EMF and Internal Resistance

Every real battery or power supply has some resistance of its own. This internal resistance affects the voltage available to the external circuit and is a key concept for A-Level Physics.

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Electromotive Force (EMF)

The electromotive force (EMF), symbol ε (epsilon), of a source is the total energy transferred per unit charge as charge passes through the source. It is measured in volts.

ε = W / Q

Despite its name, EMF is not a force — it is an energy per unit charge, just like potential difference. The key distinction is:

• EMF is the energy given to each coulomb of charge by the source (e.g., a battery converts chemical energy into electrical energy).
• Potential difference is the energy transferred by each coulomb as it passes through a component.

EMF represents the total "budget" of energy that the battery provides to each coulomb. But not all of this energy reaches the external circuit — some is wasted inside the battery itself.

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Internal Resistance

All real sources of EMF — batteries, cells, generators, solar cells — have internal resistance, symbol r. This resistance comes from the materials inside the source. For a chemical cell, the internal resistance arises from the resistance of the electrolyte and the electrodes.

When current flows through the source, some energy is dissipated inside it as heat. This energy is "lost" from the perspective of the external circuit.

The Key Equation

For a circuit with a source of EMF ε, internal resistance r, and external resistance R:

ε = V + Ir

or equivalently:

ε = IR + Ir = I(R + r)

where:

• ε = EMF of the source (V)
• V = terminal potential difference (the voltage across the external circuit) (V)
• I = current (A)
• r = internal resistance (Ω)
• R = external (load) resistance (Ω)

The term Ir is called the lost volts — the voltage "used up" inside the battery driving current through its own internal resistance. The term V = ε - Ir is the terminal p.d. — the voltage actually available to the external circuit.

Worked Example 1

A battery has an EMF of 6.0 V and an internal resistance of 0.50 Ω. It is connected to a 5.5 Ω external resistor. Calculate (a) the current in the circuit, (b) the terminal p.d., and (c) the lost volts.

Solution:

• (a) I = ε/(R + r) = 6.0/(5.5 + 0.50) = 6.0/6.0 = 1.0 A
• (b) V = ε - Ir = 6.0 - (1.0 × 0.50) = 6.0 - 0.50 = 5.5 V
• (c) Lost volts = Ir = 1.0 × 0.50 = 0.50 V (check: 5.5 + 0.50 = 6.0 V ✓)

Worked Example 2: Power Delivered to the Load

A cell has ε = 1.50 V and r = 0.80 Ω. It drives a current through a 2.2 Ω load resistor. Calculate (a) the current, (b) the terminal p.d., (c) the power dissipated in the external resistor, and (d) the power wasted internally.

Solution:

• (a) I = ε/(R + r) = 1.50/(2.2 + 0.80) = 1.50/3.0 = 0.50 A
• (b) V = ε - Ir = 1.50 - 0.50 × 0.80 = 1.50 - 0.40 = 1.10 V
• (c) P_external = I²R = (0.50)² × 2.2 = 0.25 × 2.2 = 0.55 W
• (d) P_internal = I²r = (0.50)² × 0.80 = 0.25 × 0.80 = 0.20 W
• Check: Total power = εI = 1.50 × 0.50 = 0.75 W = 0.55 + 0.20 ✓

Worked Example 3: Maximum Power Transfer

For what value of external resistance R does a battery deliver maximum power to the load?

Analysis:

• P_load = I²R = [ε/(R + r)]² × R = ε²R / (R + r)²
• Maximum power to the load occurs when R = r (this can be proven by calculus or by testing values)
• When R = r: P_max = ε² / (4r)
• At maximum power transfer, the efficiency is only 50% — half the power is wasted internally

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How Terminal p.d. Depends on Current

The equation V = ε - Ir tells us that the terminal p.d. depends on the current drawn from the battery:

• When I = 0 (open circuit, no current flowing): V = ε. The terminal p.d. equals the full EMF. This is how you can measure a battery's EMF — connect a high-resistance voltmeter when no current is being drawn.

• When I increases: The lost volts (Ir) increase, so V decreases. Under heavy load, a battery's terminal voltage drops noticeably.

• When the battery is "short-circuited" (R = 0): I = ε/r and all the EMF is wasted internally. V = 0. This is dangerous because the large current can cause overheating.

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Measuring EMF and Internal Resistance Experimentally

A standard A-Level practical for measuring ε and r uses the following method:

1. Set up a circuit with a cell, a variable resistor, an ammeter (in series), and a voltmeter (across the cell terminals).

2. Adjust the variable resistor to give different current values and record V (terminal p.d.) and I (current) for each.

3. Plot a graph of V against I.

From V = ε - Ir, comparing with y = mx + c:

• y-intercept = ε (the EMF)
• gradient = -r (the negative of internal resistance)

The Graphical Method in Detail

The V-I graph should be a straight line with a negative gradient. Extrapolating to where I = 0 gives the EMF. The magnitude of the gradient gives the internal resistance.

This is an elegant method because:

• It averages over many readings, reducing random errors.
• It does not require you to know ε or r in advance.
• The y-intercept gives the EMF directly — the value when no current flows.
• A non-linear graph would indicate the internal resistance is changing (e.g., the cell is heating up).

Worked Example 4: Interpreting Experimental Data

A student records the following data for a cell:

I (A)	V (V)
0.20	1.38
0.40	1.26
0.60	1.14
0.80	1.02
1.00	0.90

Determine the EMF and internal resistance.

Solution:

• Gradient = ΔV/ΔI = (0.90 - 1.38)/(1.00 - 0.20) = -0.48/0.80 = -0.60 V A⁻¹
• Internal resistance r = 0.60 Ω
• y-intercept: using V = ε - Ir with any data point, e.g. (0.20, 1.38): 1.38 = ε - 0.60 × 0.20 → ε = 1.38 + 0.12 = 1.50 V

Alternative Graphical Method: Plotting ε = V + Ir

You can also rearrange the equation as V = ε - Ir and plot V on the y-axis against I on the x-axis. The x-intercept gives I_max = ε/r (the short-circuit current), and the area under the line between any two current values gives information about energy transfer.

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Cells in Series and Parallel

Cells in Series (same direction)

• Total EMF = ε₁ + ε₂ + ε₃ + ...
• Total internal resistance = r₁ + r₂ + r₃ + ...

This is why we put batteries in series in a torch — to increase the total voltage.

Cells in Series (opposing)

• Total EMF = ε₁ - ε₂ (if ε₁ > ε₂)
• Total internal resistance = r₁ + r₂ (internal resistances always add)
• Current flows in the direction driven by the larger EMF
• The smaller battery is being charged

Cells in Parallel (identical cells)

• Total EMF = ε (unchanged)
• Total internal resistance = r/n (where n is the number of cells)

Connecting identical cells in parallel keeps the EMF the same but reduces the internal resistance, allowing the combination to deliver more current to a heavy load.

Worked Example 5: Cells in Opposition

A 12 V battery (r₁ = 1.0 Ω) and a 4.0 V battery (r₂ = 0.50 Ω) are connected in opposition across a 7.5 Ω resistor. Find the current and determine what happens to the 4.0 V battery.

Solution:

• Net EMF = 12 - 4.0 = 8.0 V
• Total R = R + r₁ + r₂ = 7.5 + 1.0 + 0.50 = 9.0 Ω
• I = 8.0/9.0 = 0.89 A
• The current flows in the direction driven by the 12 V battery
• The 4.0 V battery is being charged by the 12 V battery (current flows into its positive terminal)

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Practical Significance

Internal resistance explains several everyday observations:

• Why car headlights dim when you start the engine: The starter motor draws a very large current (perhaps 200 A), causing a large lost-voltage across the battery's internal resistance, which temporarily reduces the terminal p.d. available to the headlights.

• Why batteries "go flat": As a battery ages, its internal resistance increases (due to chemical changes). Even though the EMF may not change much, the increasing internal resistance means more energy is wasted internally and less is available to the external circuit.

• Why high-quality power supplies have low internal resistance: A good power supply should deliver a stable voltage regardless of the current drawn. This requires a very low internal resistance so that the lost volts remain negligible.

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Common Exam Mistakes

Mistake	Correction
Confusing EMF with terminal p.d.	EMF = total energy per coulomb from the source; terminal p.d. = voltage available after internal losses
Using V = IR with ε instead of V	ε = I(R + r), not ε = IR. Use terminal p.d. V = IR for the external circuit
Forgetting that measuring EMF requires zero current	A voltmeter across a loaded battery reads terminal p.d., not EMF
Saying "EMF is a force"	EMF has units of volts (J C⁻¹) — it is an energy per unit charge
Omitting internal resistance from total R	Total circuit resistance = R_external + r

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A-Level Deep Dive: EMF and Internal Resistance

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic 3.5 (electromotive force and internal resistance) establishes the definition of EMF $\varepsilon$ε as the energy transferred from a chemical (or other) source to each coulomb of charge passing through the source, the equation $\varepsilon = I(R + r)$ε=I(R+r) for a complete circuit, and the experimental determination of $\varepsilon$ε and $r$r from a graph of terminal p.d. $V$V against current $I$I (refer to the official Pearson Edexcel specification document for exact wording). The sub-topic builds on every prior strand of Topic 3: the definitions of charge, current and p.d. (3.1), the resistance relation $V = IR$V=IR (3.2), and the rules for combining resistors (3.4). It is examined on 9PH0-01 (Paper 1) and is the explicit subject of Core Practical 8 on 9PH0-03 (Paper 3, General and Practical Principles). The Edexcel formula booklet does list $\varepsilon = I(R + r)$ε=I(R+r) and $V = \varepsilon - Ir$V=ε−Ir, but candidates must be able to derive the second from the first and to interpret each term physically — the algebraic shortcut earns no credit on its own.

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Worked example with full mark scheme

Question (8 marks):

A student investigates a chemical cell. They connect the cell in series with a variable resistor $R$R and an ammeter, and place a voltmeter across the cell's terminals. They record the following data:

$I$I / A	$V$V / V
0.10	1.45
0.20	1.40
0.40	1.30
0.60	1.20
0.80	1.10

(a) State, in words, what is meant by the electromotive force of the cell. (2)

(b) By plotting (or otherwise analysing) $V$V against $I$I, determine the EMF $\varepsilon$ε and the internal resistance $r$r of the cell. (4)

(c) Calculate the terminal p.d. when the cell is connected to an external resistor of $3.0\ \Omega$3.0 Ω. (2)

Solution with mark scheme:

(a) B1 — EMF is the energy transferred (per unit charge) from the source to the charge passing through it. B1 — units of volts $= \text{J C}^{-1}$=J C−1, or equivalent definition referencing energy per coulomb. A bare statement "EMF is the voltage of the cell" earns 0/2 because it does not mention energy transfer per unit charge.

(b) Step 1 — recognise the linear form. $V = \varepsilon - Ir$V=ε−Ir is in the form $y = c + mx$y=c+mx with $y = V$y=V, $x = I$x=I, intercept $c = \varepsilon$c=ε and gradient $m = -r$m=−r.

M1 — using the gradient/intercept method on $V$V against $I$I (or fitting a line through the data).

Step 2 — gradient. Using two well-separated points, $(0.10, 1.45)$(0.10,1.45) and $(0.80, 1.10)$(0.80,1.10):

$m = \dfrac{1.10 - 1.45}{0.80 - 0.10} = \dfrac{-0.35}{0.70} = -0.50\ \text{V A}^{-1}$m=0.80−0.101.10−1.45​=0.70−0.35​=−0.50 V A−1

So $r = 0.50\ \Omega$r=0.50 Ω.

A1 — $r = 0.50\ \Omega$r=0.50 Ω (allow 0.45 to 0.55 $\Omega$Ω from a hand-drawn line of best fit).

Step 3 — intercept. Using $V = \varepsilon - Ir$V=ε−Ir at $(0.10, 1.45)$(0.10,1.45): $1.45 = \varepsilon - 0.50 \times 0.10$1.45=ε−0.50×0.10, giving $\varepsilon = 1.50\ \text{V}$ε=1.50 V.

M1 A1 — substitution into $V = \varepsilon - Ir$V=ε−Ir and correct $\varepsilon = 1.50\ \text{V}$ε=1.50 V.

(c) Step 1 — total circuit resistance. $R + r = 3.0 + 0.50 = 3.5\ \Omega$R+r=3.0+0.50=3.5 Ω.

Step 2 — current. $I = \varepsilon / (R + r) = 1.50 / 3.5 = 0.429\ \text{A}$I=ε/(R+r)=1.50/3.5=0.429 A.

Step 3 — terminal p.d. $V = IR = 0.429 \times 3.0 = 1.29\ \text{V}$V=IR=0.429×3.0=1.29 V (or equivalently $V = \varepsilon - Ir = 1.50 - 0.429 \times 0.50 = 1.29\ \text{V}$V=ε−Ir=1.50−0.429×0.50=1.29 V).

M1 — correct application of $\varepsilon = I(R + r)$ε=I(R+r) or $V = \varepsilon - Ir$V=ε−Ir. A1 — $V \approx 1.3\ \text{V}$V≈1.3 V to 2 s.f.

Total: 8 marks (B1 B1 + M1 A1 M1 A1 + M1 A1).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A 12 V car battery has internal resistance $r = 0.020\ \Omega$r=0.020 Ω. When the starter motor is engaged, the current drawn is $180\ \text{A}$180 A.

(a) Calculate the terminal p.d. while the starter motor is operating. (2)

(b) The car's headlights are connected directly across the battery terminals. Explain, with reference to your answer to (a), why the headlights dim noticeably when the engine is started. (2)

(c) Suggest one way in which the manufacturer could reduce the dimming effect, justifying your answer using $V = \varepsilon - Ir$V=ε−Ir. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — substituting $V = \varepsilon - Ir = 12 - 180 \times 0.020$V=ε−Ir=12−180×0.020.
• A1 (AO1.1b) — $V = 12 - 3.6 = 8.4\ \text{V}$V=12−3.6=8.4 V.

(b)

• M1 (AO2.1) — recognising that the headlights see the terminal p.d., not the EMF, and that this drops from $12\ \text{V}$12 V (negligible draw) to $8.4\ \text{V}$8.4 V (heavy draw).
• A1 (AO2.5) — concluding that headlight power $P = V^2/R$P=V2/R falls in proportion to $V^2$V2, so brightness drops sharply (a factor $(8.4/12)^2 \approx 0.49$(8.4/12)2≈0.49, roughly halving).

(c)

• M1 (AO3.1a) — identifying that reducing $r$r (e.g. larger battery, parallel cells, thicker internal connectors) reduces the lost volts $Ir$Ir for a given $I$I.
• A1 (AO3.2a) — quantitative justification: halving $r$r to $0.010\ \Omega$0.010 Ω would reduce the lost volts to $1.8\ \text{V}$1.8 V, giving a terminal p.d. of $10.2\ \text{V}$10.2 V during cranking.

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. The question is deliberately balanced across all three AOs — Edexcel uses internal-resistance contexts as a vehicle for AO3 evaluation marks because the equation $V = \varepsilon - Ir$V=ε−Ir couples a circuit calculation (AO1) to a physical interpretation (AO2) and a design recommendation (AO3) in three short sub-parts.

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Synoptic links

Connects to:

1. Topic 3 sub-topic 3.6 — Kirchhoff's laws and energy conservation: $\varepsilon = I(R + r)$ε=I(R+r) is itself a statement of Kirchhoff's voltage law applied to a single loop containing one source. The term $\varepsilon$ε represents the energy per coulomb supplied; $IR$IR and $Ir$Ir are the energies per coulomb dissipated in the external and internal resistances. Adding the equation $\varepsilon - IR - Ir = 0$ε−IR−Ir=0 around the loop is the conservation-of-energy form of KVL for this circuit.