Junction Rule (Current Law)

Kirchhoff's Current Law

When current reaches a junction in a circuit, it may split into two or more branches. When branches meet, currents may combine again.

Kirchhoff's Current Law (KCL) states:

The total current entering a junction equals the total current leaving the junction.

In simple form:

I_in = I_out

For several currents:

I₁ + I₂ + ... = I₃ + I₄ + ...

where the currents on one side enter the junction and the currents on the other side leave it.

This rule is also called:

• Kirchhoff's Current Law
• Kirchhoff's First Law
• the Junction Rule
• the Current Law

---

What Is a Junction?

A junction is a point in a circuit where three or more conducting paths meet.

At a junction, current may:

• divide into several branches
• combine from several branches
• both enter and leave through multiple paths

For example, suppose 5 A reaches a junction and divides into two branches.

If one branch carries 2 A, the other must carry:

5 = 2 + I

Therefore:

I = 3 A

---

Conservation of Charge

Kirchhoff's Current Law is based on the conservation of electric charge.

Charge cannot simply disappear when it reaches a junction, nor can extra charge suddenly appear.

Current is defined as:

I = Q/t

where:

• I = current (A)
• Q = charge (C)
• t = time (s)

Therefore, current represents the rate at which charge flows.

---

Understanding Conservation of Charge

Suppose a current of 6 A enters a junction.

Since:

1 A = 1 C/s

this means:

6 C of charge arrive each second.

If one branch carries:

2 A = 2 C/s

and another carries:

4 A = 4 C/s

then:

2 C/s + 4 C/s = 6 C/s

The amount of charge leaving each second equals the amount arriving.

Therefore:

charge is conserved.

---

Why Current Does Not Disappear

A common misconception is that some current is "used up" by components.

Current is not used up.

Electrical components transfer energy, but electric charge continues moving through the circuit.

At a junction:

charge entering = charge leaving

and therefore:

current entering = current leaving

This is why Kirchhoff's Current Law works.

---

A Simple Junction

Suppose:

I₁ = 8 A enters a junction.

Two currents leave:

I₂ = 3 A

I₃ = ?

Using KCL:

I₁ = I₂ + I₃

8 = 3 + I₃

Therefore:

I₃ = 5 A

---

Several Currents Leaving

Suppose 12 A enters a junction and three currents leave.

I₁ = 12 A

I₂ = 3 A

I₃ = 4 A

I₄ = ?

Apply KCL:

12 = 3 + 4 + I₄

12 = 7 + I₄

Therefore:

I₄ = 5 A

Check:

3 + 4 + 5 = 12 A

---

Several Currents Entering

Kirchhoff's Current Law also applies when several currents enter the same junction.

Suppose:

I₁ = 4 A enters

I₂ = 3 A enters

I₃ leaves

Then:

I₁ + I₂ = I₃

4 + 3 = I₃

Therefore:

I₃ = 7 A

The total current leaving is 7 A.

---

Several Currents Entering and Leaving

A junction can have several currents in both directions.

Suppose:

Entering:

I₁ = 6 A

I₂ = 4 A

Leaving:

I₃ = 3 A

I₄ = ?

Use:

total entering = total leaving

6 + 4 = 3 + I₄

10 = 3 + I₄

Therefore:

I₄ = 7 A

---

Current at a Branching Point

Consider a parallel circuit.

Before the junction:

I_total

At the junction, the current divides:

I_total = I₁ + I₂

If there are three branches:

I_total = I₁ + I₂ + I₃

When the branches recombine:

I₁ + I₂ + I₃ = I_total

This is why the current entering a parallel network equals the sum of the branch currents.

---

Example: Parallel Branches

A total current of 10 A reaches a junction.

The branch currents are:

I₁ = 2 A

I₂ = 3 A

I₃ = ?

Use:

I_total = I₁ + I₂ + I₃

10 = 2 + 3 + I₃

10 = 5 + I₃

Therefore:

I₃ = 5 A

---

Current Recombining

Suppose three branch currents recombine:

I₁ = 1.5 A

I₂ = 2.0 A

I₃ = 0.5 A

The current after the junction is:

I_total = 1.5 + 2.0 + 0.5

I_total = 4.0 A

The total current has been restored after the branches recombine.

---

Current Does Not Have to Divide Equally

Kirchhoff's Current Law tells us that the currents must add correctly.

It does not say that current divides equally.

Suppose:

I_total = 9 A

Possible branch currents include:

4 A and 5 A

or:

3 A and 6 A

or:

2 A, 3 A and 4 A

All satisfy:

total current entering = total current leaving

The actual current in each branch depends on factors such as branch resistance.

---

Current Division and Resistance

In parallel branches, each branch has the same voltage.

Using Ohm's Law:

I = V/R

Therefore:

lower resistance → larger current

higher resistance → smaller current

Suppose two parallel resistors are connected across 12 V:

R₁ = 4 Ω

R₂ = 6 Ω

Then:

I₁ = 12/4

I₁ = 3 A

I₂ = 12/6

I₂ = 2 A

At the junction:

I_total = 3 + 2

I_total = 5 A

Kirchhoff's Current Law is satisfied.

---

Equal Branches

If identical resistors are connected in parallel, their currents are equal.

Suppose three identical 6 Ω resistors are connected across 12 V.

Each branch current is:

I = V/R

I = 12/6

I = 2 A

Therefore:

I_total = 2 + 2 + 2

I_total = 6 A

---

Using Current Ratios

For two parallel resistors:

I₁/I₂ = R₂/R₁

Current divides inversely with resistance.

Suppose:

R₁ = 4 Ω

R₂ = 8 Ω

Then:

I₁/I₂ = 8/4

I₁/I₂ = 2

Therefore:

I₁ = 2I₂

The 4 Ω branch carries twice as much current as the 8 Ω branch.

---

Worked Current Division Problem

A total current of 6 A reaches two parallel resistors:

R₁ = 4 Ω

R₂ = 8 Ω

Since the 4 Ω branch has half the resistance, it carries twice the current.

Let:

I₂ = x

Then:

I₁ = 2x

Kirchhoff's Current Law gives:

2x + x = 6

3x = 6

x = 2 A

Therefore:

I₁ = 4 A

I₂ = 2 A

Check:

4 + 2 = 6 A

---

Using KCL with Ohm's Law

Kirchhoff's Current Law becomes especially useful when combined with:

V = IR

Suppose a 12 V supply is connected to three parallel resistors:

R₁ = 3 Ω

R₂ = 4 Ω

R₃ = 6 Ω

Calculate the branch currents.

For R₁:

I₁ = 12/3

I₁ = 4 A

For R₂:

I₂ = 12/4

I₂ = 3 A

For R₃:

I₃ = 12/6

I₃ = 2 A

Now apply KCL:

I_total = I₁ + I₂ + I₃

I_total = 4 + 3 + 2

I_total = 9 A

---

Finding an Unknown Branch Resistance

A total current of 5 A enters two parallel branches connected across 12 V.

One branch contains:

R₁ = 4 Ω

The second resistance is unknown.

First calculate I₁:

I₁ = V/R₁

I₁ = 12/4

I₁ = 3 A

Use KCL:

I_total = I₁ + I₂

5 = 3 + I₂

Therefore:

I₂ = 2 A

Now calculate R₂:

R₂ = V/I₂

R₂ = 12/2

R₂ = 6 Ω

---

Multiple Junctions

Kirchhoff's Current Law can be applied separately to every junction in a complicated circuit.

Suppose a current of 10 A reaches Junction A.

At Junction A:

3 A enters one branch.

The remaining current continues toward Junction B.

Let that current be I₁.

At Junction A:

10 = 3 + I₁

Therefore:

I₁ = 7 A

At Junction B, the 7 A divides into:

I₂ = 2 A

I₃ = ?

Therefore:

7 = 2 + I₃

I₃ = 5 A

The law can be applied one junction at a time.

---

A More Complex Junction

Consider a junction with five currents.

Entering:

I₁ = 7 A

I₂ = 5 A

Leaving:

I₃ = 4 A

I₄ = 3 A

I₅ = ?

Apply KCL:

7 + 5 = 4 + 3 + I₅

12 = 7 + I₅

Therefore:

I₅ = 5 A

Always begin by clearly identifying which currents enter and which leave.

---

Sign Convention

For more advanced circuit analysis, currents can be assigned signs.

For example:

currents entering = positive

currents leaving = negative

Then Kirchhoff's Current Law can be written as:

ΣI = 0

For example:

6 A enters

2 A leaves

4 A leaves

Then:

6 − 2 − 4 = 0

This is mathematically equivalent to:

6 = 2 + 4

Either approach is correct if the sign convention is used consistently.

---

What Does Σ Mean?

The symbol:

Σ

means sum of.

Therefore:

ΣI = 0

means that the algebraic sum of all currents at a junction is zero.

Positive and negative signs represent the chosen current directions.

This form is particularly useful for circuits containing many branches.

---

Assuming a Current Direction

Sometimes the direction of an unknown current is not obvious.

You can choose a direction and solve the equation.

For example, suppose:

5 A enters

2 A leaves

1 A leaves

and an unknown current I is initially assumed to enter.

Using the sign convention:

5 + I − 2 − 1 = 0

I + 2 = 0

I = −2 A

The negative answer tells us that the original assumed direction was wrong.

Therefore, the actual current is:

2 A leaving the junction

A negative result does not mean "negative electricity." It indicates that the current flows opposite to the direction originally assumed.

---

Junction Rule in a Complete Circuit

Consider a circuit where 8 A leaves a power supply and reaches a junction.

The current divides into:

I₁ = 5 A

I₂ = 3 A

Later the branches recombine.

At the first junction:

8 = 5 + 3

At the second junction:

5 + 3 = 8

Therefore, 8 A returns toward the source.

This demonstrates conservation of charge throughout the entire circuit.

---

Measuring Kirchhoff's Current Law

KCL can be tested experimentally using ammeters.

For a two-branch parallel circuit, measure:

• current before the junction
• current through branch 1
• current through branch 2
• current after the branches recombine

Theory predicts:

I_before = I₁ + I₂

and:

I_after = I₁ + I₂

---

Example Experimental Data

Suppose measurements give:

I_before = 0.61 A

I₁ = 0.40 A

I₂ = 0.20 A

I_after = 0.60 A

The branch currents give:

I₁ + I₂ = 0.40 + 0.20

= 0.60 A

The measurements are therefore very close to the theoretical prediction.

The small difference between 0.61 A and 0.60 A could result from measurement uncertainty.

---

Junction Rule and Charge Flow

Suppose a current of 4 A enters a junction for 5 seconds.

The charge arriving is:

Q = It

Q = 4 × 5

Q = 20 C

If one branch carries 1.5 A for the same 5 seconds:

Q₁ = 1.5 × 5

Q₁ = 7.5 C

The remaining branch must carry:

20 − 7.5

12.5 C

Therefore:

I₂ = Q₂/t

I₂ = 12.5/5

I₂ = 2.5 A

Check:

1.5 + 2.5 = 4 A

This demonstrates the direct connection between KCL and conservation of charge.

---

Steady-State Circuits

Kirchhoff's Current Law is especially straightforward in a steady-state circuit.

In steady state, charge is not continually building up at junctions.

If more current entered a junction than left it, charge would accumulate there.

If more current left than entered, the junction would continually lose charge.

In an ordinary steady circuit, this does not happen.

Therefore:

rate of charge entering = rate of charge leaving

which gives:

total current entering = total current leaving

---

Branching Circuit Problem-Solving Strategy

When solving a branching circuit:

Identify the junction

Locate where currents split or combine.

Draw current arrows

Mark the direction of each known or assumed current.

Separate entering and leaving currents

Write them clearly.

Apply KCL

Use:

total current entering = total current leaving

Solve the equation

Calculate the unknown current.

Check your answer

Make sure charge is conserved at the junction.

Repeat for additional junctions

In a multi-junction network, solve one junction at a time where possible.

---

Complete Worked Example

A current of 15 A reaches Junction A.

At Junction A:

• 4 A enters Branch 1
• the remaining current continues to Junction B

Junction A

15 = 4 + I₂

I₂ = 11 A

At Junction B, the 11 A divides into three branches:

• 2 A
• 5 A
• I₃

Apply KCL:

11 = 2 + 5 + I₃

11 = 7 + I₃

Therefore:

I₃ = 4 A

Check:

At Junction A:

4 + 11 = 15 A

At Junction B:

2 + 5 + 4 = 11 A

Both junctions satisfy Kirchhoff's Current Law.

---

Combining Junction Rules with Circuit Resistance

Suppose two parallel branches are connected across 24 V.

R₁ = 6 Ω

R₂ = 8 Ω

Calculate branch currents:

I₁ = 24/6

I₁ = 4 A

I₂ = 24/8

I₂ = 3 A

Apply KCL:

I_total = 4 + 3

I_total = 7 A

We can use this to calculate the equivalent resistance:

R_total = V/I_total

R_total = 24/7

R_total ≈ 3.43 Ω

The equivalent resistance is correctly smaller than either branch resistance.

---

Kirchhoff's Current Law and Circuit Design

The Junction Rule is important in:

• household wiring
• electronic circuits
• automotive electrical systems
• power distribution
• computer circuits
• sensor networks
• industrial electrical systems

Engineers can use KCL to determine how much current flows through different parts of a network.

It becomes especially useful when circuits are too complicated to analyze using only simple series and parallel rules.

---

Common Mistakes

Mistake: Assuming current is always divided equally.
Current divides equally only when the relevant branches have equal electrical properties.

Mistake: Adding only some of the outgoing currents.
Every branch connected to the junction must be included.

Mistake: Ignoring current direction.
Determine which currents enter and which leave before writing the equation.

Mistake: Thinking current is used up at a junction.
Charge is conserved. Current simply divides among different paths.

Mistake: Confusing current with voltage.
Kirchhoff's Current Law applies to current at junctions, not to voltage.

Mistake: Treating a negative calculated current as impossible.
A negative result normally means the actual current direction is opposite to the assumed direction.

Mistake: Applying Ohm's Law before identifying the circuit structure.
First identify branches, junctions and known quantities.

---

Key Terms

Kirchhoff's Current Law (KCL) — total current entering a junction equals total current leaving it.

Kirchhoff's First Law — another name for Kirchhoff's Current Law.

Junction Rule — rule describing conservation of current at a circuit junction.

Junction — point where three or more conducting paths meet.

Branch — individual conducting path connected to a junction.

Branch current — current flowing through one branch.

Total current — combined current entering or leaving a network.

Conservation of charge — principle that electric charge cannot be created or destroyed in an ordinary circuit.

Current — rate of flow of electric charge.

Steady state — condition in which circuit quantities remain stable and charge does not continually accumulate at junctions.

Current division — splitting of current among parallel branches.

Sign convention — consistent system for assigning positive and negative signs to current directions.

Σ (sigma) — mathematical symbol meaning "sum of."

---

Key Takeaways

• Kirchhoff's Current Law is also called the Junction Rule or Kirchhoff's First Law.
• The total current entering a junction equals the total current leaving it.
• I_in = I_out
• A more general form is ΣI = 0 when a consistent sign convention is used.
• Kirchhoff's Current Law follows from conservation of electric charge.
• Current represents the rate of charge flow.
• Charge does not disappear or appear at an ordinary circuit junction.
• Current can divide when a circuit branches.
• Branch currents do not necessarily have equal values.
• Lower-resistance parallel branches generally carry larger currents.
• When branches recombine, their currents add together.
• KCL can be applied independently to every junction in a complex network.
• Ohm's Law can be combined with KCL to calculate unknown currents and resistances.
• Current directions may be assumed when solving a circuit.
• A negative calculated current means the actual direction is opposite to the assumed direction.
• Experimental current measurements can be used to test Kirchhoff's Current Law.
• Small differences between measured currents may result from experimental uncertainty.
• KCL is fundamental to analyzing branching electrical networks.

Check Your Understanding

1. State Kirchhoff's Current Law.
2. What is meant by a junction in an electrical circuit?
3. Explain how Kirchhoff's Current Law follows from conservation of charge.
4. A current of 8 A reaches a junction. One branch carries 3 A. Calculate the current in the other branch.
5. A current of 15 A divides into three branches carrying 4 A, 6 A and I. Calculate I.
6. Two currents of 3 A and 5 A enter a junction. Calculate the total current leaving.
7. Currents of 8 A and 4 A enter a junction. Currents of 5 A and I leave. Calculate I.
8. A total current of 12 A divides into three branches. Two carry 2.5 A and 4.0 A. Calculate the third current.
9. Explain why current does not need to divide equally between parallel branches.
10. Two parallel resistors of 3 Ω and 6 Ω are connected across 12 V. Calculate each branch current.
11. For Question 10, use KCL to calculate the total current.
12. Explain why the lower-resistance branch in Question 10 carries more current.
13. A 10 A current reaches Junction A. A 3 A branch leaves, while the remainder reaches Junction B. Calculate the current reaching Junction B.
14. At Junction B in Question 13, the current divides into 2 A and two equal currents. Calculate each equal current.
15. Explain what ΣI = 0 means.
16. Why might a calculated current have a negative value?
17. A current of 2 A flows for 10 s into a junction. Calculate the total charge reaching the junction.
18. Describe an experiment using ammeters that could test Kirchhoff's Current Law.
19. Explain why charge cannot continually accumulate at a junction in a steady-state circuit.
20. Explain how Kirchhoff's Current Law and Ohm's Law can be combined to analyze a branching circuit.