Electric circuits · System
Current at a junction
Three ammeters round one split in the wire. Two of the readings always add up to the third — whatever you put in the branches, and however lopsided the split.
Start here
A river splits round an island.
A river meets an island and divides. One channel is wide and deep, the other narrow and shallow, so far more water goes one way than the other. Below the island the two channels rejoin.
How much water flows below the island, compared with above it?
The same amount, because the island neither swallows water nor makes any. The split is lopsided — far more goes down the easy channel — but the two add back to exactly what arrived. A junction in a wire behaves the same way, and for the same reason: nothing is stored at a point.
A junction is any point where a wire divides, or where two wires meet. Nothing is stored there and nothing is made there: the charge arriving has nowhere to go except out along the branches. So the current going in equals the total of the currents coming out.
That is not the same as saying the branches get equal shares. Each branch draws whatever it draws — a lamp takes more than a buzzer, an empty branch takes nothing — and the main wire simply carries the sum. Which is why the split is usually lopsided, and why you cannot guess a branch reading by halving the main one.
The rule works both ways round a parallel section. Where the branches divide, the main current shares out; where they rejoin, the branch currents add back together. The two junctions of one parallel section always carry the same total.
At the bench · one 3.0 V battery, two branches, three ammeters
Change what is in the branches.
Change a control to begin
The two branches can each hold a lamp, a resistor, a buzzer, or nothing at all. Every arrangement gives three readings, and they are always related the same way.
Commit first. A lamp draws 0.30 A and a buzzer draws 0.10 A. You put the lamp in one branch and the buzzer in the other. What does the ammeter in the main wire read?
In branch A
In branch B
Branch A carries
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—
Branch B carries
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—
The main wire carries
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—
How it splits
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Writing it down · the shape of this relationship
The current arriving at a junction is the current leaving it
Three branches: I = a + b + c
I · current in the main wire · A
a · current in the first branch · A
b · current in the second branch · A
Worked example · one step at a time
The main wire into a junction reads 0.45 A. One branch reads 0.15 A. What does the other branch carry?
Step 0 of 5
Convert
0.45 A stays 0.45 A · 0.15 A stays 0.15 A
Both meters already read in amps, so there is nothing to convert.
Formula
b = I − a
Cover b on the bar: the whole with the other part taken away.
Insert
b = 0.45 A − 0.15 A
The whole is the main wire, 0.45 A. The part you know is the first branch, 0.15 A.
Fine-tune
0.45 − 0.15 = 0.30
Amps take away amps leaves amps.
Answer
b = 0.30 A
Check it: 0.15 A and 0.30 A add to the 0.45 A in the main wire.
Worked example · one step at a time
The main wire into a junction reads 1.20 A. One branch reads 250 mA. What does the other branch carry?
Step 0 of 5
Convert
250 mA ÷ 1000 = 0.250 A
There are 1000 milliamps in an amp, so divide before you take anything away.
Formula
b = I − a
Cover b on the bar: the whole with the other part taken away.
Insert
b = 1.20 A − 0.250 A
The converted branch reading goes in. The milliamp reading never does.
Fine-tune
1.20 − 0.250 = 0.950
Amps take away amps leaves amps.
Answer
b = 0.950 A
Take 250 from 1.20 instead and you get a negative current, which no ammeter ever reads.
Your turn · the same five steps
Your junction: the main wire reads 0.40 A and branch A reads 0.30 A.
Write each line out yourself — starting by deciding whether anything needs converting. Then check your working and tick the lines you had.
The five lines, marked
Convert
0.40 A stays 0.40 A · 0.30 A stays 0.30 A
Both meters already read in amps, so nothing changes.
Formula
b = I − a
Cover b on the bar: the whole with the other part taken away.
Insert
b = 0.40 A − 0.30 A
The whole is the main wire; the part you know is branch A.
Fine-tune
0.40 − 0.30 = 0.10
Amps take away amps leaves amps.
Answer
b = 0.10 A
Check it against the meter in branch B, which reads 0.10 A.
The five lines give 0.10 A for branch B, and 0.30 A + 0.10 A is the 0.40 A in the main wire.
A junction where the main wire reads 0.80 A and branch A reads 320 mA.
This one needs the Convert line to do some work.
The five lines, marked
Convert
320 mA ÷ 1000 = 0.320 A
There are 1000 milliamps in an amp, so divide before you take anything away.
Formula
b = I − a
Cover b on the bar: the whole with the other part taken away.
Insert
b = 0.80 A − 0.320 A
The converted branch reading goes in. The milliamp reading never does.
Fine-tune
0.80 − 0.320 = 0.480
Amps take away amps leaves amps.
Answer
b = 0.480 A
Take 320 from 0.80 instead and you get a negative current, which no ammeter ever reads.
The five lines give 0.480 A in branch B, and 0.320 A + 0.480 A is the 0.80 A in the main wire.
Key fact
Charge is neither made nor stored at a junction, so the currents leaving add up to the current arriving: I = a + b. The branches do not get equal shares — each draws what it draws — and the same total passes both junctions of a parallel section.
Think again
“At a junction the current halves, because it has two ways to go.”
Only if the two branches happen to be identical. Put a lamp in one branch and a buzzer in the other and the split is 0.30 A against 0.10 A — three quarters of the charge takes the easier route. The junction does not divide anything up; it simply lets each branch take what it takes, and the main wire carries whatever that adds to. Halving is a special case, not the rule.
“Adding a second branch means less current for the first one.”
The first branch does not notice. It has the same battery across it as before, so it draws the same current as before, and the meter in it does not move when you add or remove the other branch. What changes is the main wire, which now has to carry both branch currents — so the battery works harder and goes flat sooner. It is the supply that pays, not the neighbouring branch.
Mastery ladder
Not started yet.
Rungs 3 and 4 you mark yourself.
Rung 1 · Calculate
A junction splits into three branches. The main wire carries 0.85 A, the first branch 0.30 A and the second 0.40 A. What does the third branch carry?
Rung 2 · The one that catches people
Two lamps are in parallel on a battery, each drawing 0.30 A. A student unscrews one and predicts the other will now draw 0.60 A, because it gets all the current to itself. What is right?
Rung 3 · Explain
A lamp and a buzzer are in parallel on the same battery. The lamp branch carries 0.30 A and the buzzer branch 0.10 A. Explain what the ammeter in the main wire reads, and why the split is not even.
Rung 4 · Take it somewhere new
An extension lead is marked "maximum 13 A". Someone plugs in a heater drawing 9 A, a kettle drawing 11 A and a lamp drawing 0.3 A, and points out that no single appliance is over 13 A. Explain, using the junction rule, why this is dangerous and what will happen.
Key note
A junction is a point where a wire divides or two wires meet. Charge is not made or stored there, so the currents leaving add up to the current arriving: I = a + b. The branches are not given equal shares — each draws its own current, so an easier branch takes more — and the same total passes the junction where they divide and the junction where they rejoin. Adding a branch does not steal current from the others; it makes the main wire and the battery carry more.
Going further
This rule has a name at A level: Kirchhoff's first law, published by Gustav Kirchhoff in 1845 when he was twenty-one. It is not really a law about electricity at all — it is the conservation of charge, applied to a point. Charge cannot be created or destroyed, and a junction has no room to keep any, so the books must balance every instant.
Engineers use the sum in the direction you might not expect: backwards, to size a cable. A ring main feeding a kitchen has to carry the kettle, the toaster, the fridge and the lights added together, even though each appliance only knows about itself. Underrate that main cable and it is the wire that overheats, not the appliance — which is why the fuse or breaker protecting a circuit is chosen for the total, and why adding one more heater to a loaded extension lead is the moment it becomes dangerous.
Before this lesson
Connects to
At GCSE this becomes
- Kirchhoff's current law, used with the potential difference rules to solve circuits with several loops.
Where to next
Ask Mr Badmus AI
Got two of the three readings and want to find the missing one?
The bench is a teaching model. Each component is treated as having a fixed resistance on a 3.0 V supply — lamp 10 ohms, resistor 15 ohms, buzzer 30 ohms — giving 0.30 A, 0.20 A and 0.10 A; a real filament lamp and a real buzzer both change as they run. The battery and the connecting wires are treated as having no resistance, which is why the branch readings do not sag when the other branch is added. Readings are rounded to two decimal places. Conventional current is drawn from + to − by long-standing convention; the electrons in a metal drift the other way.
Lesson content © MrBadmusAI.