Electrical Power - Physics

Calculating Power

Electrical power is the rate of energy transfer. It is measured in watts (W). One watt is one joule per second.

Power (W) = Energy transferred (J)   or P= E

                                   time (s)                               T


All electrical appliances are marked with a power rating in watts or kilowatts.

In a circuit, if the voltmeter reads 12 V and the ammeter reads 2.0 A, then the power of a lamp can be worked out as: 12 V x 2.0 A= 24W

electrical power (W) = current (A) x potential difference (V)

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A fuse is a very thin piece of wire that acts as a safety device in a mains plug.

If a fault occurs and the current becomes too high, the fuse wire will melt and break the circuit causing further damage or fire.

The fuse rating is the maximum current the fuse can carry without melting. Fuses are available at 1 A, 3 A, 5 A and 13 A.

To find the correct fuse to use, calculate the normal operating current for an appliance. For example a toaster has a power of 1100 W and is connected to a 230 V mains supply. 

Power = current X potential difference

Current = power/potential difference = 1100/230 = 4.8 A

The fuse rating must be higher than 4.8 A, so choose a 5 A fuse.

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Kilowatt - hours

Lower-power appliances use less energy in a given time, so they cost less to run.

The energy in joules used by an appliance can be calculated: energy transferred (J) = power (W) X time (s)

To calculate the energy used by a 1800 W heater in one hour:

1 hour = 60 X 60 = 3600 s

energy = 1800 X 3600 = 6 480 000 J

This number is very large, so we prefer to use the unit kilowatt-hour (kW h) for mains electricity bills. One kilowatt - hour is the energy used by a 1 kW appliance in 1 hour.

energy (kW h) = power (kW) X time (h)

The cost of electricity is calculated: cost (p) = power (kW) X time (h) X cost per kW h (p/kW h)

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Saving energy and money

Using energy-saving devices in the home has financial and environmental benefits. Examples include: compact fluorescent bulbs, motion activated lights, home insulation and standby detection devices.

The payback time of installing energy-saving devices is defined as the number of years it takes to get back the initial cost of the installation from savings on energy bills.

payback time = initial cost / annual saving

For example, if installing double-glazing costs £5000, and in one year saves £250 on heating bills, the payback time is 20 years (5000 / 250)

Sometimes other factors influence decisions oabout installingb enrgy-saving devices. For example, installing double-glazing will provide better soundproofing or increase the saleability of a home.

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