# B.5 Current and electric circuits (AHL)

> IB Physics HL · IB Physics (2025 First Assessment)
> Source: https://www.owlsprep.com/study/ib-physics-hl-u2-b-5-current-and-electric/

This AHL sub-topic extends SL DC circuit concepts to real voltage sources, multi-loop circuits, and potential dividers. Mastery of these core concepts is required for almost all complex circuit problems in IB Physics HL exams.

**Prerequisites:** [Basic DC current and resistance concepts (SL)](https://www.owlsprep.com/study/ib-physics-sl-current-resistance/)

## Learning objectives

- Derive and apply relationships for emf and internal resistance of real voltage sources
- Use Kirchhoff's laws to solve multi-loop DC circuit problems
- Analyze unloaded and loaded potential divider circuits
- Solve complex circuit problems with multiple sources and resistors

## 1. Emf and Internal Resistance

**Emf and Internal Resistance** — Emf ($\varepsilon$) 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.

*Notation:* \varepsilon, r

*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 $Ir$. This means the terminal potential difference $V$ (the voltage available to the external circuit) is given by the relationship: $V = \varepsilon - Ir$.

**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. Kirchhoff's Circuit Laws

**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. Potential Dividers

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 $\varepsilon$, the output voltage across $R_2$ is: $V_{out} = \varepsilon \frac{R_2}{R_1 + R_2}$.

**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 $R_1 = 100 \Omega$ and $R_2 = 200 \Omega$. 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. Exam Expectations for Circuit Problems

**Exam command terms**

IB exam command terms for circuit problems follow these standard expectations:

- **Derive** — Show full working starting from first principles (conservation of charge/energy) *(Derive the potential divider equation)*

- **Calculate** — Obtain a numerical answer, show all steps of working *(Calculate the internal resistance of the battery)*

- **State** — Give a brief answer without full explanation *(State Kirchhoff's junction rule)*

**Check your understanding**

Check your understanding of core concepts:

1. What happens to the terminal voltage of a battery as the current drawn from it increases?

   - Increases
   - Decreases
   - Stays the same
   - Cannot be predicted

   *Answer:* Decreases

   *Why:* Terminal voltage $V = \varepsilon - Ir$, so as current $I$ increases, the terminal voltage decreases.

2. 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

   *Answer:* Equals the sum of currents leaving

   *Why:* Kirchhoff's junction rule is a statement of conservation of charge, so charge entering equals charge leaving.

## Common pitfalls

- **Wrong:** Assuming terminal voltage always equals emf
  - Why it fails: Emf is only equal to terminal voltage for open circuits (zero current). Current flow causes a voltage drop across internal resistance.
  - Correct: Always use $V = \varepsilon - Ir$ when current is drawn from the source.
- **Wrong:** Mixing up sign conventions for potential rise/drop in Kirchhoff's loop rule
  - Why it fails: Incorrect signs lead to wrong solutions for simultaneous equations that are impossible to debug.
  - Correct: 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:** Using the unloaded potential divider formula for loaded dividers
  - Why it fails: Adding a parallel load reduces the effective resistance of the output branch, lowering output voltage.
  - Correct: Always calculate the effective parallel resistance of the output branch first before applying the divider formula.
- **Wrong:** Getting stuck trying to guess the correct current direction before solving
  - Why it fails: Many students waste time trying to assign the right direction, but the result will tell you the direction.
  - Correct: Assign any direction you want: a negative final current means the actual direction is opposite to your assumption.
- **Wrong:** Forgetting to add internal resistance to total circuit resistance
  - Why it fails: Internal resistance contributes to the total resistance of the circuit, so omitting it gives wrong current values.
  - Correct: Always add the source's internal resistance to the external resistance when calculating total circuit current.

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| Emf and terminal PD | $V = \varepsilon - Ir$ | $V$ = terminal voltage, $r$ = internal resistance |
| Kirchhoff Junction Rule | $\sum I_{in} = \sum I_{out}$ | Conservation of charge |
| Kirchhoff Loop Rule | $\sum \Delta V = 0$ | Conservation of energy around closed loop |
| Unloaded Potential Divider | $V_{out} = \varepsilon \frac{R_2}{R_1 + R_2}$ | Output across $R_2$ |
| Loaded Potential Divider | $V_{out} = \varepsilon \frac{R_{eff}}{R_1 + R_{eff}}$ | $R_{eff} = R_2 \|\| R_L$ (parallel resistance) |

## 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.

- [B.6 Heating effect of current and electric cells (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u2-b-6-heating-effect-of/)
- [Theme C: Wave behaviour](https://www.owlsprep.com/study/ib-physics-hl-u3-overview/)
- [C.1 Simple harmonic motion](https://www.owlsprep.com/study/ib-physics-hl-u3-c-1-simple-harmonic-motion/)

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