Explain junction current balance using charge conservation
Connect equal steady current sums to no net charge accumulation and distinguish charge from energy transfer.
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
- Define current as charge per time.
- Calculate entering and leaving sums.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What charge passes at constant 0.80 A for 5 s?
What you will learn
- Explain why current sums balance.
- Interpret a supplied temporary imbalance as charge accumulation.
Convert rates into charge accounts
Current is charge per unit time. Over the same interval, equal total entering and leaving currents mean equal passed charges, so the junction does not gain net charge. In steady circuit operation persistent net accumulation is absent and the sums therefore balance.
With 0.80 A entering and 0.80 A leaving for 5 s, 4 C enters and 4 C leaves. Unequal branch currents are compatible with this account; only the sum must match. Charge does not need to vanish inside a junction to explain the split.
Recognize the assumption behind the equality
If a specified transient really has 0.50 A entering and 0.20 A leaving for 2 s, the charge balance gives a 0.60 C increase at the chosen region. It cannot represent an unchanged steady state over that interval. This example exposes the condition behind the ordinary junction rule, rather than claiming such a large accumulation is a usual circuit condition.
Loads can transfer unequal amounts of energy while charge remains conserved. A junction current balance contains no assertion of equal branch powers or equal current in every branch. Keep the charge boundary and interval explicit when explaining the physical reason.
Pause and explain
0.50 A enters and 0.20 A leaves for 2 s. What net region charge change does the supplied transient imply?
Worked example
0.80 A enters and 0.80 A leaves a junction for 5 s. Explain the net charge change.
Show the worked solution
- Entering charge is 0.80 × 5 = 4 C.
- Leaving charge is also 4 C.
- The difference is zero, so no net charge accumulates.
Answer Zero net charge change; 4 C enters and 4 C leaves.
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.
Show the passed-charge account for 0.80 A entering and 0.80 A leaving for 5 s.
Give me a hint
Use the same interval for both current sums.
Compare my reasoning
- Qin = 0.80 × 5 = 4 C.
- Qout = 0.80 × 5 = 4 C.
- ΔQ = 4 − 4 = 0 C.
4 C in, 4 C out, zero net charge change.
Look for these in your work
- Use the same boundary.
- Explain the zero accumulation.
A supplied transient has 0.50 A entering and 0.20 A leaving for 2 s. Calculate the region charge increase and assess a steady-state claim.
Give me a hint
Use the excess charge-flow rate over the stated interval.
Compare my reasoning
- The excess rate is 0.30 A.
- ΔQ = 0.30 × 2 = 0.60 C.
- The non-zero increase contradicts an unchanged steady state.
+0.60 C; this is not an unchanged steady state.
Look for these in your work
- State the sign.
- Keep the transient condition.
A learner explains unequal branch currents by saying a junction destroys charge and gives every branch equal power. Correct both statements.
Give me a hint
Distinguish local charge flow from each branch energy-transfer rate.
Compare my reasoning
- Branch currents can differ while their sums balance.
- No charge destruction is needed in the steady junction account.
- Power also depends on branch p.d. and current, so equality is not guaranteed by the junction rule.
Charge is conserved with equal current sums; equal branch power does not follow.
Look for these in your work
- Explain conservation.
- Separate current balance and power.
Common mistakes
- Equal sums do not require equal branch currents.
- Charge conservation is not an equal-power rule.
- An imbalance needs an accumulating-charge account.
What can you explain now?
Why do steady junction current sums match?
Compare with the explanation
There is no net accumulation of charge.
Persistent unmatched charge-flow rates would change the junction charge.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain steady junction balance through equal passed charges and no net accumulation.
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.