Electrical Power and Ratings

Overview

This note supports the Topic 13 hub by turning the general idea of energy transfer into the formulas used in circuit questions.

The main exam use is formula selection: know which quantities are given, choose the matching power relation, and then interpret the result in context.

Related topics:

• Internal Resistance
• Current Electricity Common Exam Traps

Definition

Electrical power is the rate of energy transfer.

$$P=\frac{W}{t}$$

Where:

• $P$ = power
• $W$ = energy transferred
• $t$ = time

Why It Matters

Electrical power and ratings are used to:

• determine operating current
• compare energy use of appliances
• calculate heating in resistors and wires
• choose suitable fuses and supplies
• estimate electricity cost
• relate source output to internal losses

Key Representations

Meaning of 1 Watt

$$1\text{ W}=1{\text{ J s}}^{-1}$$

Power in Circuits

For a component with voltage $V$ and current $I$,

$$P=IV$$

This is the main circuit-power relation in Topic 13.

Power for Resistors

Using $V=IR$,

$$P=I^{2}R$$

and

$$P=\frac{V^2}{R}$$

Choosing the Correct Formula

• if current and resistance are known, use $P=I^{2}R$
• if voltage and resistance are known, use $P=V^2/R$
• if voltage and current are known, use $P=IV$

Electrical Energy Transfer

Energy transferred in time $t$ is

$$W=Pt$$

So,

$$W=VIt$$

Since

$$Q=It$$

then

$$W=VQ$$

Units

Power

• watt (W)
• kilowatt (kW)

$$1\text{ kW}=1000\text{ W}$$

Energy

• joule (J)
• kilowatt-hour (kWh)

$$1\text{ kWh}=3.6\times 10^6\text{ J}$$

Appliance Ratings

A label such as

230 V, 60 W

means the appliance is designed to operate at $230\text{V}$ and transfer energy at $60\text{W}$ under normal operation.

The operating current is

$$I=\frac{P}{V}$$

For a $230\text{V}$, $60\text{W}$ lamp,

$$I=\frac{60}{230}=0.26\text{A}$$

Heating Effect

Power dissipated as heat in a resistance is

$$P_{\text{heat}}=I^{2}R$$

This is why heating depends so strongly on current: doubling the current gives four times the heating power for the same resistance.

This matters for:

• heating elements
• power loss in cables
• fuse operation
• internal resistance of sources

Efficiency

Efficiency is useful power output divided by total power input:

$$\eta =\frac{\text{useful power out}}{\text{total power in}}$$

In source questions, the same idea becomes

$$\eta =\frac{V}{\mathcal{E}}$$

when current flows from a real source with internal resistance.

Quick Checks

• Match the formula to the known quantities.
• Use the rating to find operating current when needed.
• Keep heating questions tied to $I^{2}R$.
• Use the common-exam-traps note when the question mixes power, energy, and charge.

Links

• Current Electricity Fundamentals
• Internal Resistance
• Current Electricity Common Exam Traps
• Work, Energy and Power