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.

 

 

 

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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

 

 

 

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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

 

 

 

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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.

 

 

 

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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.

 

 

 

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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.

Loop Rule (Voltage Law)

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (KVL) describes how potential differences behave around a closed loop in an electrical circuit.

It states:

The algebraic sum of all potential differences around any closed loop is zero.

In simple terms:

total voltage supplied = total voltage dropped

For a simple circuit:

V_supply = V₁ + V₂ + V₃ + ...

The more general form is:

ΣV = 0

Kirchhoff's Voltage Law is also called:

• Kirchhoff's Second Law
• the Loop Rule
• the Voltage Law

 

 

 

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What Is a Closed Loop?

A closed loop is any continuous path through a circuit that:

• begins at one point
• follows connected components
• eventually returns to the starting point

A simple battery-and-resistor circuit contains one loop.

More complicated circuits may contain several different loops.

Kirchhoff's Voltage Law can be applied to each closed loop separately.

Conservation of Energy

Kirchhoff's Voltage Law is based on the conservation of energy.

Energy cannot be created or destroyed.

A battery or power supply transfers energy to electric charges.

Circuit components then transfer this electrical energy into other forms such as:

• thermal energy
• light
• sound
• mechanical energy

After traveling completely around a closed loop, the net energy change of each unit of charge must be zero.

Voltage as Energy per Charge

Potential difference is defined as:

V = E/Q

where:

• V = potential difference (V)
• E = energy transferred (J)
• Q = charge (C)

Therefore:

1 V = 1 J/C

A 12 V battery supplies:

12 J of energy per coulomb of charge

If that charge travels through components that transfer:

4 J/C

and:

8 J/C

then:

4 + 8 = 12 J/C

The energy supplied equals the energy transferred.

Voltage Rise and Voltage Drop

As we move around a circuit, potential can either increase or decrease.

Across a power source:

voltage rise

Across a resistor or other load:

voltage drop

For example:

battery: +12 V

resistor 1: −4 V

resistor 2: −8 V

Around the complete loop:

12 − 4 − 8 = 0

Therefore:

ΣV = 0

A Simple One-Loop Circuit

Consider a 12 V battery connected to two series resistors.

Suppose:

V₁ = 5 V

V₂ = 7 V

Kirchhoff's Voltage Law gives:

12 − 5 − 7 = 0

Therefore:

12 V = 5 V + 7 V

The energy supplied by the battery equals the energy transferred by the two resistors.

 

 

 

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Calculating an Unknown Voltage

Suppose a 20 V power supply is connected to three components.

The voltage drops are:

V₁ = 5 V

V₂ = 7 V

V₃ = ?

Use KVL:

V_supply = V₁ + V₂ + V₃

20 = 5 + 7 + V₃

20 = 12 + V₃

Therefore:

V₃ = 8 V

Check:

5 + 7 + 8 = 20 V

Another Unknown Voltage Example

A 15 V battery supplies two series resistors.

One resistor has a voltage drop of:

6 V

The second voltage drop is:

V₂ = ?

Using KVL:

15 = 6 + V₂

Therefore:

V₂ = 9 V

Kirchhoff's Voltage Law and Ohm's Law

KVL becomes especially powerful when combined with:

V = IR

Suppose a 12 V battery is connected to two series resistors:

R₁ = 2 Ω

R₂ = 4 Ω

Total resistance:

R_total = 2 + 4

R_total = 6 Ω

Current:

I = V/R

I = 12/6

I = 2 A

Voltage across R₁:

V₁ = IR₁

V₁ = 2 × 2

V₁ = 4 V

Voltage across R₂:

V₂ = IR₂

V₂ = 2 × 4

V₂ = 8 V

Apply KVL:

12 − 4 − 8 = 0

The circuit satisfies Kirchhoff's Voltage Law.

Why Larger Series Resistors Have Larger Voltage Drops

In a series circuit, the current is the same through each resistor.

Since:

V = IR

a larger resistance produces a larger voltage drop.

For example, if:

I = 2 A

R₁ = 3 Ω

R₂ = 7 Ω

then:

V₁ = 2 × 3 = 6 V

V₂ = 2 × 7 = 14 V

The total voltage is:

6 + 14 = 20 V

Choosing a Loop Direction

When applying KVL, choose a direction to travel around the loop.

You may choose:

• clockwise
• counterclockwise

Either direction works.

The important requirement is to use signs consistently.

For example, you might choose clockwise and record every voltage rise and drop encountered along that direction.

Sign Convention for Batteries

When moving through a battery:

from − terminal to + terminal → voltage rise → positive

from + terminal to − terminal → voltage drop → negative

For example, crossing a 9 V battery from negative to positive gives:

+9 V

Crossing it from positive to negative gives:

−9 V

Sign Convention for Resistors

For a resistor carrying current I:

when moving with the current through the resistor:

−IR

This represents a voltage drop.

When moving against the current:

+IR

This represents a voltage rise in the chosen traversal direction.

 

 

 

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A Reliable Sign Strategy

When moving around a loop:

Battery

− to +:

+V

• to −:

−V

Resistor

With current:

−IR

Against current:

+IR

Then add every term and set the total equal to zero:

ΣV = 0

Example Using Signs

Suppose a circuit contains:

• 10 V battery
• 2 Ω resistor
• 3 Ω resistor
• current I

Travel clockwise from the negative terminal of the battery.

Across battery:

+10

Across 2 Ω resistor with current:

−2I

Across 3 Ω resistor with current:

−3I

Apply KVL:

10 − 2I − 3I = 0

10 − 5I = 0

5I = 10

I = 2 A

The resistor voltage drops are:

V₁ = 2 × 2 = 4 V

V₂ = 2 × 3 = 6 V

Check:

10 − 4 − 6 = 0

Multiple Power Sources

Some circuits contain more than one battery or power source.

Their voltages may:

• support each other
• oppose each other

KVL allows us to determine their combined effect.

Suppose two batteries of 12 V and 4 V oppose each other.

The effective voltage is:

12 − 4

= 8 V

If the circuit resistance is 4 Ω:

I = 8/4

I = 2 A

Batteries Supporting Each Other

Suppose two batteries of:

6 V

and:

3 V

are arranged so their voltage rises act in the same direction.

Total voltage:

V_total = 6 + 3

V_total = 9 V

If total resistance is 3 Ω:

I = 9/3

I = 3 A

Battery orientation therefore matters.

 

 

 

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Loop Rule in Parallel Circuits

KVL also explains why parallel branches have the same voltage.

Suppose two resistors are connected in parallel across a 12 V battery.

Consider the loop containing the battery and resistor 1:

12 − V₁ = 0

Therefore:

V₁ = 12 V

Now consider the loop containing the battery and resistor 2:

12 − V₂ = 0

Therefore:

V₂ = 12 V

Thus:

V₁ = V₂ = V_supply

This is the familiar parallel-circuit voltage rule.

KVL and Series Circuits

Similarly, the series voltage rule follows directly from KVL.

For a battery and three series resistors:

V_supply − V₁ − V₂ − V₃ = 0

Rearranging:

V_supply = V₁ + V₂ + V₃

So the basic series circuit rule is really an application of Kirchhoff's Voltage Law.

Multiple Loops

More complicated circuits can contain several closed loops.

For example, a circuit might contain:

• one battery
• several resistors
• two junctions
• multiple branches

Different currents may flow through the different branches.

In these circuits, we often need both:

Kirchhoff's Current Law

and:

Kirchhoff's Voltage Law

 

 

 

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Current Law and Voltage Law Together

Kirchhoff's Current Law:

total current entering a junction = total current leaving

Kirchhoff's Voltage Law:

ΣV = 0 around a closed loop

Together with Ohm's Law:

V = IR

these equations allow complicated electrical networks to be solved.

Assigning Loop Currents

One common method for solving multi-loop circuits is to assign a current to each loop.

For example:

I₁ = clockwise current in the left loop

I₂ = clockwise current in the right loop

If the loops share a resistor, the actual current through that resistor depends on both loop currents.

If the currents pass through the shared resistor in opposite directions:

I_shared = I₁ − I₂

or:

I₂ − I₁

depending on the chosen reference direction.

Shared Resistors

Shared resistors are one of the most important features of multi-loop problems.

Suppose:

I₁ flows through a shared resistor in one direction.

I₂ flows through it in the opposite direction.

The net current is:

I_shared = I₁ − I₂

Therefore, the voltage across the shared resistor is:

V_shared = R(I₁ − I₂)

This term appears in the KVL equation for both loops.

Example of a Two-Loop Network

Consider two loops that share a 2 Ω resistor.

Left loop:

• 10 V battery
• 3 Ω resistor
• shared 2 Ω resistor

Right loop:

• 4 V battery
• 4 Ω resistor
• shared 2 Ω resistor

Assign clockwise currents:

I₁ = left-loop current

I₂ = right-loop current

The shared-resistor current is:

I₁ − I₂

Left Loop Equation

Travel clockwise around the left loop.

Battery rise:

+10

Voltage drop across 3 Ω resistor:

−3I₁

Voltage drop across shared resistor:

−2(I₁ − I₂)

Therefore:

10 − 3I₁ − 2(I₁ − I₂) = 0

Expand:

10 − 3I₁ − 2I₁ + 2I₂ = 0

Therefore:

5I₁ − 2I₂ = 10

Right Loop Equation

For the right loop:

4 − 4I₂ − 2(I₂ − I₁) = 0

Expand:

4 − 4I₂ − 2I₂ + 2I₁ = 0

Therefore:

−2I₁ + 6I₂ = 4

We now have two simultaneous equations:

5I₁ − 2I₂ = 10

−2I₁ + 6I₂ = 4

Solving them gives approximately:

I₁ = 2.62 A

I₂ = 1.54 A

The shared resistor current is:

I_shared = I₁ − I₂

I_shared = 2.62 − 1.54

I_shared ≈ 1.08 A

This illustrates how KVL can be used to solve networks that cannot be reduced using only simple series and parallel rules.

Negative Current Answers

Sometimes solving a multi-loop problem gives a negative current.

For example:

I = −1.5 A

This does not mean something physically impossible has happened.

It means the current actually flows opposite to the direction originally assumed.

The actual current is:

1.5 A in the opposite direction

Choosing an initial current direction is simply part of the mathematical model.

Solving Multi-Loop Circuits Systematically

A useful procedure is:

Label the currents

Assign directions to unknown currents.

Identify the junctions

Apply Kirchhoff's Current Law where necessary.

Identify independent loops

Choose closed paths through the circuit.

Select loop directions

Clockwise or counterclockwise is acceptable.

Write KVL equations

Use:

ΣV = 0

Replace resistor voltages

Use:

V = IR

Solve the simultaneous equations

Find the unknown currents.

Calculate other quantities

Use the currents to determine voltage drops, power or resistance.

Check the results

Verify both KCL and KVL.

Independent Loops

A complicated circuit may contain many possible closed paths.

Not every possible loop gives new information.

An independent loop provides an equation that cannot simply be obtained by combining the other loop equations.

For basic circuits, choose the smallest obvious loops first.

This usually produces the simplest equations.

Complete Single-Loop Example

A circuit contains:

• 18 V battery
• 2 Ω resistor
• 4 Ω resistor
• 3 Ω resistor

All components are in one loop.

Apply KVL:

18 − 2I − 4I − 3I = 0

18 − 9I = 0

Therefore:

I = 2 A

Voltage drops:

V₁ = 2 × 2 = 4 V

V₂ = 2 × 4 = 8 V

V₃ = 2 × 3 = 6 V

Check:

4 + 8 + 6 = 18 V

Therefore:

18 − 18 = 0

KVL is satisfied.

Finding an Unknown Resistance

A 24 V battery supplies a single loop containing:

R₁ = 4 Ω

R₂ = ?

Current:

I = 3 A

Apply KVL:

24 − 3(4) − 3R₂ = 0

24 − 12 − 3R₂ = 0

12 = 3R₂

Therefore:

R₂ = 4 Ω

Finding an Unknown Source Voltage

A loop contains:

• unknown battery V
• 3 Ω resistor
• 5 Ω resistor
• current of 2 A

Apply KVL:

V − 2(3) − 2(5) = 0

V − 6 − 10 = 0

Therefore:

V = 16 V

The source must provide 16 J of energy per coulomb.

Energy Interpretation

Suppose a 6 C charge moves around a circuit containing a 10 V battery.

Energy supplied by the battery:

E = VQ

E = 10 × 6

E = 60 J

If one resistor transfers:

24 J

then the remaining components must transfer:

60 − 24

= 36 J

The energy supplied and transferred must balance.

This is the physical basis of KVL.

Experimental Testing of the Loop Rule

Kirchhoff's Voltage Law can be tested using voltmeters.

For a simple series circuit, measure:

• supply voltage
• voltage across resistor 1
• voltage across resistor 2
• voltage across resistor 3

Theory predicts:

V_supply ≈ V₁ + V₂ + V₃

 

 

 

8

Example Experimental Results

Suppose:

V_supply = 12.02 V

V₁ = 3.01 V

V₂ = 4.02 V

V₃ = 4.96 V

Add the voltage drops:

V₁ + V₂ + V₃

= 3.01 + 4.02 + 4.96

= 11.99 V

The measured supply voltage is:

12.02 V

Difference:

12.02 − 11.99

= 0.03 V

The results provide strong experimental support for KVL within measurement uncertainty.

Why Experimental Values May Differ

Real measurements may not satisfy KVL perfectly because of:

• meter uncertainty
• resistor tolerance
• internal resistance of the battery
• connecting-wire resistance
• contact resistance
• temperature changes
• power-supply fluctuations
• rounding

KVL itself still reflects conservation of energy. Differences usually arise because the experimental model does not include every real component or because measurements have uncertainty.

Kirchhoff's Laws Together

The two Kirchhoff laws describe different conservation principles.

Current Law:

ΣI = 0

Voltage Law:

ΣV = 0

Together with:

V = IR

they form a powerful system for analyzing electrical circuits.

When Do We Need Kirchhoff's Laws?

Simple circuits can often be solved using:

• series resistance rules
• parallel resistance rules
• Ohm's Law

Kirchhoff's laws become particularly useful when:

• several branches are present
• several voltage sources are present
• components cannot easily be reduced to simple series or parallel combinations
• multiple loops share components
• several unknown currents must be found

They provide a systematic method that works for much more complicated networks.

Real-World Applications

Kirchhoff's Voltage Law is used in the analysis and design of:

• electronic circuits
• power supplies
• automotive electrical systems
• sensor circuits
• communication electronics
• control systems
• power-distribution networks
• circuit boards

 

 

 

8

Engineers frequently use computer software to solve large systems of Kirchhoff equations, but the underlying physical principles remain the same.

Common Mistakes

Mistake: Forgetting that KVL applies around a closed loop.
The path must return to its starting point.

Mistake: Adding all voltages as positive.
Voltage rises and drops require appropriate signs.

Mistake: Changing the sign convention halfway through a calculation.
Choose a direction and remain consistent.

Mistake: Assuming clockwise is the only correct loop direction.
Either direction works if signs are handled correctly.

Mistake: Forgetting that V = IR for a resistor.
Resistor voltage terms are usually written as IR.

Mistake: Ignoring the orientation of a battery.
Moving from − to + is a voltage rise; moving from + to − is a voltage drop.

Mistake: Using I₁ through a shared resistor when two loop currents pass through it.
The current may be I₁ − I₂.

Mistake: Thinking a negative calculated current is impossible.
It normally means the true direction is opposite to the assumed direction.

Mistake: Confusing KCL with KVL.
KCL deals with currents at junctions. KVL deals with voltages around loops.

Key Terms

Kirchhoff's Voltage Law (KVL) — the algebraic sum of potential differences around a closed loop is zero.

Kirchhoff's Second Law — another name for Kirchhoff's Voltage Law.

Loop Rule — another name for KVL.

Closed loop — continuous circuit path that returns to its starting point.

Potential difference — energy transferred per unit charge.

Voltage rise — increase in electric potential when moving through a circuit.

Voltage drop — decrease in electric potential across a component.

EMF — energy supplied per unit charge by a source.

Loop current — current assigned to a particular loop during circuit analysis.

Shared resistor — resistor that forms part of more than one loop.

Independent loop — loop that provides an independent circuit equation.

Sign convention — consistent rules for assigning positive and negative values.

Conservation of energy — principle that energy cannot be created or destroyed.

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

Key Takeaways

• Kirchhoff's Voltage Law is also called Kirchhoff's Second Law or the Loop Rule.
• KVL states that the algebraic sum of potential differences around a closed loop is zero.
• ΣV = 0
• In a simple circuit, total voltage supplied equals total voltage dropped.
• KVL follows from conservation of energy.
• Voltage represents energy transferred per unit charge.
• Batteries generally provide voltage rises.
• Resistors generally produce voltage drops in the direction of current.
• Either clockwise or counterclockwise loop directions can be used.
• Sign conventions must be applied consistently.
• For a resistor, voltage can be expressed using V = IR.
• KVL explains why voltage drops in a series circuit add to the supply voltage.
• KVL also explains why parallel branches connected across the same two points have the same voltage.
• Multiple voltage sources can either support or oppose each other.
• Multi-loop circuits may require several simultaneous equations.
• A resistor shared by two loops may carry the difference between two loop currents.
• A negative calculated current usually indicates that the actual current flows opposite to the assumed direction.
• Kirchhoff's Current Law and Voltage Law are often used together.
• KCL represents conservation of charge.
• KVL represents conservation of energy.
• Together with Ohm's Law, Kirchhoff's laws can be used to analyze complex electrical networks.

Check Your Understanding

1. State Kirchhoff's Voltage Law.
2. What is meant by a closed loop?
3. Explain how KVL follows from conservation of energy.
4. What does ΣV = 0 mean?
5. A 12 V battery is connected to two series components. One has a voltage drop of 5 V. Calculate the other voltage drop.
6. A 24 V supply is connected to three components with voltage drops of 6 V, 8 V and V₃. Calculate V₃.
7. A circuit contains a 15 V battery and two series resistors of 2 Ω and 3 Ω. Calculate the current.
8. For Question 7, calculate the voltage across each resistor and verify KVL.
9. What sign is normally assigned when crossing a battery from its negative terminal to its positive terminal?
10. What sign is normally assigned to a resistor when moving through it in the direction of conventional current?
11. Explain why either clockwise or counterclockwise can be chosen as the loop direction.
12. Two batteries of 12 V and 5 V oppose each other. What is their net voltage?
13. Two batteries of 6 V and 4 V support each other. What is their combined voltage?
14. A 20 V battery supplies a loop containing a 4 Ω resistor and an unknown resistor. The current is 2 A. Calculate the unknown resistance.
15. A circuit carries 3 A through two series resistors of 2 Ω and 5 Ω. Calculate the required source voltage.
16. Explain the difference between Kirchhoff's Current Law and Kirchhoff's Voltage Law.
17. Explain why parallel branches connected across the same two points have the same voltage using KVL.
18. Two loop currents I₁ and I₂ pass through a shared resistor in opposite directions. Explain how the current through the resistor can be represented.
19. What does a negative answer for an assumed loop current tell you?
20. Explain how Kirchhoff's Current Law, Kirchhoff's Voltage Law and Ohm's Law can be combined to solve a multi-loop circuit.

Internal Resistance

Real Batteries Are Not Ideal

In simple circuit models, a battery is often treated as an ideal voltage source. An ideal source would maintain exactly the same potential difference across its terminals regardless of the current it supplies.

Real batteries do not behave perfectly this way.

Every real battery has some resistance inside it. This is called internal resistance.

Internal resistance is represented by:

r

and is measured in:

ohms (Ω)

A useful model of a real battery is therefore:

an ideal emf source in series with an internal resistance

 

 

 

7

The internal resistance causes some of the battery's energy to be transferred into thermal energy inside the battery whenever current flows.

Electromotive Force (emf)

The electromotive force, or emf, of a source is the energy supplied by the source per unit charge.

It is represented by:

ε

and measured in:

volts (V)

Despite its name, emf is not a force.

It is an energy-per-charge quantity.

ε = E/Q

where:

• ε = emf (V)
• E = energy supplied (J)
• Q = charge (C)

Therefore:

1 V = 1 J/C

A battery with an emf of 12 V supplies approximately 12 J of energy to each coulomb of charge.

emf and Terminal Voltage

The terminal voltage is the potential difference measured across the external terminals of a battery.

It is usually represented by:

V

When no current is flowing, the terminal voltage of a battery is approximately equal to its emf.

When the battery supplies current:

terminal voltage is usually less than emf

because some energy is transferred inside the battery due to its internal resistance.

Lost Voltage

If a battery has internal resistance r and supplies current I, the voltage drop across the internal resistance is:

V_lost = Ir

This is sometimes called the:

• internal voltage drop
• lost volts
• voltage loss

Therefore:

V = ε − Ir

where:

• V = terminal voltage (V)
• ε = emf (V)
• I = current (A)
• r = internal resistance (Ω)

This is the central equation for a battery that is supplying current.

Simple Example

A battery has:

ε = 12 V

r = 0.5 Ω

I = 2 A

Internal voltage drop:

V_lost = Ir

V_lost = 2 × 0.5

V_lost = 1 V

Terminal voltage:

V = ε − Ir

V = 12 − 1

V = 11 V

Although the battery has an emf of 12 V, only 11 V appears across its terminals while it supplies this current.

Why Does Terminal Voltage Fall?

When current flows through the battery, it also passes through the battery's internal resistance.

Energy is transferred into thermal energy inside the battery.

Therefore, not all of the energy supplied by the chemical reactions reaches the external circuit.

In energy-per-charge terms:

emf = terminal voltage + internal voltage drop

or:

ε = V + Ir

 

 

 

9

An Ideal Battery

For an ideal battery:

r = 0 Ω

Therefore:

V = ε − Ir

becomes:

V = ε

So an ideal battery always has:

terminal voltage = emf

regardless of the current.

Real batteries have:

r > 0

so their terminal voltage changes with current.

Open-Circuit Voltage

If a battery is not supplying current:

I = 0

Therefore:

V = ε − Ir

V = ε − 0

V = ε

This means the terminal voltage measured when essentially no current is being drawn is approximately the battery's emf.

A voltmeter has very high resistance, so a good voltmeter draws very little current.

Therefore, measuring a disconnected battery with a voltmeter gives an approximate measurement of its emf.

 

 

 

5

Battery Connected to an External Resistance

Consider a battery with:

emf = ε

internal resistance = r

connected to an external resistor:

R

The total circuit resistance is:

R_total = R + r

Therefore, the current is:

I = ε/(R + r)

Once the current is known, the terminal voltage can be calculated using:

V = ε − Ir

or:

V = IR

Both methods should give the same answer.

Worked Example

A battery has:

ε = 12 V

r = 1 Ω

and is connected to:

R = 5 Ω

Total resistance:

R_total = 5 + 1

R_total = 6 Ω

Current:

I = ε/(R + r)

I = 12/6

I = 2 A

Internal voltage drop:

Ir = 2 × 1

Ir = 2 V

Terminal voltage:

V = 12 − 2

V = 10 V

Check using the external resistor:

V = IR

V = 2 × 5

V = 10 V

Both calculations agree.

Kirchhoff's Voltage Law

Internal resistance can also be understood using Kirchhoff's Voltage Law.

Around the complete circuit:

ε − IR − Ir = 0

Therefore:

ε = IR + Ir

Since:

V = IR

we obtain:

ε = V + Ir

and therefore:

V = ε − Ir

So the internal-resistance equation follows directly from conservation of energy around the circuit.

Energy Transfers in a Real Battery

Suppose a charge Q passes through a battery.

The battery supplies:

E_total = εQ

The external circuit receives:

E_external = VQ

Energy transferred internally is:

E_internal = IrQ

Therefore:

εQ = VQ + IrQ

Dividing by Q:

ε = V + Ir

This gives the same relationship again, but now from an energy perspective.

Power Supplied by a Battery

The total power supplied by the battery is:

P_source = εI

The useful power delivered to the external circuit is:

P_external = VI

Since:

V = IR

we can also write:

P_external = I²R

Power Lost Inside the Battery

The power transferred into thermal energy inside the battery is:

P_internal = I²r

Therefore:

P_source = P_external + P_internal

or:

εI = VI + I²r

This is an important example of conservation of energy.

Power Example

Suppose:

ε = 12 V

r = 1 Ω

I = 2 A

Total power supplied:

P_source = εI

P_source = 12 × 2

P_source = 24 W

Internal power loss:

P_internal = I²r

P_internal = 2² × 1

P_internal = 4 W

Useful external power:

P_external = 24 − 4

P_external = 20 W

Check:

V = ε − Ir

V = 12 − 2

V = 10 V

Therefore:

P_external = VI

P_external = 10 × 2

P_external = 20 W

Efficiency of a Real Battery

The efficiency of energy transfer to the external circuit can be written as:

efficiency = useful power / total power supplied × 100%

Therefore:

efficiency = VI/(εI) × 100%

Cancel I:

efficiency = V/ε × 100%

Since:

V = IR

and:

ε = I(R + r)

we can also write:

efficiency = R/(R + r) × 100%

Efficiency Example

A battery has:

ε = 10 V

terminal voltage = 9 V

Efficiency:

efficiency = 9/10 × 100%

efficiency = 90%

This means approximately 90% of the electrical energy supplied per unit charge reaches the external circuit.

Effect of Increasing Current

From:

V = ε − Ir

if ε and r remain constant:

greater I → greater Ir → lower terminal voltage

This explains why the voltage of a real battery may appear normal when very little current is drawn but fall noticeably when connected to a demanding load.

 

 

 

6

This effect is often called voltage sag.

Terminal Voltage and Current Graph

The equation:

V = ε − Ir

has the same form as:

y = c + mx

If V is plotted vertically and I horizontally:

V-intercept = ε

and:

gradient = −r

Therefore:

r = magnitude of gradient

This provides an experimental method for determining both emf and internal resistance.

Example from a V-I Graph

Suppose the relationship between terminal voltage and current produces:

V = 6.0 − 0.40I

Comparing with:

V = ε − Ir

gives:

ε = 6.0 V

and:

r = 0.40 Ω

The negative slope shows that terminal voltage decreases as current increases.

Experimental Measurement of Internal Resistance

A typical experiment uses:

• battery
• variable resistor
• ammeter
• voltmeter
• switch
• connecting wires

The ammeter measures the circuit current.

The voltmeter measures the terminal voltage.

The variable resistor allows the current to be changed.

 

 

 

9

Several pairs of V and I measurements are recorded.

Example Experimental Data

The voltage falls by:

0.2 V

for each increase of:

0.5 A

Therefore:

gradient = ΔV/ΔI

gradient = −0.2/0.5

gradient = −0.4 V/A

Since:

1 V/A = 1 Ω

the internal resistance is:

r = 0.4 Ω

The V-intercept gives:

ε = 6.0 V

Calculating Internal Resistance Directly

If emf, terminal voltage and current are known:

V = ε − Ir

Rearrange:

Ir = ε − V

Therefore:

r = (ε − V)/I

Suppose:

ε = 9.0 V

V = 8.4 V

I = 1.5 A

Then:

r = (9.0 − 8.4)/1.5

r = 0.6/1.5

r = 0.40 Ω

Calculating emf

The equation can also be rearranged:

ε = V + Ir

Suppose:

V = 11.4 V

I = 2.0 A

r = 0.30 Ω

Then:

ε = 11.4 + (2.0 × 0.30)

ε = 11.4 + 0.6

ε = 12.0 V

Calculating Current

From:

V = ε − Ir

we can rearrange:

I = (ε − V)/r

Suppose:

ε = 12 V

V = 10 V

r = 0.5 Ω

Then:

I = (12 − 10)/0.5

I = 2/0.5

I = 4 A

Short-Circuit Current

Consider what happens if the external resistance becomes extremely small.

If:

R ≈ 0

then:

I = ε/(R + r)

becomes approximately:

I_short = ε/r

This can produce a very large current if the internal resistance is small.

For example:

ε = 12 V

r = 0.10 Ω

Then:

I_short = 12/0.10

I_short = 120 A

This is one reason short circuits can cause severe heating and damage.

Internal Heating During a Short Circuit

The internal heating power is:

P_internal = I²r

Using:

I = 120 A

r = 0.10 Ω

P_internal = 120² × 0.10

P_internal = 14,400 × 0.10

P_internal = 1440 W

This shows why a battery can heat rapidly during a short circuit.

Never deliberately short-circuit a battery as an experiment.

What Causes Internal Resistance?

A battery contains:

• electrodes
• electrolyte
• separators
• electrical contacts
• chemical reaction regions

Charge moving through these materials encounters opposition.

The exact mechanisms depend on the battery chemistry, but resistance can arise from:

• resistance of the electrolyte
• resistance of electrodes
• electrical contacts
• transport of ions
• chemical reaction processes

Together, these effects contribute to the effective internal resistance.

 

 

 

8

Internal Resistance and Battery Condition

Internal resistance is not necessarily constant throughout a battery's life.

It can depend on:

• battery chemistry
• temperature
• state of charge
• age
• current
• previous use

As some batteries age, their internal resistance increases.

This can cause greater voltage drop when they supply large currents.

Why an Old Battery Can Show a Normal Voltage

A battery can sometimes show a reasonable open-circuit voltage but perform poorly when connected to a load.

With little current:

I ≈ 0

so:

Ir ≈ 0

and:

V ≈ ε

But under load, if the battery has significant internal resistance:

Ir becomes larger

and the terminal voltage falls.

Therefore, measuring only the open-circuit voltage does not always reveal how well a battery can supply current.

Battery Under a Heavy Load

Suppose two batteries both have:

ε = 12 V

Battery A:

r = 0.10 Ω

Battery B:

r = 1.0 Ω

Both supply:

I = 5 A

For Battery A:

V = 12 − (5 × 0.10)

V = 11.5 V

For Battery B:

V = 12 − (5 × 1.0)

V = 7.0 V

Although both batteries have the same emf, their behavior under load is very different.

Internal Resistance and Temperature

Temperature can affect internal resistance.

For many battery types, low temperatures reduce the rate of chemical and ionic processes and can increase effective internal resistance.

This can cause:

• larger voltage drops
• reduced available current
• poorer high-power performance

This helps explain why batteries can perform differently in cold conditions.

Charging a Battery

When a rechargeable battery is being charged, current is forced into it.

In this situation, the terminal voltage may be greater than the emf.

A simplified relationship is:

V = ε + Ir

during charging.

The sign changes because energy is being supplied to the battery rather than delivered by it.

Supplying vs Charging

Battery supplying current

V = ε − Ir

Terminal voltage is less than emf.

Battery being charged

V = ε + Ir

Applied terminal voltage must exceed the battery's emf to drive current into it.

The correct relationship depends on the direction of current and energy transfer.

Cells in Series and Internal Resistance

Suppose identical cells each have:

emf = ε

internal resistance = r

If n cells are connected in series in the same orientation:

ε_total = nε

and:

r_total = nr

For example, four cells each have:

ε = 1.5 V

r = 0.20 Ω

Then:

ε_total = 4 × 1.5

ε_total = 6.0 V

r_total = 4 × 0.20

r_total = 0.80 Ω

Internal resistances add because they are effectively in series.

Real Battery Model in Circuit Analysis

When analyzing a circuit containing a real battery, draw the battery as:

ideal emf ε + series internal resistance r

Then treat r like another resistor in the circuit.

For an external resistance R:

R_total = R + r

and:

I = ε/(R + r)

This converts the real battery into a model that can be analyzed using familiar circuit laws.

Example: Complete Real-Battery Analysis

A battery has:

ε = 15 V

r = 1 Ω

External resistor:

R = 4 Ω

Total resistance

R_total = 4 + 1

R_total = 5 Ω

Current

I = 15/5

I = 3 A

Internal voltage drop

Ir = 3 × 1

Ir = 3 V

Terminal voltage

V = 15 − 3

V = 12 V

External power

P_external = VI

P_external = 12 × 3

P_external = 36 W

Internal power loss

P_internal = I²r

P_internal = 3² × 1

P_internal = 9 W

Total power supplied

P_source = εI

P_source = 15 × 3

P_source = 45 W

Check:

36 + 9 = 45 W

Energy is conserved.

Ideal vs Real Batteries

The ideal battery model is useful for simple calculations, while the real battery model gives a more accurate description of actual circuit behavior.

Common Mistakes

Mistake: Saying emf is a force.
Despite its name, emf is energy supplied per unit charge.

Mistake: Assuming emf and terminal voltage are always identical.
They are approximately equal only when little or no current is flowing.

Mistake: Forgetting the internal voltage drop.
When a battery supplies current, V = ε − Ir.

Mistake: Adding Ir when the battery is supplying current.
For a discharging battery, the internal drop reduces terminal voltage.

Mistake: Treating internal resistance as being outside the battery.
It is a model of resistance within the source.

Mistake: Assuming current is lost inside the battery.
Current is not lost. Energy is transferred into thermal energy.

Mistake: Assuming a normal open-circuit voltage proves a battery is in excellent condition.
A battery may show substantial voltage sag under load.

Mistake: Ignoring units.
Use volts for ε and V, amperes for I, and ohms for r.

Key Terms

Internal resistance — effective resistance inside a real electrical source.

Electromotive force (emf) — energy supplied by a source per unit charge.

Terminal voltage — potential difference measured across the terminals of a source.

Internal voltage drop — voltage Ir associated with current flowing through the source's internal resistance.

Lost volts — another term for the internal voltage drop Ir.

Open-circuit voltage — terminal voltage when essentially no external current is being drawn.

Load — external circuit or device receiving electrical energy from a source.

Voltage sag — decrease in terminal voltage when significant current is drawn.

Internal power loss — rate of energy transfer inside the source, given by I²r.

Short circuit — very low-resistance connection across a source that can produce a dangerously large current.

Ideal voltage source — theoretical source with zero internal resistance.

Real voltage source — source modeled as an emf in series with internal resistance.

Key Takeaways

• Real batteries have internal resistance.
• Internal resistance is represented by r and measured in ohms.
• A real battery can be modeled as an ideal emf source in series with r.
• Emf is energy supplied per unit charge, not a mechanical force.
• Terminal voltage is the potential difference across the battery terminals.
• When little or no current flows, V ≈ ε.
• When a battery supplies current, V = ε − Ir.
• The internal voltage drop is Ir.
• Larger currents produce larger internal voltage drops.
• Internal resistance causes terminal voltage to decrease under load.
• Energy transferred internally usually appears mainly as thermal energy.
• Internal power loss is P_internal = I²r.
• Total source power is P_source = εI.
• Useful external power is P_external = VI.
• Energy conservation gives εI = VI + I²r.
• For an external resistance R, I = ε/(R + r).
• A graph of V against I has intercept ε and gradient −r.
• A battery can have a normal open-circuit voltage yet perform poorly under a heavy load.
• Battery age, temperature and state of charge can affect internal resistance.
• During charging, a simplified relationship is V = ε + Ir.
• Internal resistance is essential for understanding the behavior of real batteries and power sources.

Check Your Understanding

1. What is meant by the internal resistance of a battery?
2. What symbol is commonly used for internal resistance?
3. Define the emf of a battery.
4. Why is the term "electromotive force" potentially misleading?
5. Explain the difference between emf and terminal voltage.
6. A battery has ε = 12 V, r = 0.5 Ω and supplies 4 A. Calculate its terminal voltage.
7. A battery has ε = 9.0 V and terminal voltage 8.4 V while supplying 2.0 A. Calculate its internal resistance.
8. Explain why terminal voltage decreases as current increases.
9. A 10 V battery with internal resistance 1 Ω is connected to a 4 Ω resistor. Calculate the circuit current.
10. For Question 9, calculate the terminal voltage.
11. For Question 9, calculate the power transferred inside the battery.
12. Explain why the open-circuit terminal voltage is approximately equal to the emf.
13. What information can be obtained from the intercept of a graph of V against I?
14. What information can be obtained from the gradient of the same graph?
15. Explain why an older battery might show a normal voltage when disconnected but perform poorly when powering a device.
16. A 12 V battery has internal resistance 0.20 Ω. Calculate the theoretical short-circuit current using the simple internal-resistance model.
17. Explain why deliberately short-circuiting a battery is dangerous.
18. Explain the energy transfers that occur when a real battery supplies current.
19. Describe how a voltmeter, ammeter and variable resistor could be used to determine the emf and internal resistance of a battery experimentally.
20. Explain how internal resistance makes the behavior of a real battery different from that of an ideal voltage source.