EMF, Internal Resistance and Terminal PD

Real batteries and cells are not ideal voltage sources. They have internal resistance that affects the voltage available to the external circuit. Understanding internal resistance is essential for circuit analysis and is a key topic in AQA A-Level Physics.

Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.6 — EMF and internal resistance, covering the equation ε = I(R + r) and its rearrangements, the distinction between EMF and terminal p.d., the meaning of "lost volts", and Required Practical 6 — determining ε and r experimentally from the gradient and intercept of a V_terminal versus I graph. (Refer to the official AQA specification document for exact wording.)

Synoptic links:

• 3.5.1.2 — Potential difference and EMF: the formal distinction between energy supplied per coulomb (ε) and energy delivered per coulomb (V_terminal) introduced in Lesson 2 becomes operational here as ε − V_terminal = Ir. The definitional groundwork pays off in this lesson.
• 3.5.1.5 — Series and parallel circuits: internal resistance r is the hidden series resistor inside the cell. The combination ε = I(R + r) is the series-resistance formula with r appearing alongside R. Cells in parallel reduce r/n, an important practical reason for paralleling batteries.
• 3.4 — Mechanics, energy and momentum: the energy-balance ε = V_terminal + V_lost per coulomb mirrors the work-energy theorem in mechanics — total work done equals useful work plus dissipated energy. Conservation of energy is the underlying principle.

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The Concept of Internal Resistance

Every real source of EMF (battery, cell, generator) has some resistance within itself. This is called internal resistance (r).

When current flows through the source, some energy is dissipated as heat inside the source due to this internal resistance. This means the voltage available to the external circuit (the terminal p.d.) is less than the EMF.

The Circuit Model

A real battery can be modelled as an ideal EMF source (ε) in series with a small internal resistance (r):

Circuit description:

• Inside the battery: ideal EMF (ε) in series with internal resistance (r)
• External circuit: load resistance (R)
• Current I flows around the complete circuit

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The Key Equation

Applying conservation of energy around the circuit:

$\varepsilon = I(R + r)$ε=I(R+r)

or equivalently:

$\varepsilon = IR + Ir$ε=IR+Ir

where:

• ε = EMF of the source (V)
• I = current in the circuit (A)
• R = external (load) resistance (Ω)
• r = internal resistance (Ω)
• IR = p.d. across the external resistance (terminal p.d., V)
• Ir = "lost volts" (p.d. across internal resistance, V)

Terminal p.d. and Lost Volts

Rearranging: V_terminal = ε − Ir

Term	Meaning
ε (EMF)	Total energy supplied per coulomb by the source
V_terminal = IR	Energy per coulomb delivered to the external circuit ("useful" voltage)
Ir (lost volts)	Energy per coulomb dissipated inside the source as heat

Key Point: The terminal p.d. is what you would measure with a voltmeter connected across the battery terminals when current is flowing. It is always less than the EMF when current flows.

When is V_terminal = ε?

The terminal p.d. equals the EMF only when no current flows (I = 0):

• When the circuit is open (no external resistance connected)
• When measured with a very high-resistance voltmeter (which draws negligible current)

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Worked Examples

Worked Example 1 — Basic Internal Resistance

A battery has an EMF of 9.0 V and an internal resistance of 0.50 Ω. It is connected to a 5.5 Ω external resistor. Calculate (a) the current, (b) the terminal p.d., and (c) the lost volts.

Solution:

(a) ε = I(R + r)

I = ε / (R + r) = 9.0 / (5.5 + 0.50) = 9.0 / 6.0 = 1.5 A

(b) V_terminal = IR = 1.5 × 5.5 = 8.25 V

Or: V_terminal = ε − Ir = 9.0 − (1.5 × 0.50) = 9.0 − 0.75 = 8.25 V ✓

(c) Lost volts = Ir = 1.5 × 0.50 = 0.75 V

Check: V_terminal + lost volts = 8.25 + 0.75 = 9.0 V = ε ✓

Worked Example 2 — Finding EMF and Internal Resistance

A student connects a variable resistor to a cell and measures the following:

R (Ω)	I (A)	V_terminal (V)
10.0	0.138	1.38
5.0	0.250	1.25
2.0	0.500	1.00

Use two sets of readings to determine the EMF and internal resistance.

Solution:

Using V = ε − Ir:

From reading 1: 1.38 = ε − 0.138r ... (1) From reading 3: 1.00 = ε − 0.500r ... (2)

Subtracting (2) from (1): 0.38 = 0.362r

r = 0.38 / 0.362 = 1.05 Ω ≈ 1.0 Ω

Substituting back into (1): ε = 1.38 + (0.138 × 1.05) = 1.38 + 0.145 = 1.52 V ≈ 1.5 V

Check with reading 2: V = 1.52 − (0.250 × 1.05) = 1.52 − 0.263 = 1.26 V ≈ 1.25 V ✓

Worked Example 3 — Power Delivered to External Resistance

Using the battery from Worked Example 1 (ε = 9.0 V, r = 0.50 Ω), calculate the power delivered to the external resistor and the power wasted internally.

Solution:

I = 1.5 A (calculated above)

Power to external resistor: P_ext = I²R = (1.5)² × 5.5 = 2.25 × 5.5 = 12.4 W

Power wasted internally: P_int = I²r = (1.5)² × 0.50 = 2.25 × 0.50 = 1.125 W

Total power from source: P_total = εI = 9.0 × 1.5 = 13.5 W

Check: P_ext + P_int = 12.4 + 1.125 = 13.5 W ✓

Efficiency: η = P_ext / P_total = 12.4 / 13.5 = 91.7%

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The V-I Graph for a Source with Internal Resistance

If we plot terminal p.d. (V) against current (I) for a source with internal resistance, we get a straight line with:

• y-intercept = ε (the EMF, when I = 0)
• Gradient = −r (the negative of the internal resistance)
• x-intercept = ε/r (the short-circuit current, when V = 0)

This comes directly from the equation: V = ε − Ir, which has the form y = c + mx where m = −r.

Determining ε and r from a V-I Graph

1. Read the y-intercept → this is the EMF (ε)
2. Calculate the gradient of the line → gradient = −r, so r = −gradient

Worked Example 4 — V-I Graph Analysis

A student's V-I graph for a cell shows a straight line from (0, 1.52) to (1.2, 0.92). Determine the EMF and internal resistance.

Solution:

EMF = y-intercept = 1.52 V

Gradient = (0.92 − 1.52) / (1.2 − 0) = −0.60 / 1.2 = −0.50

r = −gradient = 0.50 Ω

Exam Tip: In the practical, you obtain ε and r from the V-I graph. Make sure you plot V on the y-axis and I on the x-axis. The y-intercept gives ε (do NOT assume the first data point is at I = 0). Draw a line of best fit and use points on the line (not data points) to calculate the gradient.

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Short Circuit

If the external resistance R = 0 (short circuit), the maximum possible current flows:

$I_{short} = \frac{\varepsilon}{r}$Ishort​=rε​

In this case, all the EMF is dropped across the internal resistance, and V_terminal = 0. All the energy is dissipated as heat inside the battery.

Worked Example 5 — Short Circuit Current

A 1.5 V AA cell has an internal resistance of 0.30 Ω. What is the short-circuit current?

Solution:

I_short = ε/r = 1.5 / 0.30 = 5.0 A

This large current would drain and potentially damage the cell quickly. This is why short circuits are dangerous — large currents cause rapid heating.

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Multiple Cells

Cells in Series

When identical cells are connected in series (same polarity):

• EMFs add: ε_total = nε (for n identical cells)
• Internal resistances add: r_total = nr

Cells in Parallel

When identical cells are connected in parallel:

• EMF stays the same: ε_total = ε
• Internal resistances combine in parallel: r_total = r/n

This is why batteries are sometimes connected in parallel — it reduces the effective internal resistance and allows larger currents to be delivered.

Worked Example 6 — Cells in Series

Four 1.5 V cells, each with internal resistance 0.40 Ω, are connected in series. They drive a current through a 10 Ω external resistor. Calculate the current.

Solution:

ε_total = 4 × 1.5 = 6.0 V

r_total = 4 × 0.40 = 1.6 Ω

I = ε_total / (R + r_total) = 6.0 / (10 + 1.6) = 6.0 / 11.6 = 0.517 A

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Worked Example: Two Cells in Parallel vs One Cell Driving a 2 Ω Load

Two identical 1.5 V cells, each with internal resistance r = 0.20 Ω, are connected in parallel and used to power an external load R = 2.0 Ω. (a) Calculate the total current drawn from the parallel cell combination and the terminal p.d. across the load. (b) Repeat the calculation for a single cell of the same type driving the same 2.0 Ω load. (c) Compare the two cases and explain when a parallel arrangement is the right design choice.

Solution — Part (a): Parallel cells driving the load.

When identical cells are connected in parallel:

• e.m.f. of the combination: ε = 1.5 V (unchanged — every cell pushes with the same e.m.f.)
• Effective internal resistance: r_eff = r / n = 0.20 / 2 = 0.10 Ω

Apply the e.m.f. equation ε = I(R + r_eff):

I = ε / (R + r_eff) = 1.5 / (2.0 + 0.10) = 1.5 / 2.10 ≈ 0.714 A

Terminal p.d. across the load: V = IR = 0.714 × 2.0 = 1.43 V

Part (b): Single cell driving the same load.

With one cell, r = 0.20 Ω.

I = ε / (R + r) = 1.5 / (2.0 + 0.20) = 1.5 / 2.20 ≈ 0.682 A

Terminal p.d. across the load: V = IR = 0.682 × 2.0 = 1.36 V

Part (c): Comparison.

Current increase: 0.714 / 0.682 ≈ 1.047, or about a 4.7% increase.

Terminal p.d. increase: 1.43 / 1.36 ≈ 1.051, also about 5%.

At first glance, the parallel arrangement looks like a modest improvement. But the more important figure is the current drawn from each individual cell: with two cells in parallel, each cell supplies only 0.714 / 2 = 0.357 A, whereas the single-cell arrangement forces 0.682 A from one cell. The current per cell almost halves. That has three practical consequences. First, the I²r heating dissipated inside each cell falls by a factor of approximately four (since 0.357² ≈ 0.127 vs 0.682² ≈ 0.465), so the cells run cooler and last longer. Second, the cells share the load, so the combined bank can deliver currents that would exceed the rated maximum for a single cell. Third, the parallel arrangement is robust against one cell failing — the remaining cell continues to supply the circuit, though at reduced performance.

When to choose parallel cells: when the load demands a current near or above any single cell's safe limit, when long battery life is more important than absolute voltage, or when redundancy matters (e.g. emergency lighting, marine applications). When to avoid parallel cells: when the cells have mismatched e.m.f.s — even small differences cause large circulating currents between cells, wasting energy and risking damage. This is why batteries should always be replaced as a matched set, not one cell at a time.

Note: A common A-Level pitfall is to claim that "putting cells in parallel doubles the current" — true only in the short-circuit limit where R → 0. For typical loads where R ≫ r, the current gain is modest because the external resistance still dominates the total circuit resistance. The big advantage of parallel cells is current-sharing and longevity, not raw current at typical loads.

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Specimen Exam Question