## Episode 118: Potential dividers

Lesson for 16-19

- Activity time 105 minutes
- Level Advanced

This episode introduces the use of a potential divider as a source of variable pd. Students will also learn to use potential dividers to detect temperature or light levels.

The potential divider circuit is a particularly useful arrangement but many students find this difficult to grasp at first. It is worth spending time to make sure they understand what is going on. It may also be worth emphasizing, especially with more able groups, that the output voltage is affected by the load resistance (many, on first meeting the device, do not realize this).

Lesson Summary

- Discussion and demonstration: The potential divider formula (15 minutes)
- Student experiment: Using potential dividers (40 minutes)
- Student questions: Using the formula (20 minutes)
- Discussion: The effect of the load on output (10 minutes)
- Student questions: The effect of the load on output (30 minutes)

#### Discussion and demonstrations: The potential divider formula

The potential divider is one of the most useful circuits your students will meet so it is particularly important that they understand how it works and how to calculate the voltage across each of its resistors. For an unloaded potential divider the current through each resistor is the same so the voltage is proportional to the resistance. This means that the pd across the pair of resistors is divided in the same ratio as the resistors themselves:

i.e.

*V*_{1}*V*_{2} = *I**R*_{1}*I**R*_{2}

*or*

*V*_{1}*V*_{2} = *R*_{1}*R*_{2}

It is worth emphasizing the practical implication of this – if *R*_{1} >> *R*_{2} then *V*_{1} is more or less the supply voltage and if *R*_{1} << *R*_{2} then *V*_{1} is close to 0 V. You could encourage them to see *V*_{source} as an input to the potential divider and *V*_{1} as an output. The circuit itself provides a way to tap off a voltage between 0 V and *V*_{source} .

This can, of course be done continuously using a rheostat or potentiometer and it is well worth demonstrating a variety of these including the rotary potentiometers used as volume controls in hi-fi systems.

The potential divider equation can be derived by rearranging the ratios above to give:

*V*_{1} = *R*_{1}*R*_{2} × *V*_{2}

#### Student experiment: Using potential dividers

Start with Parts 1-3 of these experiments.

Once they have experimented with several versions of the potential divider they can apply the ideas to a simple sensor circuit (Part 4). Set them the task of building and testing either an electronic temperature sensor or an electronic light meter (or both).

You may feel that the latter experiment as presented is sufficient to introduce the idea of a temperature or light intensity sensor, but this could be extended. The thermistor can be used as part of an electronic thermometer if it is connected into a suitable potential divider circuit and calibrated. One way to do this is to use a water bath to warm the thermistor (you can use ice with salt to get well below 0 ° C and hot water to get above 80 ° C giving a good range). However, to do this you need water-proofed thermistors (e.g. wrapped in polythene or embedded in an epoxy). If the water reaches the electrical connections then the readings will be unreliable. Beware of safety in the laboratory with both hot water and mercury thermometers (they need these for their calibration). Hot water from a kettle is safer than using a beaker on a tripod heated by a Bunsen burner. A similar approach can be used to make a light intensity meter (this can be calibrated using a lux meter).

Episode 118-1: Potential dividers (Word, 42 KB)

#### Student questions: Using the formula

Practice in potential divider calculations.

Episode 118-2: Tapping off a potential difference (Word, 44 KB)

#### Discussion: The effect of the load on output

With strong groups you might discuss the effect of loading a potential divider on its output voltage. The ideas to get across are:

Connecting a load across *R*_{1} reduces the output voltage.

This is because the effective resistance in the lower arm of the potential divider is now a parallel combination of *R*_{1} and *R*_{load} (less than *R*_{1} ) so a smaller fraction of the voltage is tapped off

.

If *R*_{load} >> *R*_{1} then there is no significant effect on the output voltage.

It is worth going back to their experiences as younger pupils tackling simple circuits, and considering what was happening when a lit bulb went out when shorted out

by a piece of wire. Pupils will often regress to a current based explanation, particularly under the pressure of examination conditions. We need to encourage them now to think in terms of potential difference, and resistance. In the shorting out

case it was not that much that all the current wanted to take the easier parallel route

, but that the low resistance of the wire in parallel reduced the combination’s total resistance, compared to the rest of the circuit.

#### Student questions

Episode 118-3: Loading a potential divider (Word, 36 KB)

Episode 118-4: Brightness of bulbs (Word, 29 KB)