Topic 1 of 4
Electrical heating and power
An appliance transfers electrical energy. Its power tells you how quickly it does so; its operating time determines how much energy it uses.
Current is charge passing per second. Potential difference is energy transferred per charge. Power is the rate of energy transfer: one watt means one joule per second.
Use the heating effect
In a resistive heating element, electrical work transfers energy into the material. Its internal energy increases as it warms. Energy then transfers from the hot element to cooler objects and surroundings.
- Electric kettle
- The heating element transfers energy to the water. Heating the water is the useful purpose; some energy also warms the kettle and surroundings.
- Electric oven
- Hot elements transfer energy to the oven and food, including by radiation and through the warmed air.
- Electric heater
- The element heats its surroundings by radiation and by heating air that can circulate through the room.
At a steady operating temperature, an element can keep receiving electrical energy while transferring energy away at the same rate. Its temperature need not keep rising. Charge continues through the circuit; it is not used up or changed into heat.
Connect voltage, current and power
If charge Q passes through a component with p.d. V, the electrical energy transferred is E = VQ. For steady current I over time t, Q = It. Combining these gives:
Use the p.d. across the component and current through that same component. Convert small units first: 1 mA = 0.001 A and 1 mV = 0.001 V.
These expressions give electrical input. They do not by themselves tell you how much becomes a particular useful output. A motor's mechanical output, for example, would need further information about the motor or its load.
Power, then energy
A heater operating for 12 minutes
A model resistive heater operates at a stated 230 V and 5.0 A, with constant power for 12 minutes.
P = VI = 230 x 5.0 = 1150 W = 1.15 kW.
The duration is 12 x 60 = 720 s, so E = Pt = 1150 x 720 = 828 000 J.
This is the electrical input during the stated operating time. The calculation uses the supplied operating values for this resistive appliance.
Determine electrical input using low voltage
Pair the component's voltage with its current
A is in series; V connects across the component's endpoints P and Q. These supplied readings are for a low-voltage d.c. circuit.
For a steady 180 s interval:
P = VI = 6.0 × 0.40 = 2.4 W.
E = Pt = 2.4 × 180 = 432 J.
The ideal voltmeter takes negligible current, so the ammeter gives the component current. These calculations find electrical input; a particular useful output may be smaller.
For this component, P = 6.0 x 0.40 = 2.4 W. Over 180 s, E = 2.4 x 180 = 432 J.
- Use a suitable low-voltage supply and component. Connect the ammeter in series and voltmeter across the component whose input you want.
- Choose meter ranges that include the expected readings and give useful resolution. Record the actual readings with their units and measure the operating time.
- Check the readings during the interval. A component that warms can change its current, so one starting reading may not represent the whole run.
- Calculate power from matching voltage and current readings, then account for the time for which that power applies.
E = Pt uses a constant power or the average power over the interval. If voltage stays fixed while current changes, the interval's average current can be used in E = VIt. If both voltage and current change, find the power for each interval and add its energy; multiplying two separate average readings is not generally enough.