Library › E.M.F & Modelling Resistance

Physics (IAL)

E.M.F & Modelling Resistance

Revise E.M.F & Modelling Resistance for Physics (IAL) — revision notes and instant AI marking. Free to start.

📖 Revision notes · preview

Edexcel IAL Physics · Unit 4

E.M.F & Modelling Resistance

Big idea: Every real battery wastes a bit of its own energy pushing charge through itself — that "waste" is internal resistance, and it's why the voltage you actually get is always a little less than the voltage the battery promises.

Summary — What This Chapter Covers

E.M.F. (ε) is the total energy a battery gives to each coulomb of charge — measured with a voltmeter when no current flows.
Internal resistance (r) is resistance inside the battery itself, which "eats" some of that energy as heat.
Lost volts (Ir) is the energy wasted inside the battery; terminal p.d. (V) is what's left over for the circuit.
The master equation ties it all together: ε = I(R + r) = V + Ir
Core Practical 8 uses a V–I graph to find ε (y-intercept) and r (negative gradient) experimentally.
Resistance rises with temperature in metals (more ion vibration = more electron collisions).
Resistance falls with temperature in thermistors (more free charge carriers released).
Resistance falls with light intensity in LDRs (more light = more freed electrons = easier flow).

1. Electromotive Force (E.M.F.)

Picture a battery as a tiny pump. It doesn't create charge — charge is already sitting in the wires — but it gives that charge a "push" of energy, like a pump lifting water uphill so it has potential energy to flow back down through your circuit and do useful work (light a bulb, turn a motor, whatever).

E.M.F. (electromotive force) is a measure of exactly how much energy the battery gives to each coulomb of charge that passes through it. It's a slightly misleading name — it isn't actually a force at all (no newtons involved), it's an energy-per-charge quantity, just like potential difference. The "force" in the name is a historical leftover from when scientists didn't fully understand what was going on.

Definition

ε = E / Q

ε = e.m.f. (volts, V) | E = energy transformed from chemical → electrical (joules, J) | Q = charge (coulombs, C)

The trick to measuring it: e.m.f. is defined as the potential difference across the battery's terminals when no current is flowing — i.e. an "open circuit" with just a voltmeter connected. Why no current? Because the moment current flows, some energy gets lost inside the battery itself (more on that in Section 2), and what you'd measure would be slightly less than the true e.m.f. With zero current, there's no internal energy loss to worry about, so the voltmeter reads the full, honest e.m.f.

┌──────( V )──────┐ │ │ │ ┤├────────┘ └────────┘ High-resistance voltmeter connected straight across the cell terminals, with NO other circuit connected. Reading = e.m.f. (ε)

Analogy

Think of e.m.f. as the total "budget" of energy the battery hands out per coulomb. Some of that budget gets spent immediately just running the battery's own internal machinery (internal resistance) — the rest is what's actually available to power your circuit.

Practice Question

A cell has an e.m.f. of 6.0 V. Explain, in terms of energy, what this value actually means.

2. Internal Resistance

No battery is a perfect, resistance-free box of energy. Inside every cell, the chemicals and electrodes themselves have some resistance to the flow of charge — this is called internal resistance (r). As current pushes through this internal resistance, energy is converted into heat right there inside the battery (which is exactly why a battery gets warm after powering something for a while, or why your phone gets hot while fast-charging).

The cleanest way to think about this: model the real, messy battery as a perfect, resistance-free e.m.f. source, connected in series with a separate resistor r that represents all of that internal resistance. This imaginary internal resistor is in series with whatever resistor (the "load", R) you've connected in the actual circuit.

Lost volts (Vr) ┌───[ r ]───┐ ┌────┤ ├────┐ │ ┤├──ε──────── │ │ (ideal e.m.f.) │ │ │ │ BATTERY │ └───────────┬───────────┘ │ ┌────┴────┐ │ R │ ← load resistor └────┬────┘ │ Terminal p.d. (VR)

Key idea

A higher internal resistance means more "lost volts" for any given current — so the battery hands over less usable voltage to your circuit. This is why old, worn-out batteries (whose internal resistance rises as they degrade) seem "weak" even though their e.m.f. hasn't changed much.

Practice Question

Why does a battery feel warm to the touch after it has been powering a torch for 20 minutes?

3. E.M.F. vs. Terminal Potential Difference

This is the part that trips people up most, so let's be very precise about the three quantities involved and how they relate:

Symbol	Name	What it means
ε	E.m.f.	Total energy per coulomb the battery produces
V_r	Lost volts	Energy per coulomb wasted overcoming the battery's own internal resistance
V_R (or V_T)	Terminal p.d.	Energy per coulomb actually delivered to the external circuit — what's "left over"

Since energy has to balance (energy in = energy out), the e.m.f. must equal the sum of the lost volts and the terminal p.d.:

The Master Equation

ε = I(R + r) = IR + Ir = V + Ir

A crucial distinction the exam loves to test: e.m.f. describes energy transferred from the power supply to charges (energy gained), while potential difference describes energy transferred from electrical form to other forms in a component like a resistor (energy lost/converted). Same units, same-looking formula, completely opposite direction of energy flow.

Exam shortcut

If a question says "a cell of negligible internal resistance," it means r = 0, so terminal p.d. = e.m.f. exactly. No lost volts, no need for the full equation — just treat the cell as ideal.

Worked Example

A battery of e.m.f. 7.3 V and internal resistance 0.3 Ω is connected in series with a 9.5 Ω resistor. Find (a) the current, and (b) the lost volts.

(a) Use ε = I(R + r), rearranged: I = ε / (R + r)

I = 7.3 / (9.5 + 0.3) = 7.3 / 9.8 = 0.745 A ≈ 0.7 A (2 s.f.)

(b) Lost volts is the voltage across the internal resistance: V_r = Ir

V_r = 0.745 × 0.3 = 0.224 ≈ 0.2 V (2 s.f.)

Practice Question

A cell of e.m.f. 9.0 V and internal resistance 1.5 Ω drives a current of 2.0 A through a resistor. Find the terminal p.d. and the resistance of the resistor.

4. Core Practical 8 — Investigating E.M.F. & Internal Resistance

Aim: investigate the relationship between e.m.f. and internal resistance by varying resistance and measuring current and voltage.

Independent variable: resistance, R (Ω) — set using a variable resistor
Dependent variables: voltage, V (V) and current, I (A)
Control variables: e.m.f. of the cell; internal resistance of the cell

Equipment

Apparatus	Purpose
1.5 V Cell	Provides an e.m.f. to the circuit
Resistor	Unknown resistance — acts as internal resistance
100 Ω Variable Resistor	Changes the values of current and voltage
Voltmeter (0–2 V)	Measures voltage — resolution 1 mV
Ammeter (0–200 mA)	Measures current — resolution 0.1 mA
Switch	Opens between readings so the battery isn't run down

Method

Connect the cell and the fixed resistor r in series — treat this pair as a single "real cell".
With the switch open, record the voltmeter reading V (this gives the e.m.f.).
Set the variable resistor to maximum, close the switch, and record V and I. Open the switch again between every reading.
Repeat for 8–10 different resistance values across the whole range of the variable resistor, recording V and I each time.

Why open the switch between readings?

Keeping the switch closed for a long time lets current continuously flow, which can heat up the cell and change its internal resistance mid-experiment — a systematic error. Opening it between readings keeps r stable and constant.

Analysing the results

Starting from ε = IR + Ir = V + Ir, rearrange to make V the subject:

Straight-line form

V = −rI + ε

Compare to y = mx + c → y = V, x = I, gradient = −r, y-intercept = ε

So if you plot a graph of V (y-axis) against I (x-axis), you get a straight line sloping downward. The y-intercept gives you the e.m.f. directly, and the gradient (negative) gives you the internal resistance.

V/V │╲ │ ╲← y-intercept = ε │ ╲ │ ╲___ │ ╲___ gradient = −r │ ╲___ │ ╲___ └───────────────────────→ I/A 0

Worked Example

Data collected: when I = 0 mA, V = 1.60 V. When I = 66.0 mA, V = 0.10 V. A line of best fit is drawn through all points.

Step 1 — gradient:

gradient = ΔV/ΔI = (1.6 − 0.1) / (0 − 66×10⁻³) = 1.5 / (−0.066) = −22.7 Ω

Step 2 — interpret:

Internal resistance, r = 22.7 Ω | E.m.f., ε = y-intercept = 1.60 V

Evaluating the experiment

Systematic errors: only close the switch for as long as it takes to take each reading — this stops the cell's internal resistance drifting during the experiment.

Random errors: use a fairly new cell (run-down batteries have unstable e.m.f. and r); wait for the meters to stabilise before reading; take at least 3 repeat readings per value and average them.

Safety: components can get hot over time — switch off immediately if you smell burning, and keep liquids away from the equipment.

Practice Question

In this experiment, why is it important to record the voltmeter reading with the switch open at the very start?

5. Resistance & Temperature

All materials resist the flow of charge to some degree. In a metal wire, free electrons drift through a lattice of positive metal ions. As they move, they inevitably collide with the ions that are in their way, transferring kinetic energy on collision — this is exactly what causes electrical heating (and resistance).

Metallic conductors

As temperature rises, the metal ions in the lattice vibrate with greater frequency and amplitude — think of them as jiggling around more violently on their fixed spots. This makes it statistically much more likely that a drifting free electron will smash into one of them. More collisions = more resistance to the flow of charge.

Low temperature (20°C) High temperature (70°C) ○ ○ ○ ○ ○●○ ●○ ○● ● ○ ● ○ ○ ●○ ○ ●○ ○ ● ○ ● → smooth flow ●○ ○● ○ ← lots of collisions ○ ○ ● ○ ○ ○● ○●○ ○ = metal ion (small vibration) ● = free electron ○● = metal ion (BIG vibration, high temp)

Rule for metallic conductors

Temperature ↑ → Resistance ↑ (and Temperature ↓ → Resistance ↓)

This follows a roughly linear trend for small temperature changes, since ρ ∝ R when the wire's length and area stay constant.

This is exactly why a filament lamp (a non-ohmic component) has a curved I–V graph rather than a straight line: as current increases, more collisions heat the filament, resistance rises, and so the current increases at a progressively slower rate — giving a graph with a decreasing gradient.

Worked Example

Explain why the temperature of a filament lamp rises as the current through it increases.

1. As current increases, the rate of flow of electrons increases.

2. This increases the number of collisions between conduction electrons and the ions in the lattice.

3. Each collision transfers kinetic energy to the ions, so the vibrations of the lattice ions increase — raising the temperature.

Thermistors

Thermistors are made of semiconductor material, which behaves in the opposite way to metals. Instead of resistance increasing with temperature, most thermistors are NTC (negative temperature coefficient) — meaning the number of charge carriers (like free electrons) available for conduction actually increases as temperature rises. More available carriers means it's easier for charge to flow, so resistance falls.

Rule for NTC thermistors

Temperature ↑ → Resistance ↓ (and Temperature ↓ → Resistance ↑)

Circuit symbol: a resistor box with a diagonal line and the letter "t", or a resistor symbol with a curved temperature indicator.

This is why thermistors are so useful in temperature-sensing circuits: ovens, fire alarms, and digital thermometers all use the predictable resistance–temperature relationship to detect temperature changes electronically.

Don't mix these up!

Metal wire: hotter = MORE resistance. Thermistor: hotter = LESS resistance. They're opposites — and exam questions love to test whether you actually understand why, not just which way the arrow points.

Practice Question

A thermistor is connected in series with a fixed resistor R and a battery. At room temperature their resistances are equal. If the temperature of the thermistor decreases, what happens to the p.d. across the thermistor?

6. Resistance & Illumination

Some semiconductors change their conductivity when light shines on them. When light energy is absorbed by the material, it frees up more electrons to become available for conduction — just like the temperature effect in a thermistor, but triggered by light instead of heat.

Low light levels High light levels ○ ○ ○ ○ ○ ●○ ○ ● ○ ● ○ ○ ●○ ○ ●○ ○ ○ ○ ○ ● → few carriers ○ ●○ ● ○● ● ○ ○ ○ ○● ○● ○ ● ○ = metal/semiconductor atom ● = free electron

Light-Dependent Resistors (LDRs)

An LDR is a non-ohmic, light-sensitive resistor. As light intensity increases, more electrons are freed for conduction, so resistance decreases.

Rule for LDRs

Light intensity ↑ → Resistance ↓

In darkness: resistance can be millions of ohms. In bright light: resistance drops to just tens of ohms.

This is exactly the mechanism behind automatic street lights: in daylight, high light intensity keeps the LDR's resistance low, which keeps the lights switched off. At night, low light intensity makes the LDR's resistance shoot up, which triggers the circuit to switch the lights on.

Worked Example

Which I–V graph shape best represents an LDR — a curve that bends upward (increasing gradient) or downward (decreasing gradient)?

1. As light intensity increases, resistance of the LDR decreases, so the p.d. across it (for a given current source) behaves such that current rises faster relative to voltage.

2. Since 1/R = I/V = gradient, and R is decreasing, the gradient (1/R) must be increasing.

3. So the correct graph is the one curving upward — an increasing gradient, the mirror image of the filament lamp's decreasing-gradient curve.

Practice Question

Explain why an LDR is useful in a garden light that switches on automatically at dusk.

What to Memorise

Term	Meaning
E.m.f. (ε)	Energy converted from chemical → electrical per unit charge (measured with no current flowing)
Internal resistance (r)	Resistance inside the power supply itself, causing energy loss as heat
Lost volts (V_r)	Voltage "used up" overcoming internal resistance = Ir
Terminal p.d. (V_R / V_T)	Voltage actually available to the external circuit = IR
NTC thermistor	Resistance falls as temperature rises (opposite of metals)
LDR	Resistance falls as light intensity rises

Formula	Use
ε = E / Q	Definition of e.m.f.
ε = I(R + r)	Master equation linking e.m.f., current, load & internal resistance
ε = V + Ir	E.m.f. = terminal p.d. + lost volts
V = −rI + ε	Straight-line form for the V–I graph (Core Practical 8): gradient = −r, y-intercept = ε

Concepts Checklist

I can define e.m.f. in terms of energy per unit charge
I know why e.m.f. is measured with no current flowing
I can explain what internal resistance is and why it causes a battery to warm up
I can distinguish e.m.f., lost volts, and terminal p.d.
I can use ε = I(R + r) to solve circuit problems
I can describe the Core Practical 8 method step-by-step
I can find e.m.f. and internal resistance from a V–I graph
I know why the switch must be opened between readings in Core Practical 8
I can explain why resistance increases with temperature in a metal (ion vibration model)
I can explain why resistance decreases with temperature in an NTC thermistor
I can explain why resistance decreases with light intensity in an LDR
I can sketch and interpret the I–V graph for a filament lamp, thermistor, and LDR

Exam Tips & Common Mistakes

"Negligible internal resistance" = shortcut

If a question states this phrase, it means r = 0. This means terminal p.d. equals e.m.f. exactly — you don't need the full ε = I(R+r) equation at all, since lost volts are zero.

Treat internal resistance as a separate resistor

Even though it isn't physically a separate component, drawing/imagining internal resistance as its own resistor in series with the load makes every calculation and circuit diagram much easier to handle correctly.

Watch your units in the graph method

In Core Practical 8, current is often measured in mA — remember to convert to amps before calculating the gradient, or your value for r will be out by a factor of 1000.

E.m.f. vs p.d. — say the right thing

Examiners specifically look for "energy transferred TO charges" for e.m.f. and "energy transferred FROM electrical TO other forms" for p.d. Mixing up the direction of energy transfer is one of the most common lost marks on this topic.

Metal vs thermistor — direction matters

Don't just say "resistance changes with temperature" — always state the direction (increases or decreases) and link it to the correct mechanism: more ion vibration (metal) vs more charge carriers (thermistor).

Systematic vs random errors

In Core Practical 8, "closing the switch too long" is a systematic error (it consistently changes r as the battery heats). Using an old battery causing inconsistent readings, or misreading fluctuating meters, are random errors. Know the difference — it's a common mark-scheme distinction.

🔓 Read the full E.M.F & Modelling Resistance note — free You're seeing the preview · free account, no card needed

Also in the full note

4. Core Practical 8 — Investigating E.M.F. & Internal Resistance
5. Resistance & Temperature
6. Resistance & Illumination
Exam Tips & Common Mistakes

What's inside

📖 Revision notes 🎯 Learn mode ✦ AI flashcards ✓ Instant AI marking 🧊 3D explorers 🧪 Experiments & simulations 📈 Progress tracking

Read the full E.M.F & Modelling Resistance notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →