of currents entering equals the sum leaving. First draw arrows or read the supplied arrows, then put every entering value on one side and every leaving value on the other. With 0.60 A entering and 0.20 A plus an unknown leaving, the unknown is 0.40 A.
More than one current can enter. If 0.30 A and 0.50 A enter while 0.20 A and I leave, 0.30 + 0.50 = 0.20 + I, giving I = 0.60 A. Page orientation is not the direction rule; arrows define entry and exit.
Interpret signs and units
Convert prefixed units before balancing. For 600 mA entering with 0.20 A leaving, use 0.60 A or convert both to milliamperes. A negative solution relative to an assumed outward arrow means the real current is inward with that magnitude; it does not mean a negative number of carriers.
The balance assumes steady charge flow with no accumulating charge at the junction. It is a local current account and does not determine every branch resistance or voltage by itself. Check that the given current directions can describe the stated circuit before making further claims.
Pause and explain
0.30 A and 0.50 A enter; 0.20 A and I leave. What is I?
Worked example
A junction has 0.30 A and 0.50 A entering, with 0.20 A and I leaving. Calculate I.
Show the worked solution
- Total entering is 0.80 A.
- 0.80 = 0.20 + I.
- I = 0.60 A leaving.
Answer 0.60 A leaving.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A junction has 0.60 A entering and 0.20 A plus I leaving. Find I.
Give me a hint
Write the complete entering and leaving sums.
Compare my reasoning
- 0.60 = 0.20 + I.
- Subtract 0.20 A from both sides.
- I = 0.40 A leaving.
0.40 A leaving.
Look for these in your work
- Use the arrow directions.
- State amperes.
300 mA and 0.50 A enter; 0.20 A and I leave. Find I.
Give me a hint
Convert 300 mA before adding currents.
Compare my reasoning
- 300 mA = 0.30 A.
- 0.30 + 0.50 = 0.20 + I.
- I = 0.60 A leaving.
0.60 A leaving.
Look for these in your work
- Convert units.
- Include both entering currents.
A junction has 0.20 A entering and 0.50 A plus an assumed outward I leaving. Solve and interpret I.
Give me a hint
Keep the assumed arrow until the signed calculation is complete.
Compare my reasoning
- 0.20 = 0.50 + I.
- I = −0.30 A relative to the outward reference.
- The actual current is 0.30 A inward.
−0.30 A on the outward reference, meaning 0.30 A inward.
Look for these in your work
- Give the signed result.
- Interpret the reversed direction.
Common mistakes
- Use arrows, not page positions.
- Convert mA and A first.
- A negative reference result reverses the assumed direction.
What can you explain now?
What steady current equality applies at a junction?
Compare with the explanation
Sum entering equals sum leaving.
The equality accounts for all connected current paths at that junction.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Balance every entering and leaving current, then interpret the sign on the stated arrow.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.