**Practical work – Graphs and their interpretation **

Robert Birke

Teaching and learning coach

Stimulating Physics Network

As a physics teacher I very rarely have the time to do whole experiments. I just get time to check the equipment is working so it was quite refreshing to have a go at some experiments under the watchful eye of Robert and reminisce about the Nuffield A level approach to practical exams.

The first practical we looked at involved using an LDR to investigate how the light level changed when a light source was moved away from it.

The LDR was connected to an ammeter and the current reading represented the light level.

**The results**

distance/cm | current/mA |

5 | 9.68 |

10 | 5.74 |

15 | 4.15 |

20 | 3.29 |

25 | 2.76 |

30 | 2.45 |

35 | 2.17 |

40 | 2 |

The results show that as the distance increases the current decreases. As current = voltage/current and the voltage is constant then we can say that as the distance increases so does the resistance.

I always get my students to plot the raw data first.

The above graph doesn’t really tell us much other than as the distance increases the current/light level decreases. With discussion students should realise that there must be an inverse relationship between the current and distance.

During the AS part of the A level course that I teach there isn’t much emphasis on the inverse square law so the students probably would go for a graph of 1/D against current.

1/distance (cm^^{-1}) |
current/mA |

0.2 | 9.68 |

0.1 | 5.74 |

0.067 | 4.15 |

0.05 | 3.29 |

0.04 | 2.76 |

0.033 | 2.45 |

0.0286 | 2.17 |

0.025 | 2.00 |

The graph below does show that current is not inversely proportional to distance because the graph is not a straight line passing through 0,0 (normally my students aren’t allowed to plot graphs on a computer but I have done it as it is easier to manipulate for this article) .

There could be two reasons for this. The first one is that there could be other things affecting the current value besides the distance moved by the light source but the second thing which my year 13 students should be able to come up with is that the current/light level is proportional 1/D^^{2} (i.e. the current is inversely proportional to the (distance)^^{2}.

In order for the inverse square law to be obeyed everything needs to be constant except distance and current.

1/(distance)^^{2} (cm^^{-2}) |
current/mA |

0.04 | 9.68 |

0.01 | 5.74 |

0.0044 | 4.15 |

0.0025 | 3.29 |

0.0016 | 2.76 |

0.0011 | 2.45 |

0.00082 | 2.17 |

0.000625 | 2.00 |

Oh dear! This graph looks even worse than the previous one. My students would get quite despondent seeing this. They find it very difficult accepting that it doesn’t mean they have failed.

I had an idea that perhaps the background light value should be subtracted from each current value.

1/(distance^^{2}) (cm^^{-2}) |
current/mA |

0.04 | 8.31 |

0.01 | 4.37 |

0.0044 | 2.78 |

0.0025 | 1.92 |

0.0016 | 1.39 |

0.0011 | 1.08 |

0.000817 | 0.8 |

0.000625 | 0.63 |

0 | 0 |

The graph below shows that this didn’t help much. It looks, on the face of it, that I haven’t proved that the current is inversely proportional to (distance)^^{2}. There is no straight line passing through 0,0.

The next stage for the students would be a discussion on why the graph does not show an inverse square relationship.

The most obvious reason is that the light intensity of the bulb wasn’t constant due to fluctuations in the power supply and increasing resistance of the bulb due to the heating effect of the current.

Other errors include the LDR not facing the bulb directly, not getting the distance measurements correct (should D be measured from the glass envelope or the filament?) and not positioning the LDR at the correct distance. I’m sure you can come up with some other reasons.

Of course if I had more time I would have taken at least three current readings for each distance and subtracted a background light current reading from each to find the average corrected current reading.

Even better I could have designed the apparatus so that no outside light could get in.

The link below gives an alternative method of doing this experiment and uses a light meter instead of an LDR.

http://www.clt.uwa.edu.au/__data/assets/pdf_file/0010/2301769/chapter09_7.pdf

The picture below shows how the light meter and bulb can be arranged to prevent external light affecting the results.

The second practical we looked at is one I have got the students to try but you really need to use fresh potatoes.

http://tap.iop.org/electricity/emf/121/page_46054.html

The practical involved measuring the emf and internal resistance of a potato.

Some of the energy given to charges by a cell (or other source of emf) is dissipated inside the cell itself, as the charges move through the cell (or other source of emf). What is left is available as a potential difference (energy per unit charge) across the “external” resistor connected to the cell. If the emf of the source is E, and its internal resistance is r, then when a current I flows the potential difference V is V = E – I r.

Collect this apparatus:

2 digital multimeters

potato

0.5 cm x 2 cm copper sheet, 0.5 cm x 2 cm zinc sheet

2 pairs of crocodile clips

resistance substitution box

5 x 4 mm leads

http://tap.iop.org/electricity/emf/121/page_46054.html

1. Assemble your cell. Use the copper and zinc sheets as electrodes, inserting one in each end of the potato. Use crocodile clips to make connections to the circuit.

2. Connect the cell to a resistance substitution box and add an ammeter and voltmeter to measure the current through the external resistor and a voltmeter across the cell

3. Initially set the resistance substitution box to 4.7 kW as a load for the cell and record both the current and the pd.

4. Record as many current/pd measurements as you can for different load resistances.

The formula V = E – Ir can be used to plot a graph of V against I which should give a gradient equal to –r and an intercept on the V axis equal to emf E.

Resistance/ohm | p.d/V | Current/microA) |

500 | 0.096 | 55 |

1000 | 0.118 | 52 |

1500 | 0.138 | 50 |

2000 | 0.157 | 48 |

2500 | 0.174 | 46 |

3000 | 0.19 | 45 |

3500 | 0.204 | 43 |

4000 | 0.219 | 42 |

4700 | 0.24 | 41 |

The second graph is more useful as it shows the intercept on the x axis and gives a value of the emf E as about 0.63 V.

The gradient of the graph is I/V but we actually want V/I. So the internal resistance will be 1/gradient.

1/gradient = (1.5 – 0.5)/-100 microA = -r so r = 10000 ohms which seems reasonable to me. Of course I would expect my students to take more than one reading for each measurement and they would have to do an error analysis.

Thank you so much… Really helpful

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