Heating up wires
As charge flows through wires they heat up.
Electrical energy carried by the charge is converted to heat energy.
This lets a kettle heat water, or an electric fire heat a room.
Electric power
Electric power is the amount of electrical energy converted into other forms of energy in one second.
It be calculated using the equation:
electric power = current Ă— voltage
P = VI
where:
P = electric power in watts, W
I = current in amperes, A
V = voltage in volts, V
| \({P} = {VI}\) | \({P} = {V}\times{I}\) |
| \({I} =\frac{\text{P}}{\text{V}}\) | \({I} = {P} \div {V}\) |
| \({V} = \frac{\text{P}}{\text{I}}\) | \({V} = {P} \div {I}\) |
| \({P} = {VI}\) |
| \({P} = {V}\times{I}\) |
| \({I} =\frac{\text{P}}{\text{V}}\) |
| \({I} = {P} \div {V}\) |
| \({V} = \frac{\text{P}}{\text{I}}\) |
| \({V} = {P} \div {I}\) |
Remember that one watt is equal to one The unit of work or energy, written as J. per second (1 W = J/s).
Example
What is the power of a small electric motor if a current of 2 A flows when connected to a 12 V power supply?
P = VI
I = 2 A
V = 12 V
P = 2 x 12
P = 24 W
The power of the electric motor is 24 W.
In the example above the power of the electric motor is 24 W, but what does a power of 24 W mean?
It means that the electric motor converts 24 J of electric energy into other forms of energy (Energy that an object possesses by being in motion., heat energy, sound energy), every second.
Question
A light bulb has 60 W, 240 V printed on it. What does that mean?
It means that when the voltage across the bulb is 240 V, the bulb has a power of 60 W.
The light bulb converts 60 J of electrical energy into light and heat energy every second.
Question
A kettle has a power of 2.2 kW and is connected to mains voltage of 240 V.
- What current flows when the kettle is operating normally?
- What is the resistance of the kettle’s heating element?
1.
\({I} =\frac{\text{P}}{\text{V}}\)
To use this equation the power must be in watts, W.
The question gives the power in kilowatts, and so this must be converted into watts.
1 kW = 1000 W.
P = 2.2 kW = 2.2 x 1000 = 2200 W
V = 240 V
\({I} =\frac{\text{2200}}{\text{240}}\)
I = 9.17 A
2.
Use The rule that states that the current (I) flowing through a resistor (R) is directly proportional to the voltage (V) across the resistor, provided the temperature remains constant. to calculate the resistance of the kettle’s heating element
\({R} =\frac{\text{V}}{\text{I}}\)
V = 240 V
I = 9.17 A
\({R} =\frac{\text{240}}{\text{9.17}}\)
R = 26.2 \(\Omega\)
The resistance of the kettle’s heating element is 26.2 \(\Omega\)