# Electrical Energy and Power

This topic is part of the HSC Physics course under the section *Electric Circuits.*

### HSC Physics Syllabus

**investigate quantitatively and analyse the rate of conversion of electrical energy in components of electric circuits, including the production of heat and light, by applying `P = VI` and `E = Pt` and variations that involve Ohm’s Law (ACSPH042)**

**investigate quantitatively the application of the law of conservation of energy to the heating effects of electric currents, including the application of `P = VI` and variations of this involving Ohm’s Law (ACSPH043)**

### Electrical Power Explained for HSC Physics

Electrical power is a fundamental concept in physics that relates to the rate at which electrical energy is converted to other forms of energy in an electric circuit.

### Electrical Power

Recall that the total voltage of an electric circuit is defined as work done per unit charge, and is given by:

$$V = \frac{W}{q}$$

$$V = \frac{E}{q}$$

and current is given by:

$$I = \frac{q}{t}$$

When we multiply voltage and current:

$$VI = \frac{E}{q} \times \frac{q}{t}$$

$$VI = \frac{E}{t}$$

Therefore, electrical power is given by:

$$P = VI$$

where

- power is measured in watts (W)
- voltage in volts (V)
- current in amperes (A)

### Electrical Power and Ohm's Law

Ohm's Law states that:

$$$V = IR$$$

Combining Ohm’s Law with the power formula, we can derive expressions for power involving resistance:

$$$P = (IR)(I) = I^2R$$$

$and$

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

These variations show that power depends on both the square of the current and the resistance, or the square of the voltage and the inverse of the resistance.

So how can power be proportional and inversely proportional to resistance at the same time?

Let's look at each equation case by case:

$$P = I^2R$$

This equation is used when voltage is variable and current is kept constant. When there's a large resistance, a greater voltage (and thus work done) is required to produce the same magnitude of current. This means the battery needs to deliver more power to produce the same amount of current. Vice versa, when resistance is small, a smaller voltage and work is required.

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

This equation is used when the voltage is constant e.g. fixed voltage from a battery. As resistance of a circuit increases, the magnitude of current decreases. This means the electrical power delivered by the battery also decreases.

### Conservation of Energy in Electric Circuits

The power in an electrical circuit is the rate at which electrical energy is used by a component. By the law of conservation of energy, the electrical energy used by a circuit component is transformed into other forms of energy, such as heat and light. The heating effect of electric currents, for example, is applied in appliances like toasters and electric heaters.

#### Light Energy

The power dissipated by a light bulb can be calculated using the same formulas as any other resistive component in a circuit. For a given bulb (or resistor), the power dissipation dictates the amount of light (and heat) produced. A higher power dissipation results in a brighter bulb, assuming the efficiency of converting electrical energy to light remains constant.

#### Resistive Heat Production

When current flows through a resistor, power is dissipated as heat. This is because resistors impede the flow of electrons, causing them to collide with atoms within the resistor and transfer kinetic energy to the atoms, thus heating the material.

### Calculation Example 1

An electrical kettle when connected to 120 V supply draws 13 A of current. How much energy (in Joules) does the kettle use over 2 minutes?

*Solution:*

The electrical power is given by:

$$P = VI$$

$$P = 120 \times 13 = 1560 \text{ W}$$

Energy is given by:

$$E = Pt$$

$$E = 1560 \times (2 \times 60) = 187200 \text{ J}$$

### Calculation Example 2

A 4.5 V battery is connected to a 25 Ω resistor.

(a) How much electrical power is delivered by the battery?

(b) How much power is lost as heat by the resistor?

*Solution to part (a):*

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

$$P = \frac{4.5^2}{25} = 0.81 \text{ W}$$

*Solution to part (b):*

****By law of conservation of energy, the electrical power delivered by the battery should equal to the power dissipated as heat by the resistor. Therefore, 0.81 Joules of heat is lost per second.