Electrical energy and electrical power

Cambridge·Physics·4.2.5·Extended

Energy transfer in electric circuits

An electric circuit transfers energy from a source, such as an electrical cell or the mains supply, to its components. The source supplies energy to the circuit, and the components transfer that energy to their surroundings.

Different components transfer energy in different ways. A heater transfers energy by heating, a lamp produces light and also transfers energy by heating, and a motor can cause motion.

Components do not create the energy they transfer: the energy comes from the source. The circuit provides a way for energy to be transferred from the source to the components and then to the surroundings.

Electrical power

Electrical power is the rate at which electrical energy is transferred. A component with a higher power transfers more energy each second. Power is measured in watts ($$\mathrm{W}$$): one watt is one joule of energy transferred per second.

The power of a component depends on the current through it and the potential difference across it. Current is the rate at which charge flows; potential difference is the energy transferred per unit of charge. Multiplying them gives the energy transferred per second:

P=IVP = IV

Here, $$P$$ is power in watts ($$\mathrm{W}$$), $$I$$ is current in amperes ($$\mathrm{A}$$), and $$V$$ is potential difference in volts ($$\mathrm{V}$$).

For example, if the current through a component is $$2,\mathrm{A}$$ and the potential difference across it is $$12,\mathrm{V}$$, its power is $$2 \times 12 = 24,\mathrm{W}$$. It transfers $$24,\mathrm{J}$$ of electrical energy each second.

Rearranging the power equation

When power and one of the other quantities are known, the equation can be rearranged to find the missing quantity:

I=PVV=PII = \frac{P}{V} \qquad V = \frac{P}{I}

Divide power by potential difference to find current, or divide power by current to find potential difference.

Electrical energy calculation

A component’s power tells us how quickly it transfers energy. The energy transferred therefore depends on both the power and the time for which the transfer takes place:

E=PtE = Pt

Here, $$E$$ is energy in joules (J), $$P$$ is power in watts (W), and $$t$$ is time in seconds (s). One watt is one joule transferred per second.

Electrical power is also given by $$P = IV$$. Substituting this into $$E = Pt$$ gives:

E=IVtE = IVt

In this equation, $$I$$ is current in amperes (A) and $$V$$ is potential difference in volts (V). Use the time in seconds to obtain energy in joules.

Calculating energy transferred

A component has a potential difference of $$12,\text{V}$$ across it and a current of $$2,\text{A}$$ through it for $$5$$ minutes. First, convert the time to $$300,\text{s}$$. Then:

E=2×12×300=7200 JE = 2 \times 12 \times 300 = 7200\,\text{J}

Finding an unknown quantity

The equation can be rearranged to find any one quantity when the other three are known:

I=EVtV=EItt=EIVI = \frac{E}{Vt} \qquad V = \frac{E}{It} \qquad t = \frac{E}{IV}

For example, if the component transfers $$7200,\text{J}$$ at $$12,\text{V}$$ over $$300,\text{s}$$, its current is $$I = 7200/(12 \times 300) = 2,\text{A}$$.

Kilowatt-hours and electricity cost

A kilowatt-hour ($$\mathrm{kW,h}$$) is the energy transferred when a power of one kilowatt is maintained for one hour. It is a unit of energy, not power. One kilowatt-hour equals $$3.6 \times 10^6\ \mathrm{J}$$.

Energy used by an appliance

The energy an appliance transfers depends on its power and how long it is used. To find the energy in kilowatt-hours, multiply the power in kilowatts by the time in hours:

E=PtE = Pt

Here, $$E$$ is energy in $$\mathrm{kW,h}$$, $$P$$ is power in $$\mathrm{kW}$$ and $$t$$ is time in hours.

If the power is given in watts, divide it by $$1000$$ to convert it to kilowatts before calculating the energy. For example, $$2000\ \mathrm{W} = 2\ \mathrm{kW}$$.

Cost of electricity

The cost depends on both the energy used and the price charged per kilowatt-hour:

cost=energy used in kW h×price per kW h\text{cost} = \text{energy used in kW h} \times \text{price per kW h}

For example, a $$2000\ \mathrm{W}$$ appliance used for $$1.5$$ hours transfers $$2\ \mathrm{kW} \times 1.5\ \mathrm{h} = 3\ \mathrm{kW,h}$$ of energy. At $$30$$ pence per $$\mathrm{kW,h}$$, it costs $$3 \times 30 = 90$$ pence to use.