Increasing the power in a pathway – resistance is added in parallel to a simple circuit loop
Wrong Track: When the resistor is added in parallel, there is more resistance, so energy will be shifted from the cell more slowly.
Right Lines: When the resistor is added in parallel, an extra current loop is provided and energy is shifted from the cell at a greater rate.
Using quantitative examples to help keep the thinking on track
Thinking about the teaching
As with the previous challenge, using physical quantities offers a direct way to sort out an answer. So, you might invent some figures along the following lines:
Teacher: Suppose we say that the cell is a 6 volt supply and that the resistor has a resistance of 12 ohm.
Teacher: We can then work out the current in the loop using I = VR, in this case 6 volt12 ohm, so the current works out at 0.5 A.
Teacher: So, 6 joule for each coulomb, 0.5 coulomb arriving each second, this predicts that 3 joule of energy is shifted in the resistor each second
Teacher: OK, what happens when we add a second resistor in parallel?
Teacher: 6 volt is the potential difference across each resistor.
Teacher: Nothing has changed from the original circuit. The current through each resistor is:I = VR, in this case 6 volt12 ohm, so the current works out at 0.5 A.
As a result, 3 joule of energy is shifted in each resistor per second.
Teacher: Alternatively we can say that if each of the two bulbs is shifting energy at the rate of 3 joule second -1, then the store associated with the cell must be emptying at the rate of 6 joule second -1.
Teacher: So, adding an equal resistor in parallel doubles the rate at which the cell's store of energy is depleted.
The physical picture to leave the students with is one of there being a circulation of charge through each of the two parallel loops. The flow of charge through the cell is doubled and therefore the rate at which energy is shifted is increased.