### Reducing the power in the pathway: from one to two bulbs in series

**Wrong Track: **When the resistor is added in series, there is more resistance so energy will be shifted from the cell more quickly.

**Right Lines: ** When the resistor is added in series, the extra resistance reduces the current and energy is shifted from the cell at a lower rate.

### Using quantitative examples to help keep the thinking on track

**Thinking about the teaching**

This is the kind of question where using physical quantities (numbers and units) 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*_{1} = V*R*_{1}, 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 means that 3 joule of energy is shifted by the resistor each second.

Teacher: OK, what happens when we add a second resistor in series?

Teacher: The total resistance is now 24 ohm and 3 volt is the potential difference across each resistor.

Teacher: The current in the loop is 6 volt across 24 ohm, so 0.25 ampere. This makes sense, because doubling the resistance halves the current.

Teacher: So, 0.25 coulomb arriving each second and 3 joule shifted by each coulomb, this means that 0.75 joule of energy is shifted by each resistor every second.

Teacher: Alternatively we can say that, if each of the two bulbs is shifting energy at the rate of 0.75 joule second ^{-1}, then the cell must be emptying at the rate of 1.5 joule second ^{-1}.

Teacher: So, adding an equal resistor in series halves the rate at which the chemical store of the cell is depleted of energy.

The physical picture to leave the students with is one of the circulation of charge being reduced and therefore the rate of shifting energy being reduced. It's also worth emphasising that the rate of shifting energy at each resistor is reduced by a factor of 4 (from 3 joule second ^{-1} to 0.75 joule second ^{-1}) when the resistance of the loop is doubled since both the current through the resistor and the potential difference across the resistor are halved.

### Prepare for teaching using these links

kitset path represent & reason outline decide suggest issues evidence