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Power

Spec 6.2.4.1 📗 Foundation
📖 In-Depth Theory

Electrical Power Equations

ELECTRICAL POWER is the rate of energy transfer by an electrical component.
P = V × I
P = I² × R
P = power (W)
V = potential difference (V)
I = current (A)
R = resistance (Ω)
Use P = VI when you know V and I.
Use P = I²R when you know I and R.
EXAMPLE:
Lamp: 12 V, 2 A.
P = 12 × 2 = 24 W
Check: R = 12/2 = 6 Ω; P = 2² × 6 = 4 × 6 = 24 W ✓

Energy and Charge

E = P × t (energy in J, time in seconds)
E = V × Q (energy = pd × charge)
EXAMPLE:
60 W lamp for 5 minutes:
E = 60 × 300 = 18,000 J
Charge link:
150 C through 12 V component:
E = 12 × 150 = 1800 J

Choosing the Correct Fuse

Step 1: I = P ÷ V
Step 2: Choose fuse rated JUST ABOVE operating current.
Common ratings: 1 A, 3 A, 5 A, 13 A.
EXAMPLE 1 — 500 W TV at 230 V:
I = 500 ÷ 230 ≈ 2.2 A → 3 A fuse.
EXAMPLE 2 — 2300 W kettle at 230 V:
I = 2300 ÷ 230 = 10 A → 13 A fuse.
Fuse too low → blows in normal use.
Fuse too high → won't blow in a fault → dangerous.
⚠️ Common Mistake

In P = I²R, CURRENT is squared, not resistance. Students often write P = I × R² by mistake. To find I from power: use I = P ÷ V (not P = I²R unless R is known and V is not).

📐 Variables
PPower (P) is measured in watts (W)
VPotential difference (V) is measured in volts (V)
ICurrent (I) is measured in amperes (A)
RResistance (R) is measured in ohms (Ω)
EEnergy (E) is measured in joules (J)
QCharge (Q) is measured in coulombs (C)
📐 Key Equations
P = V × I
P = I² × R
E = P × t
E = V × Q
📌 Key Note

P = VI. P = I²R. E = Pt. E = VQ. I = P/V for fuse selection — choose just above. UK mains = 230 V.

🎯 Matching Activity — Power Calculations

Match each scenario to the correct answer. — drag the symbols on the right to match the component names on the left.

24 W
Drop here
100 W
Drop here
3 A fuse
Drop here
13 A fuse
Drop here
690 W appliance on 230 V — I = 3 A
V = 12 V, I = 2 A — P = 12 × 2
I = 10 A, R = 1 Ω — P = 10² × 1
2300 W kettle on 230 V — I = 10 A
⚽ FIFA Worked Examples
Electrical Power

A lamp operates at 240 V and draws 0.25 A. Calculate its power.

F

P = V × I

I

V = 240 V, I = 0.25 A

F

P = 240 × 0.25

A

P = 60 W

🎯 Test Yourself
Question 1 of 2
1. A resistor (R = 8 Ω) carries 3 A. What is the power dissipated?
2. A 1380 W iron on 230 V mains. Which fuse?
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