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

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

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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

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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%

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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.

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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.

This effect is often called voltage sag.

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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.

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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.

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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.

Several pairs of V and I measurements are recorded.

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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

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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

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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

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.