Resistance

1 Plan and risk assessment

Resistance is how hard it is for a current to flow: resistance = potential difference ÷ current. Every centimetre of the same wire adds the same resistance, so a longer wire has more. In series, resistances add; in parallel, the total is less than the smallest.

V = I R

Prediction

2 Apparatus

• ✓Low-voltage power pack (0–12 V DC) Drives a current round the circuit.
• ✓Digital ammeter (multimeter on the 10 A range) (resolution 0.01 A) Measures the current through the wire, in series.
• ✓Digital voltmeter (multimeter on the 20 V range) (resolution 0.01 V) Measures the potential difference across the wire, in parallel.
• ✓Resistance wire taped to a metre rule (resolution 1 mm) The wire under test; the rule measures its length.
• ✓Two crocodile clips One fixed at 0 cm, one moved along the wire to change the length.
• ✓Five 4 mm leads Connect the circuit.
• ✓Resistors A (6.8 Ω) and B (10 Ω) on holders For the series and parallel part.

3 Method: your choices

• Which version are you doing?: AQA required practical 3: a wire’s length, then resistors in series and parallelResistance against length for a wire, then two resistors alone, in series and in parallel.
• Put the fixed clip on the 0 cm mark: DoneThe length you write down is read from the rule. If the clip is past the zero, every length of wire is longer than you record.
• Set the power pack: 2 V, a low pd, so the wire hardly warmsCurrents of about 0.2–1.5 A. The wire only gets warm. At the long lengths the current is small, so the ammeter’s last digit (0.01 A) is a 2–5 % uncertainty in the resistance.
• Switch off between readings: Unplug lead 3 after each pair of readings, and plug it in only for the nextThe wire only carries current while you read, so it stays close to room temperature. Its resistance stays the same (a control variable), and it does not burn you.
• Repeat each length: Read both meters twice at each length, clipping on afresh the second timeTwice as many readings, but the mean resistance has less random scatter from the clip’s contact and the meters’ last digit, and an anomaly stands out when the two readings disagree.

• 0.0 Struck through: the anomaly, retaken; the new reading is beside it.

Anomaly: Current at 40 cm, retaken. Likely cause: the sliding clip was not biting the wire properly, so its poor contact added extra resistance and the current was too small.

6 Analysis

1. Resistance of 20 cm of wire

   resistance = potential difference ÷ current (V = I R)

   Calculate the resistance of the 20 cm length of wire.

   0.908 Ω Right first time

2. Shape of the graph

   What does your graph show?

   A straight line: the resistance goes up by the same amount for every extra centimetre. It crosses the resistance axis a little above zero (the clips’ contact resistance), so the wire’s own resistance is directly proportional to its length. Right first time

3. Gradient (resistance per cm)

   gradient = change in y ÷ change in x

   0.0415 Ω/cm Right first time

4. y-intercept

   The extra resistance of the clips’ contacts, and any zero error in where the fixed clip sits

   0.0877 Ω Right first time

5. Resistance per metre

   4.15 Ω/m Right first time

6. Inside a longer wire

   Mechanism explained on the first choice Right first time

7. A and B in series

   resistance = potential difference ÷ current (V = I R)

   Calculate the resistance of A and B joined in series.

   18.3 Ω Right first time

8. The rule for series

   Your resistances: A = 7.35 Ω, B = 10.3 Ω, A and B in series = 18.3 Ω. Which rule fits?

   R total = R A + R B Right first time

9. A and B in parallel

   resistance = potential difference ÷ current (V = I R)

   Calculate the resistance of A and B joined in parallel.

   4.16 Ω Right first time

10. Resistors in parallel

    Your resistances: A = 7.35 Ω, B = 10.3 Ω, A and B in parallel = 4.16 Ω. What do they show?

    The total resistance of two resistors in parallel is less than the smaller of the two. Right first time

7 Conclusion and evaluation

Conclusion

Sources of error

Improvement

Repeatable or reproducible

9 Exam questions on your data

On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 30/30.

1. 01.1[2 marks]

   A student sets up a circuit to measure the resistance of a length of wire.

   Which two statements about how the meters are connected are correct?

   Tick (✓) two boxes.

   • ✓ The ammeter is connected in series with the wire. (your choice)
   • The ammeter is connected in parallel with the wire.
   • ✓ The voltmeter is connected in parallel with the wire. (your choice)
   • The voltmeter is connected in series with the wire.
   • The voltmeter is connected in series with the power supply.

   Your answer

   The ammeter is connected in series with the wire.; The voltmeter is connected in parallel with the wire.

   2/2

   • The ammeter is connected in series with the wire.
   • The voltmeter is connected in parallel with the wire.

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: The ammeter is connected in series with the wire. The voltmeter is connected in parallel with the wire.

2. 01.2[1 mark]

   Write down the equation that links current (I), potential difference (V) and resistance (R).

   Your answer

   potential difference = current × resistance

   1/1 self-marked

   • potential difference = current × resistance, V = I R (or any correct rearrangement, in words or symbols)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: potential difference = current × resistance V = I R

3. 01.3[2 marks]

   Give one hazard in this investigation and describe how the risk from it can be reduced.

   Your answer

   A high current makes the wire get very hot, which could cause burns. Use a low current and switch off between readings.

   2/2 self-marked

   • Hazard: a high current makes the wire overheat or get hot (which can burn the skin)
   • Use a low current (low pd), or switch off between readings, or do not touch the wire

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: A large current makes the wire very hot, which could burn your skin. Use a low potential difference so the current is low, and switch off between readings.

4. 01.4[2 marks]

   You switched the current off between readings.

   Explain why the current should be switched off between readings.

   Your answer

   The current heats the wire, so switching off keeps the temperature of the wire constant. The resistance of the wire increases as its temperature increases.

   2/2 self-marked

   • The current heats the wire / keeps the temperature of the wire constant (the same)
   • The resistance of the wire changes (increases) as its temperature increases

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: A current heats the wire. Switching off keeps the temperature of the wire constant, because the resistance of the wire increases when its temperature increases.

5. 02.1[2 marks]

   Your first readings at 50 cm were a potential difference of 1.58 V and a current of 0.73 A.

   Calculate the resistance of 50 cm of the wire.

   Your answer

   2.16

   Working: R = V ÷ I R = 1.58 ÷ 0.73 R = 2.16 Ω

   2/2

   • R = 1.58 ÷ 0.73
   • R = 2.16 Ω

   Full marks. Clear and precise.

   Model answer: R = V ÷ I R = 1.58 ÷ 0.73 R = 2.16 Ω

6. 02.2[1 mark]

   Table 1 shows your first readings for each length of wire.

   Which result is anomalous?

   Select one row.

   Table 1

   Length in cm	Potential difference in V	Current in A	Resistance in Ω
   20	1.18	1.31	0.90
   30	1.39	1.01	1.38
   40	1.48	0.67	2.21
   50	1.58	0.73	2.16
   60	1.64	0.64	2.56
   70	1.69	0.57	2.96
   80	1.74	0.51	3.41
   90	1.76	0.46	3.83
   100	1.80	0.42	4.29

   Your answer

   the reading at 40 cm

   1/1

   • The reading at 40 cm: its resistance does not fit the pattern of the others

   Full marks. Your answer meets all the points for this question. At 40 cm the resistance is well above the steady rise of the other lengths.

   Model answer: The reading at 40 cm. Its resistance is much bigger than the pattern of the others suggests.

7. 02.3[1 mark]

   A graph of resistance against length gives a straight line, but it does not pass through the origin. At zero length there is still a small resistance.

   Suggest one reason why.

   Your answer

   There is contact resistance where the crocodile clips touch the wire.

   1/1 self-marked

   • Resistance of the crocodile clips / contact resistance (or the fixed clip not at 0 cm: zero error; or the leads)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: The crocodile clips have some contact resistance where they touch the wire (or the fixed clip was not exactly on the 0 cm mark).

8. 02.4[3 marks]

   Your line of best fit passes through (20 cm, 0.92 Ω) and (100 cm, 4.24 Ω).

   Determine the resistance of each metre of the wire. Use the gradient of the line.

   Your answer

   4.15

   Working: gradient = (4.24 − 0.92) ÷ (100 − 20) gradient = 3.32 ÷ 80 = 0.0415 Ω/cm resistance per metre = 0.0415 × 100 = 4.15 Ω/m

   3/3

   • Change in resistance = 3.32 Ω over 80 cm
   • Gradient = 0.0415 Ω/cm (or 4.15 Ω/m)
   • 4.15 Ω/m (allow their gradient × 100)

   Full marks. Well done.

   Model answer: gradient = (4.24 − 0.92) ÷ (100 − 20) gradient = 3.32 ÷ 80 = 0.0415 Ω/cm resistance per metre = 0.0415 × 100 = 4.15 Ω/m

9. 02.5[2 marks]

   Another student fixed the crocodile clip 1.0 cm past the 0 cm mark, so each length of wire was 1.0 cm longer than the length they recorded.

   Explain the effect this has on the gradient of their graph of resistance against length.

   Your answer

   The gradient is unchanged. The wire is 1.0 cm longer than recorded each time, so every resistance is bigger than it should be by the same amount and the line moves up: it is a zero error.

   2/2 self-marked

   • The gradient is unchanged (the same)
   • For each recorded length the resistance is bigger than it should be by the same amount (the resistance of the extra 1.0 cm), so the line moves up (a zero error / systematic error)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The gradient stays the same. The wire is always 1.0 cm longer than the recorded length, so each resistance is bigger than expected by the same amount (a zero error) and the line just moves up.

10. 02.6[1 mark]

    A student repeated the reading at each length three times. The three resistances at each length were very close to each other.

    Which word describes these results?

    Tick (✓) one box.

    • Accurate
    • ✓ Precise (your choice)
    • Valid
    • Reproducible

    Your answer

    Precise

    1/1

    • Precise

    Full marks. Your answer meets all the points for this question.

    Model answer: Precise: the repeat results show little variation (spread). They could still all be wrong by the same amount, so they are not necessarily accurate.

11. 03.1[3 marks]

    A 15 Ω resistor and a 33 Ω resistor are connected in series to a 6.0 V supply.

    Calculate the current in the circuit.

    Your answer

    0.125

    Working: R total = 15 + 33 = 48 Ω I = V ÷ R = 6 ÷ 48 I = 0.125 A

    3/3

    • Total resistance = 15 + 33 = 48 Ω
    • I = 6 ÷ 48 (allow their total resistance)
    • I = 0.125 A

    Full marks. That is exactly what the mark scheme is looking for.

    Model answer: R total = 15 + 33 = 48 Ω I = V ÷ R = 6 ÷ 48 I = 0.125 A

12. 03.2[2 marks]

    In the same series circuit, the potential difference across the 33 Ω resistor is bigger than the potential difference across the 15 Ω resistor.

    Explain why.

    Your answer

    The current is the same through both resistors. The potential difference is shared in proportion to the resistance, so the 33 Ω resistor has a bigger potential difference.

    2/2 self-marked

    • The current is the same through both resistors (series)
    • So V = I R is bigger across the bigger resistance / the pd is shared in proportion to the resistance

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: The current is the same through both resistors because they are in series. V = I R, so the 33 Ω resistor, with the bigger resistance, has the bigger potential difference: the pd is shared in proportion to the resistance.

13. 03.3[2 marks]

    A second resistor is connected in parallel with a resistor in a circuit.

    Explain why the total resistance of the circuit decreases.

    Your answer

    The current has two paths, so the total current is bigger for the same potential difference, and the total resistance decreases.

    2/2 self-marked

    • The current has more than one path (another path / branch)
    • So the total current is bigger for the same potential difference (R = V ÷ I is smaller)

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: The current has another path to flow through, so the total current from the supply increases for the same potential difference. Since R = V ÷ I, the total resistance is smaller.

14. 04[6 marks]

    A student wants to investigate how the resistance of a wire depends on its length. The wire is taped to a metre ruler.

    Describe a method the student could use. Your answer should include a hazard and how to reduce its risk.

    Your answer

    1. Connect the wire in series with an ammeter and a power supply, with a voltmeter in parallel across the wire. 2. Fix one crocodile clip at the 0 cm end of the wire, which is taped to a metre ruler. 3. Put the other crocodile clip at 20 cm and measure the length between the clips with the ruler. 4. Switch on and record the current from the ammeter and the potential difference from the voltmeter. 5. Calculate the resistance using R = V ÷ I. 6. Move the clip to change the length in 10 cm steps up to 100 cm and repeat. 7. Repeat the readings for each length and calculate a mean resistance. 8. Plot a graph of resistance against length. Hazard: a high current can make the wire overheat and cause burns, so use a low current and switch off between readings.

    6/6 self-marked

    • Ammeter in series with the wire; voltmeter in parallel (across the wire)
    • Read the length of wire between the two clips on the metre ruler
    • Length varied by moving a crocodile clip
    • Current measured with the ammeter
    • Potential difference measured with the voltmeter
    • Calculate the resistance for each length using V = I R (R = V ÷ I)
    • Take each length more than once and average the readings (or the resistances)
    • Plot a graph of resistance against length
    • Hazard: high current makes the wire overheat (melt) and could cause burns; use low currents (switch off between readings)

    Self-marked: Level 3, 6 of 6 marks. You ticked 9 of 9 indicative points. This is your own judgement against the mark scheme.

    Model answer: 1. Connect the wire in series with an ammeter and a power supply, with a voltmeter in parallel across the wire. 2. Fix one crocodile clip at the 0 cm end of the wire, which is taped to a metre ruler. 3. Put the other crocodile clip at 20 cm and measure the length between the clips with the ruler. 4. Switch on and record the current from the ammeter and the potential difference from the voltmeter. 5. Calculate the resistance using R = V ÷ I. 6. Move the clip to change the length in 10 cm steps up to 100 cm and repeat. 7. Repeat the readings for each length and calculate a mean resistance. 8. Plot a graph of resistance against length. Hazard: a high current can make the wire overheat and cause burns, so use a low current and switch off between readings.

Unseen data: the same skills, new apparatus

1. 01[1 mark]

   A student hangs masses on a spring and measures the extension of the spring for each force. Figure 1 shows the results and the student's line of best fit.

   The student says: “The extension of the spring is directly proportional to the force.”

   Is the student right?

   Tick (✓) one box.

   • Yes: the graph is a straight line through the origin.
   • No: the graph is a straight line, but it does not pass through the origin.
   • ✓ Only for the smaller forces: for the largest forces the graph curves. (your choice)
   • Yes: as the force increases, the other quantity increases.

   Your answer

   Only for the smaller forces: for the largest forces the graph curves.

   1/1

   • Only for the smaller forces: for the largest forces the graph curves.

   Full marks. Your answer meets all the points for this question.

   Model answer: Only for the smaller forces: for the largest forces the graph curves. The line is straight through the origin up to about 4 N, then curves upwards: beyond the limit of proportionality the extension is no longer proportional to the force.

2. 02.1[3 marks]

   A student draws a thick line on paper with a soft pencil. She measures the resistance of the graphite line between a fixed clip at one end and a metal probe pressed on the line at different distances from the clip, using an ohmmeter.

   Figure 2 shows her results and her line of best fit.

   Use Figure 2 to determine the gradient of the line.

   Give the unit.

   Your answer

   0.95 kΩ/cm

   Working: (9.84 - 2.24) / (10 - 2)

   3/3

   • Change in resistance ÷ change in length, from two points far apart on the line
   • Gradient = 0.95 (allow 0.9 to 1)
   • Unit: kΩ/cm (allow Ω/cm with the value in ohms)

   Full marks. Well done.

   Model answer: From the line: 9.8 kΩ at 10 cm and 2.2 kΩ at 2 cm. Gradient = (9.8 − 2.2) ÷ (10 − 2) = 0.95 kΩ/cm

3. 02.2[2 marks]

   On another day the student draws a new line and repeats the experiment. Every resistance is about 21 % lower than before.

   Which two variables were most likely not controlled?

   Tick (✓) two boxes.

   • The resolution of the ohmmeter.
   • ✓ How many times the line was drawn over (how much graphite is on the paper). (your choice)
   • ✓ The width of the pencil line. (your choice)
   • The length of line between the clip and the probe.
   • The colour of the paper.

   Your answer

   The width of the pencil line.; How many times the line was drawn over (how much graphite is on the paper).

   2/2

   • The width of the pencil line.
   • How many times the line was drawn over (how much graphite is on the paper).

   Full marks. Well done.

   Model answer: The width of the line, and how many times it was drawn over. Both change the cross-section of graphite the current passes through, so they change the resistance of every length: a wider or thicker line has less resistance. The length is the independent variable, and the ohmmeter’s resolution cannot shift every reading by 21 %.