Electrical Energy, Work and Power

Overview

By the principle of conservation of energy, the energy given to the charges by the cell must equal the sum of the energies given by the charges to the various components.
- This means that the e.m.f. across the cell is equal to the sum of the p.d.s round the circuit.
- For example, the total voltage across the cell equals the sum of the voltages across the lamps.
Batteries and power supplies give energy to the charges in a circuit. Similarly, we can think about other components in a circuit.
- A small lamp may have a p.d. of 1.5 V across it. This means that each coulomb of charge does 1.5 J of electrical work to pass through the lamp and will transfer 1.5 J of energy to the lamp.

Electrical power is both: $power (W) = \frac{energy transferred (J)}{time taken (s)}$ $¶ = \frac{Δ E}{t}$
and: $power (W) = current (A) \times p.d. (V)$ $P = V \times I$
Electrical power is the rate at which energy is transferred per unit time.
One watt is one joule per second.

Remember that the e.m.f. of a supply tells us how much energy it transfers to charges flowing around the circuit.
- The greater the current flowing around the circuit, the faster that energy is transferred.
The rate at which energy is transferred in the circuit (the power, $P$ ) depends on both:
- the e.m.f, $E$ of the supply
- the current, $I$ , that it pushes round the circuit.

Since $energy transferred = power \times time$ , the equation $P = V I$ can be used to give an equation for electrical energy transferred $E$ in terms of current and voltage: $energy transferred (J) = current (A) \times p.d. (V) \times time (s)$ $E = I V t$

Like other stores of energy, we could calculate the amount of energy transferred electrically in joules (J), but it is more convenient to use kWh.
- This is because 1 kW = 1000 W and 1 h = 3600 s, so 1 kWh = 1000 W $\times$ 3600 s = $3.6 \times 1 0^{6} J$ .
1 kWh is a unit of energy.