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v0.1.1040
NotesPhysicsTopic 2.5Electrical power and energy
Back to Physics Topics
2.5.42 min read

Electrical power and energy

IB Physics • Unit 2

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Contents

  • What electrical power is
  • Working out the power
  • Exam-style question
The big idea: Power is how fast electrical energy is turned into other forms — heat, light, motion.

When a current flows through a component, the component transfers energy every second. That rate is its power.

Unit: the watt (W) — 1 watt = 1 joule of energy every second.

[Diagram: phys-circuit] - Available in full study mode

[Diagram: phys-formula-triangle] - Available in full study mode

Spot it: Bigger current or bigger voltage → more power.

The simplest form is P = I × V (current times voltage). A 12 V supply driving 2 A delivers 12 × 2 = 24 W.

Power is current × voltage. Using Ohm's law (V = IR) you can swap V or I out, giving three equal forms of the same equation:

Given in the data booklet. Three forms of the same power equation — pick the one matching the quantities you know.
electrical power (watts, W)
current (amperes, A)
potential difference / voltage (volts, V)
resistance (ohms, Ω)
Which form do I use?: Choose the form using the two quantities you already know, so you don't have to find a third first:
You know…Use this formWhy
I and VP = IVboth are given directly
I and RP = I²Rno need to find V first
V and RP = V²/Rno need to find I first
Ohm's law links them: All three forms come from P = IV combined with V = IR (resistance). V = IR is also given in the data booklet.
Ohm's law — voltage = current × resistance. Given in the data booklet.
potential difference / voltage (volts, V)
current (amperes, A)
resistance (ohms, Ω)

Worked example — power from V and R

A 24 Ω heater element is connected across a 12 V supply. Find the power it dissipates.

Solution

  1. You know V and R, so pick the matching given form:
  2. Put in the numbers (V = 12, R = 24):
  3. Work it out — keep the unit:

Final answer

P = 6.0 W.

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How this is tested: On Paper 1A, power is usually a 'determine' question — and almost always a ratio, not a one-off number.

- Paper 1A: compare the power of two components when something changes — e.g. a wire twice as long, or a switch opened/closed that changes the voltage or current. - Paper 2: plug numbers into P = IV / I²R / V²/R, or find the energy E = Pt and the cost of running an appliance.

Classic trap: picking the wrong form. If the voltage stays fixed, use P = V²/R (P ∝ 1/R); if the current stays fixed, use P = I²R (P ∝ R). They pull opposite ways.
Resistance of a wire: For a wire of the same metal and thickness, resistance is proportional to its length: a wire twice as long has twice the resistance (R ∝ L). Combine that with the right power form to get the ratio.

IB-style question — same supply, a longer wire

Two heating wires are made of the same metal with the same thickness, and each is connected across the same 6.0 V supply. Wire 2 is twice as long as wire 1. Determine the ratio of the power dissipated in wire 2 to that in wire 1.

Solution

  1. The voltage is the same for both, so use the form with V and R:
  2. Same V means P depends only on R, and twice the length means twice the resistance (R ∝ L), so R₂ = 2R₁:
  3. Put in R₂ = 2R₁:

Final answer

P₂ : P₁ = 1 : 2 — the longer wire has more resistance, so at a fixed voltage it dissipates HALF the power.

Try an IB Exam Question — Free AI Feedback

Test yourself on Electrical power and energy. Write your answer and get instant AI feedback — just like a real IB examiner.

A phone charger delivers a steady 5.0 W to a phone while charging it.

The phone takes 2.5 hours to charge fully.

the electrical energy delivered to the phone during one full charge, giving your answer in joules.
[2 marks]

Related Physics Topics

Continue learning with these related topics from the same unit:

2.1.1Internal energy and the particle model
2.1.2Specific heat capacity
2.1.3Latent heat and calorimetry
2.1.4Conduction, convection and radiation
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