Electrical Power and Energy

Resistors convert electrical energy into heat. Lamps convert it into heat and light. Motors convert it into mechanical work. In each case, energy flows from the battery into the component at a rate determined by the electrical power.

This lesson derives three equivalent forms of the power equation, shows how to compute the total energy transferred over time, and applies them to practical examples from household appliances to fuses and heating elements.

OCR A-Level Physics A Module 4.2.2 (Energy, power and resistance).

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Defining Electrical Power

Recall that the pd across a component is energy per unit charge:

V = W / Q

and current is charge per unit time:

I = Q / t

Power is energy per unit time. Combining these:

P = W / t = (V Q) / t = V × (Q / t) = V × I

So the power dissipated by a component is the product of the pd across it and the current through it:

P = V I

• P in watts (W)
• V in volts (V)
• I in amperes (A)

The watt is defined as 1 W = 1 J s⁻¹ = 1 V A.

Quick Example

A 12 V lamp carries a current of 2.0 A. Its power dissipation is:

• P = V I = 12 × 2.0 = 24 W

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Three Equivalent Forms of the Power Equation

Using V = IR (the definition of resistance) in the power equation P = VI, we get three equivalent formulas:

P = V I = I² R = V² / R

Derivation:

• Substitute V = IR: P = (IR) × I = I² R
• Substitute I = V/R: P = V × (V/R) = V² / R

Each form is useful in different circumstances:

Form	Known quantities	Best for
P = VI	V and I	Any direct measurement
P = I²R	I and R	Circuits with a known resistor and measured current
P = V²/R	V and R	Devices with a known supply voltage

Worked Example 1 — Three Forms

A 60 W lamp is designed for a 230 V mains supply. Calculate (a) the current it draws; (b) its operating resistance.

(a) P = VI → I = P/V = 60 / 230 = 0.26 A ≈ 260 mA

(b) P = V²/R → R = V²/P = 230² / 60 = 52 900 / 60 = 880 Ω

Notice we used two different forms of the same law — this is typical in exam questions.

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Energy Transferred Over Time

If a component dissipates power P for a time t, the total energy transferred is:

W = P × t

and expanding the power:

W = V I t = I² R t = V² t / R

• W in joules (J)
• t in seconds (s)

Worked Example 2 — Total Energy

A 2.0 kW kettle runs on the 230 V mains for 3.0 minutes. Calculate (a) the current drawn; (b) the total energy converted.

(a) I = P/V = 2000 / 230 = 8.7 A

(b) W = P × t = 2000 × (3.0 × 60) = 2000 × 180 = 360 000 J = 360 kJ

This is enough energy to raise about 1 litre of water from room temperature to boiling, which is roughly what a kettle does in three minutes.

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The Kilowatt-Hour

Electricity bills use a non-SI unit, the kilowatt-hour (kWh):

1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3.6 × 10⁶ J = 3.6 MJ

If electricity costs (say) 30 p/kWh, then running that 2 kW kettle for 3 minutes costs:

• Energy = 2 kW × (3/60) h = 0.10 kWh
• Cost = 0.10 × 30 = 3 p

You are expected to be able to convert between joules and kWh at A-Level. The kilowatt-hour is not used in physics calculations but is helpful in applied questions.

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Worked Example 3 — Choosing a Fuse

A 240 V hair dryer rated at 1200 W is plugged into a UK 13 A socket. Is a 3 A fuse, 5 A fuse or 13 A fuse appropriate?

• I = P/V = 1200 / 240 = 5.0 A
• The 3 A fuse would blow immediately (current exceeds rating).
• The 5 A fuse would be marginal — likely to blow under normal use.
• A 13 A fuse is appropriate (nearest standard size above the working current).