Electrical power and energy

IB Physics SL· Topic 2: Electricity, The particulate nature of matter unit· 12 min read

1. Core Definitions and Fundamental Formulas★★☆☆☆⏱ 10 min

Electrical energy describes the total work done by an electric field to push charge carriers through a potential difference across a component. When a charge $Q$ moves through a potential difference $V$ , the total energy transferred is given by $E = Q V$ . Since current $I$ is defined as charge per unit time, we can substitute $Q = I t$ to get $E = V I t$ , the universal formula for electrical energy transferred over time $t$ .

📘 Definition

Electrical Power

P

The rate at which electrical energy is dissipated or transferred by a circuit component, measured in watts (W), where 1 W = 1 J s⁻¹.

P = \frac{E}{t} = VI

📐 Worked Example

A 5 A current flows through a resistor connected to a 12 V battery for 30 seconds. Calculate the total energy dissipated and the power rating of the resistor.

1
First, identify given values: $V = 12$ V, $I = 5$ A, $t = 30$ s
2
$E = VIt = 12 \times 5 \times 30$
3
Calculate total energy transferred
4
$E = 1800 \text{ J}$
5
Use the universal power formula to find the power rating
6
$P = 12 \times 5 = 60 \text{ W}$

✓ Quick check

Test your understanding of base formulas

What is the power of a component that transfers 200 J of energy in 10 seconds?
- 10 W
- 20 W
- 200 W
- 2000 W
Reveal answer
20 W —
Power is energy divided by time: 200 / 10 = 20 W.

2. Deriving Alternative Power Formulas for Resistive Circuits★★★☆☆⏱ 12 min

🔬 Derivation

Goal:

Derive P = I²R and P = V²/R for ohmic resistors

Starting from:

Universal power formula P = VI and Ohm's law V = IR

1
Substitute Ohm's law V = IR into the universal P = VI formula
2
$P = (IR) \times I = I^2 R$
3
Rearrange Ohm's law to I = V/R and substitute into P = VI
4
$P = V \times \frac{V}{R} = \frac{V^2}{R}$

Result:

These two derived formulas only apply to ohmic components that obey Ohm's law at constant temperature.

📐 Worked Example

A 10 Ω resistor is connected across a 6 V battery. Use two different power formulas to confirm the power dissipated by the resistor.

1
Method 1: Use P = V²/R directly from the supply voltage and known resistance
$P = \frac{6^2}{10} = \frac{36}{10} = 3.6 \text{ W}$
2
Method 2: First find current via Ohm's law I = V/R, then use P = I²R
$I = 6/10 = 0.6 \text{ A}, P = 0.6^2 \times 10 = 3.6 \text{ W}$

Exam tip:

IB exam questions often ask you to verify two power formulas produce the same result to confirm you understand their valid use cases.

3. Power Dissipation in Series and Parallel Circuits★★★☆☆⏱ 10 min

In series circuits, current is identical across all resistors, so P = I²R is the most convenient formula to compare power dissipation: larger resistors dissipate more power in series. In parallel circuits, potential difference is identical across all branches, so P = V²/R is optimal: smaller resistors dissipate more power in parallel.

Circuit Type	Shared Quantity	Most Useful Formula	Power vs Resistance Relationship
Series	Current I	P = I²R	P ∝ R
Parallel	Potential Difference V	P = V²/R	P ∝ 1/R

📐 Worked Example

Two resistors of 2 Ω and 4 Ω are connected first in series, then in parallel across a 12 V supply. Compare the power dissipated by each resistor in both configurations.

1
Series case: Total resistance = 6 Ω, total current I = 12/6 = 2 A
$P_{2Ω} = 2^2 \times 2 = 8 \text{ W}, P_{4Ω} = 2^2 \times4 =16 \text{ W}$
2
Parallel case: Voltage across both resistors is equal to the 12 V supply
$P_{2Ω} = 12^2 / 2 =72 \text{ W}, P_{4Ω}=12^2 /4=36 \text{ W}$

4. Real-World Energy Calculations: Kilowatt-Hour★★☆☆☆⏱ 8 min

📘 Definition

Kilowatt-hour (kWh)

Non-SI commercial unit of electrical energy, equal to the energy transferred by a 1000 W device operating for 1 full hour, where 1 kWh = 3.6 × 10⁶ J.

📐 Worked Example

A 1500 W electric heater runs for 4 hours. Calculate the total energy used in kWh, and the cost if 1 kWh costs 0.25 USD.

1
Convert power to kW: 1500 W = 1.5 kW
2
$E = P \times t = 1.5 \times 4 = 6 \text{ kWh}$
3
$Total cost = 6 \times 0.25 = \$1.50$

5. Common Pitfalls

Wrong move:

Using P = V²/R for a non-ohmic LED circuit

Why:

This formula assumes constant resistance, which does not hold for non-ohmic components

Correct move:

Always use the universal P = VI formula for any non-ohmic device

Wrong move:

Forgetting to convert time to seconds when calculating energy in joules

Why:

Power in watts is defined as joules per second, so time in minutes/hours will produce incorrect values

Correct move:

Convert all time inputs to SI units (seconds) before substituting into E = VIt

Wrong move:

Comparing power of resistors in series using P = V²/R with total supply voltage

Why:

Voltage is not equal across series resistors, so you cannot use the full supply voltage for individual components

Correct move:

Use P = I²R for series circuits where current is identical across all resistors

Wrong move:

Classifying kilowatt-hour as a unit of power

Why:

The word 'watt' in the name leads students to misclassify it as a power unit

Correct move:

Remember kWh = power (kW) × time (h), so it is a product of power and time, hence energy

Wrong move:

Using total circuit voltage to calculate power for a single resistor in series

Why:

The single resistor only has a fraction of the total supply voltage across it

Correct move:

First calculate the voltage drop across the individual resistor, or use the shared series current to find power

6. Quick Reference Cheatsheet

Formula	Valid For	Units
P = VI	All circuit components	P in W, V in V, I in A
P = I²R	Only ohmic resistors	P in W, I in A, R in Ω
P = V²/R	Only ohmic resistors	P in W, V in V, R in Ω
E = VIt	All circuit components	E in J, V in V, I in A, t in s
1 kWh = 3.6 × 10⁶ J	Energy unit conversion	N/A

7. Frequently Asked

Can I use P = V²/R for non-ohmic components like LEDs?

No. P = V²/R only applies to ohmic resistors with constant resistance. The universal formula P = VI works for all circuit components, including non-ohmic devices.

Is a kilowatt-hour a unit of power or energy?

It is a unit of energy: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J, representing total energy used by a 1 kW device running for 1 hour.

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.

2024 · Paper 1
Multiple choice power calculation
2023 · Paper 2
Heater energy dissipation problem
2022 · Paper 2
Solar cell power output task

Going deeper

practice_questionsIB Physics SL Electrical Power Question BankCurated past paper questions with official mark schemes

What's Next

You now have a full grasp of electrical power and energy calculations, a core foundational skill for all subsequent IB Physics SL electricity topics. Mastering these formulas will let you tackle more complex problems involving electrical efficiency, domestic energy use, and power dissipation in household circuits. This concept is also frequently combined with thermal energy transfer questions in Paper 2 structured exams, so ensure you can quickly select the correct power formula for any given circuit context.