4.2.4 Energy Transfers

In this section, we explore how electrical appliances transform energy from one form to another and how we can mathematically calculate the amount of work being done by the current.

1. Energy Transfers in Everyday Appliances

Electrical appliances are designed to transfer energy. The "work done" in a circuit is when charge flows against a resistance. Common examples include:

1. Kinetic Energy: Motors (e.g., in a washing machine or fan).
2. Thermal Energy: Heating elements (e.g., in a kettle, toaster, or iron).

Some energy is always dissipated (wasted) as thermal energy to the surroundings due to the resistance of the wires.

2. Calculating Energy Transferred

The amount of energy an appliance transfers depends on how long it is used and its power rating. There are two primary formulas you need to master for AQA:

Formula 1: Energy and Power

The energy transferred is equal to the power multiplied by the time.

$E=P\times t$

Where:

1. E is energy transferred in joules (J)
2. P is power in watts (W)
3. t is time in seconds (s)

Formula 2: Energy and Charge Flow

Energy is also transferred when a charge flows through a potential difference.

$E=Q\times V$

Where:

1. E is energy transferred in joules (J)
2. Q is charge flow in coulombs (C)
3. V is potential difference in volts (V)

3. Electrical Power

Power is the rate at which energy is transferred. For any electrical component (or appliance), the power is related to the potential difference across it and the current through it.

The Power Equations

1. P = V * I
2. P = $I^2\times R$

Why $I^2$ ? The second formula ($P=I^{2}R$ ) is crucial for explaining why the National Grid uses high voltages. By keeping current (I) low, we significantly reduce the power lost as heat, because power loss is proportional to the square of the current.

4. Mathematical Example

Question: An electric motor is connected to a 12V supply. A charge of 300C passes through the motor. Calculate the energy transferred.

Step 1: Identify the variables

1. V = 12V
2. Q = 300C

Step 2: Choose the correct formula

Since we have charge and potential difference, we use:

E = Q * V

Step 3: Substitute and solve

E = 300 * 12 = 3,600J

Answer: The energy transferred is 3,600J (or 3.6kJ).

5. Edge Cases and Unit Conversions

1. Time: Students often forget that t must be in seconds. If a question says an appliance is on for 5 minutes, you must convert: 5 * 60 = 300s.
2. Power: Large appliances are often rated in kilowatts (kW). 1kW = 1,000W.
3. Efficiency: Remember that Energy Transferred is the "Total Energy." To find "Useful Energy," you would need to know the efficiency of the device.

6. Summary of Power and Energy Equations