3.7 - Kirchhoff’s Junction Rule

The Junction Rule

Kirchhoff’s junction rule comes straight out of the conservation of electric charge.

A junction (also called a node) is a point where three or more wires meet, so it’s a spot where current can either split up or merge back together.

Statement of the Junction Rule

The total charge flowing into a junction per unit time has to equal the total charge flowing out of it per unit time.

Since current is just the rate of charge flow (

I = \frac{\Delta q}{\Delta t}

), that means:

\sum I_{\text{in}} = \sum I_{\text{out}}

or written out:

I_1 + I_2 + \ldots = I_3 + I_4 + \ldots

where the currents on the left are coming in and the ones on the right are heading out.

The junction rule isn’t printed on the AP equation sheet, so know it from the idea behind it: charge conservation.

Another Way to Write It

You can also state it as:

\sum I = 0

where currents coming in count as positive and currents going out count as negative (or the other way around, as long as you’re consistent).

What It Means

Charge can’t pile up or vanish at a junction, since charge is conserved. Whatever flows in has to flow back out.

Think of water at a pipe junction. The rate of water coming in has to match the rate going out, otherwise water would either stack up or disappear.

Current isn’t used up. A bulb or resistor converts electric potential energy into other forms, but it doesn’t destroy charge. The current going into a bulb equals the current coming out of it. What drops across the bulb is the electric potential, not the current.

How It Works with the Loop Rule

Kirchhoff’s two rules together let you crack complex circuits:

Junction rule (from charge conservation): connects the currents at branching points.

Loop rule (from energy conservation): connects the voltages around closed loops.

For circuits with several loops and junctions, you usually need junction-rule equations at each independent junction plus loop-rule equations for each independent loop, then you combine them all to solve for the unknown currents and voltages.

Example: a 12 V battery is connected to three resistors in parallel:

R_1 = 6.0

Ω,

R_2 = 4.0

Ω, and an unknown

R_3

. An ammeter next to the battery reads 6.0 A. Find

R_3

Loop rule first: each branch makes a loop with the battery, so each resistor has the full 12 V across it.

I_1 = \frac{12 \text{ V}}{6.0\ \Omega} = 2.0 \text{ A}, \qquad I_2 = \frac{12 \text{ V}}{4.0\ \Omega} = 3.0 \text{ A}

Then the junction rule where the battery current splits:

I_{\text{battery}} = I_1 + I_2 + I_3 \quad\Rightarrow\quad 6.0 \text{ A} = 2.0 \text{ A} + 3.0 \text{ A} + I_3 \quad\Rightarrow\quad I_3 = 1.0 \text{ A}

R_3 = \frac{\Delta V}{I_3} = \frac{12 \text{ V}}{1.0 \text{ A}} = 12\ \Omega

Answer:

R_3 = 12

Ω. Check:

\frac{1}{R_{\text{eq}}} = \frac{1}{6.0} + \frac{1}{4.0} + \frac{1}{12} = \frac{6}{12}

, so

R_{\text{eq}} = 2.0

Ω and

I = \frac{12 \text{ V}}{2.0\ \Omega} = 6.0

A, matching the ammeter.

Practice Problems

1. Three wires meet at a junction. Current

I_1 = 2.0

A comes in, and current

I_2 = 0.5

A also comes in. Current

I_3

leaves. Find

I_3

2. Four wires meet at a junction. Currents

I_1 = 3.0

A and

I_2 = 1.5

A come in, while currents

I_3 = 2.0

A and

I_4

leave. Find

I_4

3. At a junction, two currents come in:

I_1

and

I_2 = 0.8

A. Three currents leave:

I_3 = 0.5

I_4 = 0.6

A, and

I_5 = 0.9

A. Find

I_1

4. In a circuit, current

I_1 = 4.0

A comes out of the battery. It splits at a junction into two parallel branches. If one branch carries

I_2 = 1.5

A, what’s the current

I_3

through the other branch? After the branches merge again at another junction, what’s the current heading back to the battery?

5. A junction has five wires. If you know the currents in four of them, explain why you can always figure out the fifth.

6. At a junction, a student measures three currents coming in (1.2 A, 0.8 A, and 1.5 A) and two currents leaving (2.0 A and 1.0 A). The student thinks something’s off. What might have gone wrong?

Solutions

1. Apply the junction rule:

\sum I_{\text{in}} = \sum I_{\text{out}}

I_1 + I_2 = I_3

2.0 + 0.5 = I_3

I_3 = 2.5 \text{ A}

Answer:

I_3 = 2.5

A leaves the junction.

2. Apply the junction rule:

\sum I_{\text{in}} = \sum I_{\text{out}}

I_1 + I_2 = I_3 + I_4

3.0 + 1.5 = 2.0 + I_4

4.5 = 2.0 + I_4

I_4 = 2.5 \text{ A}

Answer:

I_4 = 2.5

A leaves the junction.

3. Apply the junction rule:

\sum I_{\text{in}} = \sum I_{\text{out}}

I_1 + I_2 = I_3 + I_4 + I_5

I_1 + 0.8 = 0.5 + 0.6 + 0.9

I_1 + 0.8 = 2.0

I_1 = 1.2 \text{ A}

Answer:

I_1 = 1.2

A enters the junction.

4. At the first junction, where the current splits:

I_1 = I_2 + I_3

4.0 = 1.5 + I_3

I_3 = 2.5 \text{ A}

Answer: the other branch carries

I_3 = 2.5

At the second junction, where the currents merge back:

I_2 + I_3 = I_{\text{return}}

1.5 + 2.5 = 4.0 \text{ A}

Answer: the current going back to the battery is 4.0 A, the same as the current that left it, exactly what you’d expect for a single-loop equivalent.

5. The junction rule,

\sum I_{\text{in}} = \sum I_{\text{out}}

, is one equation. With five wires meeting at a junction you have five currents but only that one independent equation.

So if you know four of the five currents (both size and direction), the junction rule pins down the fifth. That last current is fully locked in by charge conservation, since it has to be whatever value makes the total in equal the total out.

In general, at a junction with

n

wires, knowing

n-1

of the currents lets you solve for the

n

-th.

6. Check whether the junction rule holds:

Total current in:

1.2 + 0.8 + 1.5 = 3.5

Total current out:

2.0 + 1.0 = 3.0

Since

3.5 \neq 3.0

, the junction rule is broken.

Possible explanations: - Measurement error: one or more ammeters gave a bad reading. - Wrong current direction: the student may have mixed up which currents were coming in versus going out. - Non-ideal ammeter effects: if the ammeters have resistance, they could be tweaking the circuit. - A missed wire: there might be another wire at the junction that wasn’t measured. - Arithmetic slip: a plain calculation mistake.

The most likely culprit is a measurement error or a mislabeled current direction. The junction rule has to hold, since charge conservation is a fundamental law.

Common Mistakes

Confusing the directions. Carefully figure out which currents come in and which go out. Drawing arrows on the diagram helps. A wrongly labeled direction throws off the whole answer.

Sign errors. Stay consistent with your sign convention. Either go with “in is positive, out is negative,” or keep “in” on one side of the equation and “out” on the other.

Using it at the wrong spots. The junction rule only applies where three or more wires meet. It doesn’t apply to a plain two-wire series connection.

Missing junctions. In complex circuits, hunt down every point where current splits or merges. A missed junction means a missed equation.

Assuming the current is the same everywhere. Current only stays the same in the series parts of a circuit. At junctions it splits, so different branches carry different amounts.

Forgetting to use it. When solving multi-loop circuits, people often remember the loop rule but forget the junction rule, which leaves them short on equations.

Double-counting junctions. In a circuit with several junctions, not every junction equation is independent. For a circuit with $n$ junctions, only $n-1$ give you independent equations.

Breaking charge conservation. If your calculated currents don’t satisfy $\sum I_{\text{in}} = \sum I_{\text{out}}$ at every junction, you’ve made a mistake somewhere. Charge conservation is absolute.

Mixing it up with the loop rule. The junction rule deals with currents at a point. The loop rule deals with voltages around a closed path. Keep them straight.

Units. Current is always in amperes (A). Sanity-check your answer, since a negative current just means your assumed direction was backwards.

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