Episode 120: Energy transfer in electric circuits
Lesson for 16-19
- Activity time 55 minutes
- Level Advanced
This episode revises the idea of EMF, and introduces the idea of internal resistance and energy dissipation.
- Demonstration and discussion: Revising ideas about energy in the context of electric circuits (20 minutes)
- Student experiment: Leading to the idea of
lost volts(30 minutes)
- Demonstration: (Optional) Drop in terminal PD of a source on load (5 minutes)
Discussion and demonstrations: Revising ideas about energy in the context of electric circuits
Here you have the opportunity to check and reinforce students’ basic ideas about current, voltage and energy in electrical circuits. This could be truncated or skipped with an able group.
Use any suitable battery-driven device as a prop – e.g. toy car, MP3 player, torch – and ask the class a number of questions. (For the sake of illustration, we have assumed a supply voltage of 3.0 V from two AA cells, but this should be adapted to the particular example used.)
Some questions to ask:
- How is energy stored in the batteries? (Chemically.)
- This is powered by two 1.5 V AA cells in series – what is the supply voltage? (3.0 V)
- What is the EMF of the supply? (3.0 V – remind them that EMF is the name given to the potential difference across the supply when it is not loaded.)
- After a while the battery needs to be replaced; why? (Steer them away from the idea that it
runs out of charge– you are looking for an answer based on chemical reactions. The reactions that allow the battery to do work on charge have completed – like burning fuel.)
- What determines how quickly it runs down? (Look for answers linked to power, the rate at which energy is transferred and link this to current drawn and supply voltage.)
- How could the design be modified so it runs for longer before batteries need replacing? (Use two pairs of two AA batteries in parallel – effectively more
chemical fuelbut running at the same voltage so each cell provides half the previous power and current and lasts twice as long. This is unwise in practice though, if one pair of cells runs down faster than the other current can be forced backwards through the cells and they get very hot.)
- What determines how much current is drawn from the supply? (Load resistance – the lower the load resistance the greater the current and power and the sooner the battery needs to be replaced.)
Now open up the discussion to include what is happening within the cells. Point out that current flows through all parts of the circuit, including the cells. Since the cells themselves are made of material with resistance there will be some heating in the cell. This will increase with current, so there is more dissipation at higher currents. Although the chemicals that produce the current are the same ones that are responsible for the cell’s internal resistance (this term can be introduced quite naturally through discussion) it is convenient to separate these two roles so that they end up with an idea of heating as a way that energy is dissipated.
We represent a
real cell as two components in a circuit diagram:
idealcell, with no internal resistance, and
- a small resistor, representing the internal resistance.
It can make things clearer if you draw a box or circle around these two, to indicate that they are aspects of the same component.
Energy is conserved, so the energy that was stored chemically that is transferred by the current must be equal to the energy dissipated by the components in the circuit, both th external load and the internal resistance of the cell.
Clearly the heating in the cell is not doing useful work and so is in a sense
lost. This sets up the idea of
lost volts that will be referred to later, though the energy is not
lost, merely dissipated.
Student experiment: Leading to the idea of
This could be carried out as a demonstration but, if resources are available, the point comes across more strongly as a quick class practical. It can be advisable to do these experiments using rechargeable cells otherwise it can be quite expensive! However, rechargeable cells have a very low internal resistance so the experiment doesn’t work well. One could always cheat and place a rechargeable cell in a box with a series resistor.
The idea is simple. Students add lamps in parallel to a suitable battery. Introduce the idea of terminal voltage (terminal pd) and get them to measure it. They could also measure the current drawn from the cell, although this is less important.
Ask them to record the terminal voltage of the battery with no lamps connected and then as the number of lamps in parallel is increased.
Can they explain what is happening? Draw out the following ideas:
- As the number of parallel lamps increases, the terminal voltage drops and the current increases.
- This is because there is more heating of the internal resistance of the cell.
- The terminal voltage of the cell is equal to only its EMF when no current is drawn (no lamps attached).
- The difference between the battery EMF and its terminal voltage is a
lost voltageacross the internal resistance of the battery, given by Ir.
- The more current that is drawn from the battery the larger the
Class discussion after the experiment should prepare the way for the theory session which follows. It is worth pointing out that the EMF of a cell can be measured by connecting a high resistance voltmeter directly across its terminals (
high here means high compared to the internal resistance of the cell).
Demonstration: (optional) Drop in terminal pd of a source on load (5 minutes)
This shows the effect of a high internal resistance and an EHT supply.