B.6 Heating effect of current and electric cells (AHL)

IB Physics HL· Theme B: The particulate nature of matter, AHL Topic B.6· 45 min read

1. Joule Heating and Power Dissipation★★☆☆☆HL only⏱ 15 min

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📘 Definition

Joule Heating

Heating effect of current

The process by which electrical energy is converted to thermal energy when current flows through a resistive conductor, caused by collisions between moving charge carriers and the atomic lattice.

🔬 Derivation

Goal:

Derive expressions for power dissipated in an Ohmic resistor

Starting from:

Power is defined as work done per unit time: $P = \frac{Δ W}{Δ t}$

1
Work done to move charge $q$ across potential difference $V$ is $Δ W = q V$ . Current is defined as $I = \frac{q}{Δ t}$ , so $q = I Δ t$ .
2
Substitute into power:
3
$P = \frac{(I\Delta t V)}{\Delta t} = IV$

Result:

Using Ohm's law $V = I R$ , we get two equivalent forms for power: $P = I^{2} R = \frac{V ^{2}}{R}$

📐 Worked Example

A 6.0 Ω resistor connected to a 12 V battery carries a constant current. Calculate the power dissipated as heat in the resistor.

1
We know the potential difference across the resistor is 12 V, so use $P = \frac{V ^{2}}{R}$ for simplicity:
2
$P = \frac{(12)^2}{6.0} = \frac{144}{6} = 24 \text{ W}$
3
Cross-check with $P = I V$ : First find $I = \frac{V}{R} = \frac{12}{6} = 2 A$ , so $P = 12 \times 2 = 24 W$ , which matches.

2. Emf and Internal Resistance of Real Cells★★★☆☆HL only⏱ 15 min

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All real cells have internal resistance, arising from the resistance of the electrolyte and electrodes inside the cell. We model a real cell as an ideal source of emf $ε$ connected in series with an internal resistor $r$ .

📘 Definition

Electromotive Force (emf)

$ε$

The total energy per unit charge supplied by the cell, equal to the terminal potential difference when no current is drawn (open circuit).

\varepsilon = V_{\text{terminal}} + Ir

📐 Worked Example

A cell has an emf of 3.0 V and internal resistance of 0.2 Ω. It is connected to an external 2.8 Ω resistor. Calculate the current in the circuit.

1
Total resistance for the series circuit is the sum of external resistance $R$ and internal resistance $r$ :
2
$R_{\text{total}} = R + r = 2.8 + 0.2 = 3.0 \text{ } \Omega$
3
Apply Ohm's law to the whole circuit, using emf as the total potential difference:
4
$I = \frac{\varepsilon}{R_{\text{total}}} = \frac{3.0}{3.0} = 1.0 \text{ A}$

3. Terminal Potential Difference★★★☆☆HL only⏱ 15 min

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Terminal potential difference $V$ is the actual potential difference available across the external load connected to the cell. When current flows, energy is dissipated across the internal resistance, so $V < ε$ .

V = \varepsilon - Ir = IR_{\text{load}}

📐 Worked Example

For the cell in the previous example ( $ε = 3.0$ V, $r = 0.2$ Ω, $R = 2.8$ Ω), calculate the terminal potential difference and power lost in the internal resistance.

1
We found $I = 1.0$ A, so use the terminal pd formula:
2
$V = \varepsilon - Ir = 3.0 - (1.0)(0.2) = 2.8 \text{ V}$
3
Power lost across internal resistance is given by $P = I^{2} r$ :
4
$P = (1.0)^2 (0.2) = 0.2 \text{ W}$
5
Cross-check: $V = I R = (1.0) (2.8) = 2.8$ V, which confirms our result.

4. Cell Types and Combinations★★☆☆☆HL only⏱ 15 min

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📘 Definition

Primary vs Secondary Cells

Primary cells are non-rechargeable: their chemical reaction is irreversible, so they are discarded once exhausted. Secondary cells are rechargeable: their chemical reaction can be reversed by applying an external current, restoring the cell's capacity.

Example:

Primary: alkaline dry cell; Secondary: lithium-ion, lead-acid car battery

Series combination (n identical cells): Total emf $= n ε$ , total internal resistance $= n r$ (used to get higher total emf)
Parallel combination (n identical cells): Total emf $= ε$ , total internal resistance $= \frac{r}{n}$ (used to reduce total internal resistance and deliver higher current)

📐 Worked Example

Three identical cells (each $ε = 2.0$ V, $r = 0.6$ Ω) are connected in parallel across a 0.1 Ω external resistor. Calculate the total current in the external load.

1
For parallel identical cells, total emf equals the emf of one cell, and total internal resistance is divided by the number of cells:
2
$\varepsilon_{\text{total}} = 2.0 \text{ V}, \quad r_{\text{total}} = \frac{0.6}{3} = 0.2 \text{ } \Omega$
3
Total resistance is sum of internal and external resistance:
4
$R_{\text{total}} = 0.2 + 0.1 = 0.3 \text{ } \Omega$
5
Calculate total current:
6
$I = \frac{\varepsilon_{\text{total}}}{R_{\text{total}}} = \frac{2.0}{0.3} \approx 6.7 \text{ A}$

5. Common Pitfalls

Wrong move:

Concluding higher resistance always gives higher power dissipation

Why:

The relationship between resistance and power depends on whether current or voltage is constant. $P = I^{2} R$ applies for constant current, $P = V^{2} / R$ applies for constant voltage.

Correct move:

First check which quantity is held constant, then select the appropriate power formula to use.

Wrong move:

Treating emf as a force, equal to terminal potential difference when current flows

Why:

The name 'electromotive force' is misleading. Emf is energy per unit charge (units of volts), and only equals terminal pd when no current is drawn.

Correct move:

Remember: $V = ε - I r$ , so terminal pd is always less than emf when current flows through the cell.

Wrong move:

Adding emfs of identical parallel cells to get total emf

Why:

Parallel combination of identical cells does not increase total emf, it only reduces total internal resistance.

Correct move:

For n identical cells in parallel, total emf equals the emf of a single cell, total internal resistance is $r / n$ .

Wrong move:

Ignoring internal resistance when calculating current for a real cell

Why:

All real cells have non-zero internal resistance, which contributes to total circuit resistance, especially for large currents.

Correct move:

Always add internal resistance to external resistance when calculating total current from a real cell.

6. Quick Reference Cheatsheet

Concept	Formula	Key Note
General power dissipation	$P = I V$	Always true for any conductor
Power for Ohmic resistors	$P = I^{2} R = V^{2} / R$	Match to constant variable
Real cell equation	$ε = V + I r$	Ideal emf + series internal resistance
Terminal potential difference	$V = ε - I r$	$V < ε$ when $I > 0$
n identical cells in series	$ε_{total} = n ε, r_{total} = n r$	Higher total emf
n identical cells in parallel	$ε_{total} = ε, r_{total} = r / n$	Lower total internal resistance

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Power dissipation calculation
2022 · 2
Internal resistance experiment
2021 · 1
Cell terminal pd comparison

Going deeper

practical guideMeasuring internal resistance of a cellCommon IA experiment for IB Physics

What's Next

This sub-topic is a foundational AHL topic for all advanced circuit and energy transfer concepts in IB Physics HL. Understanding internal resistance and power dissipation is critical for solving problems in electromagnetic induction, alternating current circuits, and renewable energy systems. It is also one of the most common topics for IB Physics internal assessment practical investigations, so mastering these calculations will help you with your IA. Next, you will explore the magnetic effects of electric current, which builds directly on your understanding of current and energy in circuits.