# Voltages across parts of a loop(Teaching tip)

### Sum of component voltages is equal to the battery voltage

Pupils will be familiar with the idea that when a second identical bulb is added in series to a circuit the two bulbs become equally dim. In episode 02, the point was made that the current in both bulbs was identical. The power dissipated in each bulb is also identical (you can see that–they're equally bright). We now have a way of accounting for this by noticing that the voltage supplied by the battery is shared between the two bulbs–equally if the bulbs are identical.

Having introduced voltage as a measure of power in the circuit and emphasised the point that energy must be conserved:

Teacher: So, the battery is rated at 6 volt. What does this tell us?

Wyn: The amount of energy it can supply.

Teacher: That's nearly right. It supplies 6 joule of energy every second (a power of 6 watt) for while there is a current of one ampere in the battery. You need to know both the current and the voltage to find the power.

Now it should not be too big a step to establish that the sum of the bulb voltages (when connected in series) is the battery voltage.

The simplest approach is for pupils to make their own predictions and measurements of voltage values, starting with simple battery/bulb circuits and moving on to circuits with extra bulbs in series. Pupils will be able to reason with a rope loop in order to be able to make semi-quantitative predictions, which should be enough here.

### Voltages across equal and unequal resistances

The idea to get over here is that if two resistors are connected in series to a battery, the greater share of the battery voltage is dropped across the bigger resistance. A full understanding of the circuit involves recognising that:

• The sum of the voltage drops across the two resistors must equal the battery voltage.
• More volts are dropped across the bigger resistance.
• The current is the same through both resistances.

This is an interesting problem, which might be set as a practical extension exercise for some pupils. Encourage active modelling of the situation, with two pupils gripping the rope to model the two resistors.

Here is a place where the loop model really does come into its own. It's very simple to model series connections. Simply grasp the rope with two hands rather than one to model two resistors instead of one. If you grasp equally firmly with both hands, you'd expect both hands to warm equally–what else could they do? It wouldn't even matter if you reversed the direction in which the rope was pulled.

It's also easy to model unequal resistances: simply grasp the rope more firmly with one hand than with the other. One hand will warm more quickly than the other. This predicts exactly what will happen in the electrical loop: the larger resistance will warm more than the smaller resistance.

The current in both resistors will be identical, just as the flow of rope through both hands is identical. It is the case that the larger resistor has a larger voltage across it, and this is the reason why it gets warmer. Similarly, the hand which grasps the rope more firmly shifts energy to the thermal store at a greater rate.

The power is largest where the current is largest and the voltage is largest.