Internal Resistance and EMF

Spec mapping: OCR H556 Module 4.2 — Energy, power and resistance (electromotive force $\varepsilon$ε; internal resistance $r$r of a real source; terminal pd $V = \varepsilon - Ir$V=ε−Ir; the relationship $\varepsilon = I(R + r)$ε=I(R+r); experimental determination of $\varepsilon$ε and $r$r from a graph of terminal pd against current; PAG 4 anchor practical). Refer to the official OCR H556 specification document for exact wording.

In lesson 4 we distinguished EMF (energy supplied per unit charge by a source) from potential difference (energy delivered per unit charge to a component). For an ideal source the two would be equal across the source's terminals. For a real source they are not — every battery, cell, generator or thermocouple has some internal resistance $r$r that dissipates a fraction of the EMF as heat inside the source itself. The pd measured at the source's terminals is therefore always less than the EMF when current is drawn.

This lesson derives the central equation $\varepsilon = I(R + r)$ε=I(R+r), introduces the concept of "lost volts" $Ir$Ir dropped inside the cell, develops the graphical method for extracting $\varepsilon$ε and $r$r from a single $V$V-vs-$I$I plot (the PAG 4 anchor practical for this topic), and explores the maximum-power-transfer condition $R = r$R=r as a teaser for impedance-matching theory. By the end you should be able to compute terminal pd, lost volts, and external/internal power for any real cell + load combination, and analyse $V$V-vs-$I$I data with confidence.

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The Model of a Real Cell

A real cell is modelled as an ideal EMF source $\varepsilon$ε in series with a fixed internal resistance $r$r. The "internal resistance" is not a separate physical resistor you can remove — it is the mathematical embodiment of all losses inside the cell: ionic conduction through the electrolyte, electrode-electrolyte contact resistance, electrode material resistivity. The single equivalent value $r$r captures them all.

flowchart LR
  subgraph Cell["Cell (terminals A and B)"]
    EMF["ε (ideal EMF)"] --- IR["r (internal resistance)"]
  end
  Cell --> R["External load R"]
  R --> Cell

When the cell drives current $I$I through external load $R$R, the circuit is a single series loop. By Kirchhoff-2 (lesson 11, intuitively here):

$\varepsilon = IR + Ir = I(R + r)$ε=IR+Ir=I(R+r)

The quantity $Ir$Ir is voltage dropped inside the cell, called the lost volts. It represents electrical energy per coulomb that is dissipated as heat inside the cell rather than delivered to the external load.

The terminal pd $V$V — what an ideal voltmeter (infinite $R$R, draws no current) would read across the cell's external terminals — is:

$\boxed{\; V = \varepsilon - Ir \;}$V=ε−Ir​

When $I = 0$I=0 (no current drawn): $V = \varepsilon$V=ε exactly. As $I$I increases, the terminal pd falls below the EMF by the lost-volts amount.

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Worked Example 1 — Full Energy Audit

A cell of EMF $\varepsilon = 1.50$ε=1.50 V and internal resistance $r = 0.50$r=0.50 $\Omega$Ω is connected to an external resistor $R = 4.5$R=4.5 $\Omega$Ω.

(a) The current. $R + r = 5.0$R+r=5.0 $\Omega$Ω, so $I = \varepsilon/(R+r) = 1.50/5.0 = \mathbf{0.30}$I=ε/(R+r)=1.50/5.0=0.30 A.

(b) The terminal pd. $V = \varepsilon - Ir = 1.50 - 0.30 \times 0.50 = \mathbf{1.35}$V=ε−Ir=1.50−0.30×0.50=1.35 V. (Cross-check: $V = IR = 0.30 \times 4.5 = 1.35$V=IR=0.30×4.5=1.35 V ✓.)

(c) The lost volts. $V_{\text{lost}} = Ir = 0.30 \times 0.50 = \mathbf{0.15}$Vlost​=Ir=0.30×0.50=0.15 V.

(d) The power dissipations.

• External load: $P_{\text{ext}} = I^2 R = 0.30^2 \times 4.5 = \mathbf{0.405}$Pext​=I2R=0.302×4.5=0.405 W.
• Internal (wasted as heat in the cell): $P_{\text{int}} = I^2 r = 0.30^2 \times 0.50 = \mathbf{0.045}$Pint​=I2r=0.302×0.50=0.045 W.
• Total supplied by cell: $P_{\text{total}} = \varepsilon I = 1.50 \times 0.30 = \mathbf{0.450}$Ptotal​=εI=1.50×0.30=0.450 W.

Check: $P_{\text{ext}} + P_{\text{int}} = 0.405 + 0.045 = 0.450$Pext​+Pint​=0.405+0.045=0.450 W $= \varepsilon I$=εI ✓. Energy conservation: total power supplied by the EMF equals power delivered to the load plus power wasted inside the cell.

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Dependence on Load Resistance — A Table of Cases

As the load $R$R changes (with $r = 0.50$r=0.50 $\Omega$Ω, $\varepsilon = 1.50$ε=1.50 V), the current and terminal pd change too:

$R$R ($\Omega$Ω)	$I = \varepsilon/(R+r)$I=ε/(R+r)	$V = \varepsilon - Ir$V=ε−Ir	Comment
$\infty$∞ (open circuit)	$0$0	$1.50$1.50 V ($=\varepsilon$=ε)	No current, no lost volts
100	0.0149 A	1.493 V	$R \gg r$R≫r: efficient, low loss
10	0.143 A	1.429 V	$R = 20r$R=20r: small loss
1	1.00 A	1.00 V	$R = 2r$R=2r: ~50% efficient
0.5	1.50 A	0.75 V	$R = r$R=r: maximum power transfer, 50% efficient
0 (short circuit)	3.00 A	0 V	All voltage dropped internally; $I_{\max} = \varepsilon/r$Imax​=ε/r

Observations:

• Open circuit ($R \to \infty$R→∞): $I \to 0$I→0, lost volts $\to 0$→0, $V \to \varepsilon$V→ε. The open-circuit voltage equals the EMF — this is the only way to measure $\varepsilon$ε directly without drawing current.
• Short circuit ($R \to 0$R→0): all the EMF is dissipated inside the cell, $V = 0$V=0, and the current is $\varepsilon/r$ε/r — the maximum current the cell can deliver. For this example, $I_{\max} = 3.0$Imax​=3.0 A.
• In between: the terminal pd falls smoothly from $\varepsilon$ε toward $0$0 as the load decreases.

Exam Tip. "Why does the terminal pd of a car battery fall when the headlights are switched on?" Answer: switching on the headlights closes a circuit that draws more current $I$I; the lost volts $Ir$Ir rise proportionally; $V = \varepsilon - Ir$V=ε−Ir falls. This is the cause of the brief "dim" you see in dashboard lights when the starter motor turns the engine over — the starter draws hundreds of amps, the lost volts $Ir$Ir become substantial, and $V$V momentarily drops.

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Maximum Power Transfer — A Beautiful Result

A surprisingly deep theorem: the power delivered to the external load $P_{\text{ext}} = I^2 R$Pext​=I2R is maximised when $R = r$R=r.

Derivation: $P_{\text{ext}} = I^2 R = (\varepsilon/(R+r))^2 \, R = \varepsilon^2 R/(R+r)^2$Pext​=I2R=(ε/(R+r))2R=ε2R/(R+r)2. Differentiating with respect to $R$R and setting $dP/dR = 0$dP/dR=0:

$\frac{dP_{\text{ext}}}{dR} = \varepsilon^2 \, \frac{(R+r)^2 - R \cdot 2(R+r)}{(R+r)^4} = \varepsilon^2 \, \frac{(R+r) - 2R}{(R+r)^3} = \varepsilon^2 \, \frac{r - R}{(R+r)^3} = 0$dRdPext​​=ε2(R+r)4(R+r)2−R⋅2(R+r)​=ε2(R+r)3(R+r)−2R​=ε2(R+r)3r−R​=0

gives $R = r$R=r.

At maximum power transfer:

• $I = \varepsilon/(2r) = \varepsilon/(2R)$I=ε/(2r)=ε/(2R) (half the short-circuit current).
• $P_{\text{ext, max}} = \varepsilon^2 r / (2r)^2 = \varepsilon^2/(4r)$Pext, max​=ε2r/(2r)2=ε2/(4r).
• $P_{\text{int}} = I^2 r = \varepsilon^2/(4r)$Pint​=I2r=ε2/(4r) (same as external).
• Efficiency = 50% (half goes to load, half wasted inside the cell).

This is why batteries designed to deliver significant power (car batteries, lithium-polymer drone batteries, capacitor banks) have deliberately low internal resistance — to achieve high $P_{\text{ext, max}} = \varepsilon^2/(4r)$Pext, max​=ε2/(4r). For $R > r$R>r the efficiency rises above 50% but raw power falls. For a torch battery (small $I$I, long lifetime) you want $R \gg r$R≫r — efficiency, not maximum power.

Audio engineering footnote: the impedance-matching condition $Z_{\text{load}} = Z_{\text{source}}$Zload​=Zsource​ is the AC analogue of $R = r$R=r — it maximises power transfer at the cost of 50% efficiency. Modern amplifier output stages avoid this and use low-output-impedance drive (an ideal voltage source) rather than matched impedance, because efficiency at scale matters more than absolute power delivered to one channel.

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PAG 4 Anchor Practical — Determining $\varepsilon$ε and $r$r from a $V$V-vs-$I$I Graph

The OCR H556 specification requires students to perform the experimental determination of EMF and internal resistance of a cell. This is a PAG 4 anchor practical.

Apparatus

• Cell under test (e.g. a fresh AA cell).
• Variable resistor (rheostat $0$0 to $\sim 10$∼10 $\Omega$Ω or a decade box).
• Digital ammeter ($0$0 – $1$1 A range).
• Digital voltmeter ($0$0 – $2$2 V range, high input impedance).
• Switch.
• Connecting wires.

Circuit

flowchart LR
  CELL["Cell (ε, r) under test"] --> SW["Switch"]
  SW --> AM["Ammeter (in series)"]
  AM --> RH["Variable resistor (rheostat)"]
  RH --> CELL
  CELL -.-> VM["Voltmeter (in parallel with cell terminals)"]

Procedure

1. Connect the cell to the variable resistor via the ammeter and switch. Connect the voltmeter directly across the cell's terminals (in parallel).
2. Close the switch only while taking each reading. This minimises battery drain and prevents $r$r drifting upward due to internal heating.
3. Starting with the rheostat at its maximum resistance (lowest current), record the pair $(I, V)$(I,V).
4. Reduce the rheostat resistance in small steps, recording $(I, V)$(I,V) at each setting. Get 6-8 data points spanning the safe current range. Do not exceed the cell's safe current rating.
5. Plot $V$V on the $y$y-axis against $I$I on the $x$x-axis.

The Key Graph

Starting from $V = \varepsilon - Ir$V=ε−Ir, this is a straight line of the form $y = c + mx$y=c+mx with:

• $y$y-intercept (at $I = 0$I=0) $= \varepsilon$=ε (the EMF).
• Gradient $= -r$=−r (negative of the internal resistance).

So the $y$y-intercept of the $V$V-$I$I graph gives the EMF, and the negative of the gradient gives the internal resistance. One graph, two quantities.

flowchart LR
  ORIGIN["I = 0"] --> Y1["V = ε (intercept)"]
  Y1 --> SLOPE["gradient = -r"]
  SLOPE --> X1["V falls linearly as I rises"]

Why Not Just Measure Open-Circuit Voltage?

You could measure $\varepsilon$ε directly with an ideal voltmeter on an open circuit ($I = 0$I=0, $V = \varepsilon$V=ε). But that gives you only $\varepsilon$ε, not $r$r. The graphical method extracts both in a single experiment, and the gradient is a robust measurement that averages random scatter across multiple data points.

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Worked Example 2 — Graphical Analysis

A student measures the following $V$V-$I$I data for a cell:

$I$I (A)	$V$V (V)
0.10	1.45
0.20	1.40
0.30	1.35
0.40	1.30
0.50	1.25

Determine $\varepsilon$ε and $r$r.

Intercept (extrapolate to $I = 0$I=0): $V = 1.50$V=1.50 V $\Rightarrow \mathbf{\varepsilon = 1.50 \text{ V}}$⇒ε=1.50 V.

Gradient: $\Delta V / \Delta I = (1.25 - 1.45)/(0.50 - 0.10) = -0.20/0.40 = -0.50$ΔV/ΔI=(1.25−1.45)/(0.50−0.10)=−0.20/0.40=−0.50 V A$^{-1}$−1.

So $\mathbf{r = 0.50 \text{ }\Omega}$r=0.50 Ω.

Verification — substitute a data point: at $I = 0.30$I=0.30 A, $V = \varepsilon - Ir = 1.50 - 0.30 \times 0.50 = 1.35$V=ε−Ir=1.50−0.30×0.50=1.35 V ✓.

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Sources of Error in the Practical

• Heating of the internal resistance. As current is drawn, $r$r drifts slightly upward (Joule heating inside the cell), which curves the $V$V-$I$I line. Mitigation: keep current small, close the switch only while reading.
• Voltmeter not ideal. If the voltmeter's resistance is too low, it draws a measurable current of its own, distorting both $V$V and $I$I. Use a high-impedance digital voltmeter ($> 10^7$>107 $\Omega$Ω).
• Ammeter resistance. A real ammeter has a small non-zero resistance, which is added to the external load. Use a low-resistance digital ammeter; if necessary correct for it.
• Contact resistance at terminals and crocodile clips — typically small ($\sim 0.01$∼0.01 $\Omega$Ω) but non-zero, and gets worse with oxidation.

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Why Internal Resistance Matters in Real Devices

• Battery health. Old or cold cells have higher $r$r (chemistry has degraded, less electrolyte conductivity). A failing car battery "cranks slowly" because the starter motor draws $\sim 200$∼200 A and lost volts $Ir = 200 \times r$Ir=200×r can drop $V$V below the 9 V threshold the starter solenoid needs.
• High-current applications. Speakers, motors, camera flashes, electric vehicle motors all draw large peak currents. Low-$r$r chemistries (lithium-polymer, lithium-iron-phosphate) are essential — alkaline cells simply cannot deliver the current without massive voltage sag.
• Energy efficiency. $P_{\text{int}} = I^2 r$Pint​=I2r is wasted inside the cell as heat. This is why fast-discharging batteries get warm. For an electric vehicle, internal-resistance losses are typically 5-10% of charge-discharge cycle energy.
• Safety. A cell with very low $r$r short-circuited can deliver enormous currents — a $1.5$1.5 V AA with $r \sim 0.5$r∼0.5 $\Omega$Ω shorts at $3$3 A (mild). A $3.7$3.7 V lithium-ion 18650 cell with $r \sim 50$r∼50 m$\Omega$Ω shorts at $\sim 70$∼70 A — enough to start a fire from the resulting $I^2 r$I2r dissipation in the wire or terminals.

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Worked Example 3 — Two Cells in Series

Two identical cells, each EMF $1.5$1.5 V, internal resistance $0.40$0.40 $\Omega$Ω, connected in series with a $2.2$2.2 $\Omega$Ω external resistor.

• Cells in series: $\varepsilon_{\text{total}} = 2 \times 1.5 = \mathbf{3.0}$εtotal​=2×1.5=3.0 V.
• Internal resistances in series: $r_{\text{total}} = 2 \times 0.40 = \mathbf{0.80}$rtotal​=2×0.40=0.80 $\Omega$Ω.
• Total circuit resistance: $R + r_{\text{total}} = 2.2 + 0.80 = 3.0$R+rtotal​=2.2+0.80=3.0 $\Omega$Ω.
• Current: $I = \varepsilon_{\text{total}}/3.0 = \mathbf{1.0}$I=εtotal​/3.0=1.0 A.
• Terminal pd across the two-cell pack: $V = \varepsilon_{\text{total}} - I r_{\text{total}} = 3.0 - 1.0 \times 0.80 = \mathbf{2.2}$V=εtotal​−Irtotal​=3.0−1.0×0.80=2.2 V (check: also $= IR = 1.0 \times 2.2$=IR=1.0×2.2 ✓).

Series cells: EMFs add, internal resistances add. Doubles the EMF and doubles the internal resistance — net effect is usually higher current to the load (if $R$R is the dominant resistance).

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Worked Example 4 — Two Cells in Parallel

Two identical cells of EMF $1.5$1.5 V and internal resistance $0.4$0.4 $\Omega$Ω connected in parallel (correct polarity — positive to positive, negative to negative). The combination drives a $2.2$2.2 $\Omega$Ω load.

• EMF of the parallel combination: still $1.5$1.5 V (cells with same EMF in parallel give that EMF, not double).
• Internal resistance: two resistors of $0.4$0.4 $\Omega$Ω in parallel: $r_{\text{parallel}} = 0.4/2 = 0.20$rparallel​=0.4/2=0.20 $\Omega$Ω.
• Total circuit resistance: $R + r_{\text{parallel}} = 2.2 + 0.20 = 2.4$R+rparallel​=2.2+0.20=2.4 $\Omega$Ω.
• Current: $I = 1.5/2.4 = \mathbf{0.625}$I=1.5/2.4=0.625 A.
• Each cell contributes half this: $0.313$0.313 A.

Key insight: parallel cells halve the internal resistance without changing the EMF. Useful for high-current applications where lost volts must be minimised. Parallel cells must have the same EMF and correct polarity — otherwise the higher-EMF cell discharges into the lower-EMF cell (as in the Kirchhoff-2 specimen question of lesson 11).

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