# Electrical power and energy

> IB Physics SL · IB Physics SL
> Source: https://www.owlsprep.com/study/ib-physics-sl-u2-electrical-power-and-energy/

We cover core definitions, formula derivations, series/parallel power rules and real-world kWh calculations for IB Physics SL exam scenarios.

**Prerequisites:** [Ohm's Law and basic resistance calculations](https://www.owlsprep.com/study/ib-physics-sl-u2-ohms-law/); [Series and parallel circuit rules](https://www.owlsprep.com/study/ib-physics-sl-u2-series-parallel-circuits/)

## Learning objectives

- Define electrical energy as total work done by electric fields on charge carriers
- Derive and distinguish between universal and resistive-only electrical power formulas
- Calculate power dissipation across resistors in series and parallel configurations
- Convert between joules and kilowatt-hours for domestic energy billing problems

## Core Definitions and Fundamental Formulas

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 $E = QV$. Since current $I$ is defined as charge per unit time, we can substitute $Q = It$ to get $E = VIt$, the universal formula for electrical energy transferred over time $t$.

**Electrical Power** — 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.

*Notation:* P

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

**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. $$P = 12 \times 5 = 60 \text{ W}$$

**Check your understanding**

Test your understanding of base formulas

1. What is the power of a component that transfers 200 J of energy in 10 seconds?

   - 10 W
   - 20 W
   - 200 W
   - 2000 W

   *Why:* Power is energy divided by time: 200 / 10 = 20 W.

## Deriving Alternative Power Formulas for Resistive Circuits

**Derivation:** 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. $$P = V \times \frac{V}{R} = \frac{V^2}{R}$$

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

> **warning**
>
> Never use P = I²R or P = V²/R for non-ohmic components like diodes or LEDs, as their resistance is not constant across operating conditions.

**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.

## Power Dissipation in Series and Parallel Circuits

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

   $$P_{2Ω} = 12^2 / 2 =72 \text{ W}, P_{4Ω}=12^2 /4=36 \text{ W}$$

## Real-World Energy Calculations: Kilowatt-Hour

**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.

> **mnemonic**
>
> To calculate kWh directly: convert power to kW and time to hours before multiplying, no need to convert to joules first.

**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. $$Total cost = 6 \times 0.25 = \$1.50$$

## Common pitfalls

- **Wrong:** Using P = V²/R for a non-ohmic LED circuit
  - Why it fails: This formula assumes constant resistance, which does not hold for non-ohmic components
  - Correct: Always use the universal P = VI formula for any non-ohmic device
- **Wrong:** Forgetting to convert time to seconds when calculating energy in joules
  - Why it fails: Power in watts is defined as joules per second, so time in minutes/hours will produce incorrect values
  - Correct: Convert all time inputs to SI units (seconds) before substituting into E = VIt
- **Wrong:** Comparing power of resistors in series using P = V²/R with total supply voltage
  - Why it fails: Voltage is not equal across series resistors, so you cannot use the full supply voltage for individual components
  - Correct: Use P = I²R for series circuits where current is identical across all resistors
- **Wrong:** Classifying kilowatt-hour as a unit of power
  - Why it fails: The word 'watt' in the name leads students to misclassify it as a power unit
  - Correct: Remember kWh = power (kW) × time (h), so it is a product of power and time, hence energy
- **Wrong:** Using total circuit voltage to calculate power for a single resistor in series
  - Why it fails: The single resistor only has a fraction of the total supply voltage across it
  - Correct: First calculate the voltage drop across the individual resistor, or use the shared series current to find power

## 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 |

## 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.

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