# Potential difference, resistance and Ohm's law

> IB Physics SL · IB Physics SL
> Source: https://www.owlsprep.com/study/ib-physics-sl-u2-potential-difference-resistance-and-ohm/

We cover core definitions of potential difference, resistance, Ohm's law, V-I graph analysis, and step-by-step circuit calculations for IB SL Physics.

**Prerequisites:** [Electric current basics](https://www.owlsprep.com/study/ib-physics-sl-u2-electric-current/); [Basic circuit diagram notation](https://www.owlsprep.com/study/ib-physics-sl-u2-circuit-diagrams/)

## Learning objectives

- Define electric potential difference and its SI unit of measurement
- State the definition of electrical resistance and calculate it from measured values
- Apply Ohm's law to solve quantitative circuit problems for ohmic conductors
- Distinguish between ohmic and non-ohmic conductors using V-I characteristic graphs

## Defining Potential Difference

Potential difference (pd) describes the amount of energy transferred per unit charge as charge carriers move between two points in a circuit. It is measured with a voltmeter connected in parallel across the two points, as this setup ensures the voltmeter draws negligible current and samples the full energy difference between the points.

**Potential Difference** — The work done (energy transferred) per unit positive charge moving between two points

*Notation:* V

*Example:* A 1.5 V AA battery transfers 1.5 J of energy to every 1 C of charge passing through it

**Worked example:** Calculate the potential difference across a filament bulb that transfers 600 J of light and heat energy to 200 C of charge passing through it.

1. Start with the formal definition of potential difference
2. $$V = \frac{E}{Q}$$
3. Substitute the given values E = 600 J, Q = 200 C
4. $$V = \frac{600}{200} = 3 \text{ V}$$
5. Final answer: potential difference across the bulb is 3 V

## Electrical Resistance

Resistance quantifies how much a circuit component opposes the flow of electric current. For a fixed applied potential difference, higher resistance results in lower current flowing through the component. Resistance in metallic conductors arises from collisions between moving free electrons and the vibrating positive ions of the metal lattice.

**Resistance** — The ratio of the potential difference across a component to the current flowing through it at a specific operating point

*Notation:* R

*Example:* A 10 Ω fixed resistor allows 0.1 A of current when 1 V is applied across its terminals

**Worked example:** A current of 0.4 A flows through a fixed resistor when 12 V is applied across it. Calculate the resistance of the resistor.

1. Rearrange the resistance definition formula to isolate R
2. $$R = \frac{V}{I}$$
3. Substitute the measured values V = 12 V, I = 0.4 A
4. $$R = \frac{12}{0.4} = 30 \ \Omega$$
5. Final answer: resistance of the resistor is 30 Ω

## Ohm's Law

Ohm's law is an empirical relationship that applies to most metallic conductors at constant temperature. It states that the current flowing through a conductor is directly proportional to the potential difference applied across it, as long as all physical conditions including temperature and pressure remain unchanged.

**Derivation:** Derive the standard mathematical form of Ohm's law

*Starting from:* Observed proportionality between current I and potential difference V for constant temperature

1. Write the proportionality relationship: I ∝ V
2. Introduce a constant of proportionality equal to 1/R, where R is fixed resistance
3. $$I = \frac{V}{R}$$
4. Rearrange to isolate V on the left hand side

*Conclusion:* The standard form of Ohm's law is V = IR, which holds for all ohmic conductors at constant temperature

**Exam command terms**

IB Physics SL exam questions use specific command terms for Ohm's law problems:

- **Show that** — You must show full substitution of values to arrive at the given result, no skipped steps

- **Calculate** — State the formula explicitly, substitute values, and give the final answer with correct SI unit

**Worked example:** A 20 Ω fixed resistor is connected directly to a 9 V battery. Calculate the current flowing through the resistor.

1. Start with the standard Ohm's law formula V = IR
2. Rearrange the formula to solve for unknown current I
3. $$I = \frac{V}{R} = \frac{9}{20}$$
4. Final answer: current flowing through the resistor is 0.45 A

## V-I Characteristic Graphs

A V-I characteristic graph plots potential difference on the y-axis against measured current on the x-axis. For an ohmic conductor at constant temperature, this graph is a perfectly straight line passing through the origin. Non-ohmic components such as filament lamps and diodes produce curved V-I graphs, as their resistance changes with operating temperature or current direction.

> **tip**
>
> To find resistance at a specific point on a non-ohmic V-I curve, never use the gradient of the curve: always calculate R = V/I directly from the coordinates of that exact point.

**Check your understanding**

Test your understanding of V-I graph properties:

1. What shape is the V-I graph for an ohmic conductor held at constant temperature?

   - Curved line through origin
   - Straight line through origin
   - Horizontal line
   - Vertical line

   *Why:* Ohmic conductors have constant resistance, so V is directly proportional to I, producing a straight line that passes through the origin.

## Common pitfalls

- **Wrong:** Using the gradient of a non-ohmic V-I curve to calculate resistance at a point
  - Why it fails: The gradient gives dV/dI, not the actual definition of resistance V/I that IB exam mark schemes require
  - Correct: Divide the V coordinate by the I coordinate at the exact operating point you are measuring
- **Wrong:** Forgetting to state that Ohm's law only applies at constant temperature
  - Why it fails: Heating a metallic conductor increases its resistance, breaking the proportionality between V and I
  - Correct: Always explicitly mention constant temperature when proving a component is ohmic in exam answers
- **Wrong:** Connecting a voltmeter in series with a circuit component
  - Why it fails: Voltmeters have extremely high internal resistance, so almost no current will flow in the rest of the circuit
  - Correct: Always connect voltmeters in parallel across the two points you are measuring potential difference between
- **Wrong:** Using mA directly instead of converting to A in Ohm's law calculations
  - Why it fails: 1 mA = 0.001 A, so using mA directly will produce an answer 1000 times larger than the correct value
  - Correct: Convert all current values to amperes before substituting into V=IR

## Cheatsheet

| Quantity | Symbol | SI Unit | Core Formula |
| --- | --- | --- | --- |
| Potential Difference | V | Volt (V) | $V = E/Q$ |
| Current | I | Ampere (A) | $I = V/R$ |
| Resistance | R | Ohm (Ω) | $R = V/I$ |
| Ohm's Law | - | - | $V = IR$ |

## What's next

Now that you have mastered the core definitions of potential difference, resistance and Ohm's law, you are ready to apply these concepts to more complex circuit arrangements. You will learn how to calculate total resistance for series and parallel resistor combinations, how to measure unknown resistance using practical circuits, and how to analyse the behaviour of non-ohmic components like diodes and thermistors in exam scenarios. These skills make up 15-20% of the IB Physics SL Paper 2 mark allocation, so regular practice with calculation questions will significantly boost your exam performance.

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