B.5 Current and electric circuits (AHL)

IB Physics HL· Theme B, B.5· 25 min read

1. 1. Emf and Internal Resistance★★☆☆☆⏱ 6 min

📘 Definition

Emf and Internal Resistance

\varepsilon, r

Emf ( $ε$ ) is the total energy per unit charge supplied by a source, equal to terminal potential difference when no current flows. Internal resistance ( $r$ ) is the inherent resistance of the source to current flow.

Example:

A 1.5 V AA battery typically has an internal resistance of ~0.5 Ω

When current flows through a source, the internal resistance causes a potential drop equal to $I r$ . This means the terminal potential difference $V$ (the voltage available to the external circuit) is given by the relationship: $V = ε - I r$ .

📐 Worked Example

A battery with emf 12 V is connected to an external 22 Ω resistor. The terminal voltage of the battery is measured as 11.2 V. Calculate the internal resistance of the battery.

1
Find the current through the external resistor using Ohm's law:
2
$I = \frac{V_{ext}}{R} = \frac{11.2}{22} = 0.509 \text{ A}$
3
Rearrange the emf equation to solve for internal resistance $r$ :
4
$\varepsilon = V + Ir \implies r = \frac{\varepsilon - V}{I}$
5
Substitute the known values to get the result:
6
$r = \frac{12 - 11.2}{0.509} = \frac{0.8}{0.509} \approx 1.6 \text{ }\Omega$

Exam tip:

Emf is constant for a given source, but terminal potential difference always decreases as the current drawn from the source increases.

2. 2. Kirchhoff's Circuit Laws★★★☆☆⏱ 9 min

📘 Definition

Kirchhoff's Junction and Loop Rules

Junction rule (conservation of charge): The sum of currents entering a junction equals the sum of currents leaving the junction. Loop rule (conservation of energy): The sum of potential differences around any closed loop is zero.

These laws allow you to solve multi-loop circuits with multiple voltage sources that cannot be reduced to simple series-parallel combinations. You can assign any initial direction to currents in each branch: a negative result just means the actual current direction is opposite to your assumption.

📐 Worked Example

A 10 V source in series with a 2 Ω resistor is connected in parallel with a 6 V source in series with a 1 Ω resistor, across a 5 Ω load. Find the current through the 5 Ω load.

1
Label currents: $I_{1}$ from 10 V source, $I_{2}$ from 6 V source, $I_{3}$ through 5 Ω load. Apply junction rule:
2
$I_1 + I_2 = I_3$
3
Apply loop rule to the 10 V loop (energy conservation):
4
$10 - 2I_1 - 5I_3 = 0 \implies 2I_1 + 5I_3 = 10$
5
Apply loop rule to the 6 V loop:
6
$6 - I_2 - 5I_3 = 0 \implies I_2 + 5I_3 = 6$
7
Substitute $I_{3} = I_{1} + I_{2}$ and solve the simultaneous equations to get:
8
$I_3 \approx 1.16 \text{ A}$

Exam tip:

Consistent sign conventions are the most important part of applying Kirchhoff's laws. Mark all current directions and potential drops clearly before solving.

3. 3. Potential Dividers★★☆☆☆⏱ 6 min

A potential divider divides an input source voltage into a smaller output voltage using two or more series resistors. For an unloaded divider with resistors $R_{1}$ and $R_{2}$ across emf $ε$ , the output voltage across $R_{2}$ is: $V_{o u t} = ε \frac{R _{2}}{R _{1} + R _{2}}$ .

📘 Definition

Loaded Potential Divider

A potential divider with an external load resistor connected in parallel with one of the divider resistors. The parallel combination reduces the effective resistance of the output branch, changing the output voltage.

📐 Worked Example

A 9 V source is connected to a divider with $R_{1} = 100Ω$ and $R_{2} = 200Ω$ . A 300 Ω load is connected in parallel across $R_{2}$ . Calculate the output voltage across the parallel combination.

1
Calculate the effective resistance of the parallel combination of $R_{2}$ and the load $R_{L}$ :
2
$\frac{1}{R_{eff}} = \frac{1}{200} + \frac{1}{300} = \frac{5}{600} \implies R_{eff} = 120 \text{ }\Omega$
3
Calculate total resistance of the full divider:
4
$R_{total} = R_1 + R_{eff} = 100 + 120 = 220 \text{ }\Omega$
5
Apply the potential divider rule to find output voltage:
6
$V_{out} = 9 \times \frac{120}{220} \approx 4.9 \text{ V}$

Exam tip:

Compare your loaded output voltage to the unloaded value (6 V in the example above): it should always be lower, so this is a good check for your working.

4. 4. Exam Expectations for Circuit Problems★★★☆☆⏱ 4 min

✓ Quick check

Check your understanding of core concepts:

What happens to the terminal voltage of a battery as the current drawn from it increases?
- Increases
- Decreases
- Stays the same
- Cannot be predicted
Reveal answer
1 —
Terminal voltage $V = ε - I r$ , so as current $I$ increases, the terminal voltage decreases.
According to Kirchhoff's junction rule, what is the sum of currents entering a junction?
- Equals the sum of currents leaving
- Equals zero
- Equals the total source current
- Is always positive

5. Common Pitfalls

Wrong move:

Assuming terminal voltage always equals emf

Why:

Emf is only equal to terminal voltage for open circuits (zero current). Current flow causes a voltage drop across internal resistance.

Correct move:

Always use $V = ε - I r$ when current is drawn from the source.

Wrong move:

Mixing up sign conventions for potential rise/drop in Kirchhoff's loop rule

Why:

Incorrect signs lead to wrong solutions for simultaneous equations that are impossible to debug.

Correct move:

When travelling around a loop, mark emf from negative to positive as positive, and resistor current in the direction of travel as a negative drop.

Wrong move:

Using the unloaded potential divider formula for loaded dividers

Why:

Adding a parallel load reduces the effective resistance of the output branch, lowering output voltage.

Correct move:

Always calculate the effective parallel resistance of the output branch first before applying the divider formula.

Wrong move:

Getting stuck trying to guess the correct current direction before solving

Why:

Many students waste time trying to assign the right direction, but the result will tell you the direction.

Correct move:

Assign any direction you want: a negative final current means the actual direction is opposite to your assumption.

Wrong move:

Forgetting to add internal resistance to total circuit resistance

Why:

Internal resistance contributes to the total resistance of the circuit, so omitting it gives wrong current values.

Correct move:

Always add the source's internal resistance to the external resistance when calculating total circuit current.

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
Emf and terminal PD	$V = ε - I r$	$V$ = terminal voltage, $r$ = internal resistance
Kirchhoff Junction Rule	$\sum I_{in} = \sum I_{o u t}$	Conservation of charge
Kirchhoff Loop Rule	$\sum Δ V = 0$	Conservation of energy around closed loop
Unloaded Potential Divider	$V_{o u t} = ε \frac{R _{2}}{R _{1} + R _{2}}$	Output across $R_{2}$
Loaded Potential Divider	$V_{o u t} = ε \frac{R _{e f f}}{R _{1} + R _{e f f}}$	$R_{e f f} = R_{2} ∣∣ R_{L}$ (parallel 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 · 2
Multi-loop circuit calculation
2022 · 1
Loaded potential divider problem
2021 · 2
Internal resistance experiment analysis

What's Next

This sub-topic is the foundation for all advanced circuit concepts in IB Physics HL. Mastery of Kirchhoff's laws and potential dividers is essential for solving any complex circuit problem in both Paper 1 and Paper 2 exams, and these concepts frequently appear in 4-8 mark long answer questions. Next, you can extend your knowledge to capacitors in DC circuits, which use many of the same circuit rules you learned here, or explore related concepts like semiconductors and sensing circuits. Consistent practice with multi-loop problems will help you build speed for exam conditions.