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Energy in Circuits

Electrical appliances transfer energy. As an electric current flows through a circuit, energy is transferred from the battery or power supply to the components in the circuit. For example, electrical energy is transferred to light and heat by bulbs. An electric current is a flow of charge. The charge is carried by electrons flowing around the circuit.When charge flows through any resistor, some electrical energy is transferred as heat.

Power

Power is the amount of energy an appliance transfers every second. The more powerful it is, the more energy it transfers per second.

Power is measured in joules per second, or watts (W).

So, 1 J/s = 1W

A kilowatt (kW) is 1000 watts.

A kettle transfers 2300 joules of energy every second. It has a power rating of 2300 W, or 2.3 kW.A radio only transfers about 12 joules per second. It has a power rating of 12 W.

Electrical power can be calculated using the following equation:

 

power

=

potential difference

x

current

 

 or

P

 

 V

 

 I

 

 

  (watt, W)

(volt, V)

         (ampere, A)

 

For example, if a 230 V fan heater takes a current of 5 A,

P = V I

 

P = 230 x 5

 

P = 1150 W

 

Most appliances have their power in watts and the voltage of the supply they need printed on them. From this we can calculate their current and so the size of fuse that should be fitted in a plug. The fuse should have a higher value, but as close as possible to, the current through the appliance when it is working normally.

For example, a 230 V kettle has a power rating of 2300 W.

 

P = V I can be rearranged to:

I =

P

 

 

 

 

V

 

 

 

I =

2300

 

 

 

 

 230

 

 

 

I =

10 A

 

The closest fuse above this is 13 A.


Charge

Electrical charge is measured in coulombs (C). A higher voltage gives more power, because each electron carries more energy. More energy is transferred for a given amount of charge which flows. This is shown in the equation:

 

energy transferred

=

potential difference

x

charge

 

or

E

=

V

 

Q

 

 

(joule, J)

 

(volt, V)

 

(coulomb, Q)

 

If a current of 1 ampere flows for 1 second then 1 coulomb of charge is passing.

This is shown in the equation:

 

charge

=

current

x

time

 

 or

Q

=

 I

 

t

 

 

(coulomb, C)

         (ampere, A)

           (second, s)

 

These two equations can be put together. If I x t is substituted for Q into the first equation we get:

  energy transferred =

potential difference

x

current

x

time

or

E =

 V

 

 I

 

t

(joule, J)

(volt, V)

          (ampere, A)

                           (second, s)


For example, if a 250 V fan takes 2 A of current, how much energy will it transfer in 10 minutes?

 E = V I t

 

Time needs to be in seconds so,

 

 t = 10 x 60

 

   = 600 s

 

E = 250 x 2 x 600

 

E = 300000 J

 

 

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