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Course: SQA National 5 Physics
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SQA National 5 Physics

1. Dynamics

0/22

2. Space

0/12

3. Electricity

0/16

4. Properties of Matter

0/11

5. Waves

0/8

6. Radiation

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Text lesson

3.5.2 Electrical Power Relationships

3.5.2 Electrical Power Relationships: Use P = IV, P = I²R and P = V²/R and choose the best equation for the information given.

3.5.2 Electrical Power Relationships

Combining P = IV with Ohm's law gives two further useful relationships.

Power and current
  • P = IV, where P is power in watts, I is current in amperes and V is voltage in volts.
Using current and resistance
  • Substituting V = IR gives P = I²R, useful when the voltage is not known.
Using voltage and resistance
  • Substituting I = V ÷ R gives P = V²/R, useful when the current is not known.
Same power
  • All the equations give the same power; the choice depends on the data supplied.
Known quantities Equation to use
Current and voltage P = IV
Current and resistance P = I²R
Voltage and resistance P = V²/R
Energy and time P = E ÷ t

Selecting the equation correctly saves time and avoids mistakes.

1

List the two quantities you are given and the one you need to find.
2

Pick the equation that contains exactly those three quantities.
3

Rearrange if needed to get the unknown on its own.
4

Substitute with consistent units and calculate carefully.
5

Check the result by using a different equation if time allows.

These relationships explain how appliances are rated and chosen.

Applications
Current drawn
  • I = P ÷ V gives the current taken by an appliance from its rated power and the mains voltage.
Heating
  • A heater with a larger power rating draws more current and transfers more heat each second.
Cable heating
  • Because P = I²R, a large current can heat the wiring significantly even when its resistance is small.
Efficiency of choice
  • Appliances for 230 V mains are rated by their power, so comparing ratings shows which transfers energy fastest.

Question: A 230 V heater has a power rating of 2300 W. Calculate its current and resistance, and check the power using P = I²R.

  1. Current I = P ÷ V = 2300 ÷ 230 = 10 A.
  2. Resistance R = V ÷ I = 230 ÷ 10 = 23 Ω.
  3. Check: P = I²R = 10² × 23 = 2300 W.

Answer: The current is 10 A and the resistance is 23 Ω.

3.5.2 – Key Points for the Exam

  • P = IV: Power equals current multiplied by potential difference.
  • P = I²R: Power equals current squared multiplied by resistance.
  • P = V²/R: Power equals potential difference squared divided by resistance.
  • Rating: The power and voltage at which an appliance is designed to operate.

Common Mistakes

  • Forgetting to square the current or voltage in P = I²R or P = V²/R.
  • Using the wrong equation for the data given and then having to guess a missing quantity.
  • Mixing up P (power) with the symbol for potential difference.
  • Leaving out the unit watts, or confusing it with joules.
YOUR STUDY PATH

3.5.2 Electrical Power Relationships

SQA National 5 Physics / 3. Electricity

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