AQA AS Level Physics Paper 1, June 2022: Question 8
14 marks · Medium difficulty · Practical Techniques & Data Analysis
Calculate electrical resistance, current, power, and discuss circuit configurations involving a variable resistor, lamp, and potential divider.
Practise this questionQuestion
Question text
08 Figure 11 shows a variable resistor that has a maximum resistance of 25 Ω.
A sliding contact P is mounted on a thick copper bar. P can be set to any position
between X and Y.
Figure 11
08.1 Figure 12 shows the variable resistor being used to investigate the variation of current
with voltage for a filament lamp.
The normal operating voltage of the lamp is 12 V.
The 12 V battery has negligible internal resistance.
Figure 12
The position of P is adjusted so that the reading on the voltmeter is at its minimum
value of 0.75 V.
Calculate the resistance of the lamp when the voltmeter reading is 0.75 V.
[2 marks]
21 resistance = Ω
08.2 Figure 13 shows the variation of current with voltage for the lamp between 2 V
and 12 V.
Figure 13
Calculate the resistance of the lamp when the voltage across the lamp is 8.0 V.
[2 marks]
resistance = Ω
08.3 Explain, in terms of electron movement, why the resistance of the filament lamp
changes as the voltage changes as shown in Figure 13.
[3 marks]
08.4 Figure 14 shows an alternative circuit used to investigate the variation of current with
voltage for the lamp.
Figure 14
The circuit components are the same as in Figure 12.
When the voltage across the lamp is 12 V its resistance is 6.0 Ω.
P is moved to position Y.
Calculate the total resistance of the circuit.
[2 marks]
total resistance = Ω
08.5 Calculate the power transferred by the battery when P is at position Y.
[2 marks]
23 power = W
08.6 A student wants to control the brightness of the lamp.
*22* He gives two reasons why the circuit in Figure 14 is better than the circuit in
Figure 12 for controlling the brightness. The two reasons are:
• the Figure 14 circuit can achieve a greater range of voltages across the lamp
• the Figure 14 circuit is more efficient at transferring energy to the lamp.
Discuss, without calculation, whether either of these two reasons is correct.
[3 marks]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
08.1 (pd across the variable resistor) = 11.25 (V) seen For Max 1: 2 1 x AO1
OR 1 x AO2
R1 Condone mix up of R1and R2
Use of V0 = ×Vin
R1 + R2
OR
𝑉𝑉 𝑅𝑅 Condone V=12 V and R=25 Ω leading to an
use of V1: V2 = R1 : R2 or 1 = 1
answer of 1.56 Ω or 1.6 Ω
𝑉𝑉2 𝑅𝑅2
OR
Condone V=12 V and R=25 Ω leading to
(I =) 0.45 A I= 0.48 A and an answer of 1.56 Ω or 1.6 Ω
Accept a correctly rounded answer to 2 or
(R =) 1.7 (Ω) c.a.o more significant figure.
(Calculator displays 1.66666666)
Within ½ square of accuracy (1.65 A to 1.75 A)
08.2 Clear read-off seen on graph of I= 1.7 A 2
Or
use of V = IR Accept = 4.8(4) or = 4.5(7) in
1.65 1.75
use of V=IR
(condoning ‘read-off’ to within 1 square of
accuracy)
Don’t need to see read-off for use of V=IR 1 x AO1
8 × 1.7 = 13.6 would be insufficient as use of 1 x AO2
and in this case MP1 can only be scored
where read-off is seen.
Allow answer in range 4.57 to 4.85
(R =) 4.7 (Ω)
Do not accept 1 significant figure in answer
08.3 As voltage increases the current increases / as the voltage 3 3 x AO1
increases more electrons move through the wire (per
second)
More collisions (per second) between the (conduction)
electrons and the lattice ions / Vibration of the lattice ions Allow vibration of the ions in the filament / wire
increases / metal increases
Accept rate of collisions for number of
(Rate of) vibration of the lattice ions increases causing a collisions per second.
greater number of collisions per second causing increased Talk out on MP3 where current decreases
resistance
08.4 1 1 2 1 x AO1
11 1 𝑅𝑅1 ×𝑅𝑅2 allow use of + seen without subject
use of = + or 𝑅𝑅 = 6 25
RT R1 R2 𝑇𝑇 1 x AO2
𝑅𝑅1+ 𝑅𝑅2 Alternative MP1:
(IT = Ilamp + IXY= 2.48 A) and use of V = IR
Accept a correctly rounded answer to 2 or
more significant figure.
(R =) 4.8(4) (Ω) 29
(Calculator displays 4.838709677)
V 2
08.5 use of P = by substitution of V = 12 V and R= 4.8 Ω Ecf from 08.4 for MP1 and MP2 2 1 x AO1
R MP1:
1 x AO2
Condone use of R = 6 Ω or R = 25 Ω in this
substitution for MP1 (where not ecf from 08.4)
OR
Condone use of P=IV or use of P=I2R by
substitution of their (battery) I and ecf R from
08.4. Must have clearly identified I in working in
08.4 or by use of I = 12 here
𝑒𝑒𝑒𝑒𝑒𝑒 𝑅𝑅
30 Ecf answer must be
𝑅𝑅 𝑜𝑜𝑛𝑛 𝑎𝑎𝑛𝑛𝑎𝑎𝑎𝑎𝑒𝑒𝑟𝑟 𝑙𝑙𝑖𝑖𝑛𝑛𝑒𝑒 𝑖𝑖𝑛𝑛 08.4
(P =) 30 (W)
(Calculator display for non-rounded answer
29.76)
Penalise answers with more than two digits
that have been rounded to 1 significant
figure.
08.6 Wider range in Figure 14’s circuit and lower efficiency in 3 3 x AO3
Figure 14’s circuit
Details: Condone referring to Figure 12 as Figure 13. 31
Voltage range is wider 0–12 V (in Figure 14’s circuit) Allow ‘can get zero volts in Figure 14’
compared to 0.75 V – 12 V (in Figure 12’s circuit) / can’t get
voltages between 0 and 0.75 V In Figure 12 / wider range
when using XY as a potentiometer
OR
bulb won’t light at lower voltages, so control is unaffected
At any particular voltage across lamp more power dissipated
Current splits in Figure 14 is insufficient
in circuit in Figure 14 / any voltage across the lamp there is
always 12 V across the resistor in Figure 14’s circuit which
produces more heating (whereas only the remaining portion of
12 V is across the resistor in Figure 12’s circuit) / for any
current in the lamp there is always more current in Figure 14’s
circuit which produces more heating
Total 14
How to answer it
Analysis of Potential Divider Circuits & Filament Lamps
What this question tests
This question assesses your mastery of direct current (DC) circuits, specifically potential dividers, potential drop calculations, graphical analysis of non-ohmic components (filament lamps), microscopic explanations of electrical resistance, parallel resistor combinations, electrical power, and circuit efficiency evaluations.
Question 08.1
✅ Correct Answer
1.7 Ohm (Acceptable range: 1.66 to 1.76 Ohm depending on working steps)
💡 Key Knowledge
In Figure 12, the variable resistor is connected as a potential divider (potentiometer). The total resistance of the coil is 25 Ohm. When the voltmeter reads its minimum value (0.75 V) across the lamp, contact P is at terminal X, placing the entire 25 Ohm resistor across the supply in series with the lamp.
📐 Step-by-Step Calculation
- Find p.d. across the variable resistor: 12 V - 0.75 V = 11.25 V
- Calculate circuit current using variable resistor values ( R = 25 Ohm ): I = 11.25 / 25 = 0.45 A
- Calculate lamp resistance: R = V / I = 0.75 / 0.45 = 1.666... Ohm -> round to 1.7 Ohm .
❌ Common Errors
Students often incorrectly assume the lamp voltage is the full supply voltage, or confuse series/parallel rules when combining component resistances.
Question 08.2
✅ Correct Answer
4.7 Ohm (Acceptable range: 4.57 to 4.85 Ohm)
🧠 Exam Technique
When using an IV graph for a non-ohmic component, always read values straight off the grid with high precision. For V = 8.0 V , find the corresponding current I from Figure 13 (around 1.7 A ) before applying R = V / I .
❌ Common Errors
Giving answers to only 1 significant figure (e.g., 5 Ohm) is penalised. Read-offs must be within half a square of accuracy.
Question 08.3
💡 Key Knowledge (Microscopic Model)
As voltage increases, current increases, meaning more electrons pass per second. Greater current/temperature increases the rate of vibration of metal lattice ions, causing more frequent collisions between conduction electrons and lattice ions, thus increasing resistance.
🧠 Exam Technique
To secure all 3 marks, structure your answer chronologically: 1. Higher voltage -> higher current -> more electrons moving per second. 2. Increased thermal energy -> ions vibrate with greater amplitude/frequency. 3. Increased collision rate -> impedes charge flow more -> resistance rises.
Question 08.4
✅ Correct Answer
4.8 Ohm (Exact calculator value: 4.838... Ohm)
📐 Step-by-Step Calculation
- Identify component values: Lamp resistance = 6.0 Ohm , Variable resistor maximum = 25 Ohm (connected in parallel in Figure 14).
- Apply parallel resistance formula: 1 / R_T = (1 / 6.0) + (1 / 25)
- R_T = (6.0 × 25) / (6.0 + 25) = 150 / 31 = 4.84 Ohm
Question 08.5
✅ Correct Answer
30 W (Allow ECF from 08.4)
📐 Step-by-Step Calculation
- Power equation: P = V² / R_T or P = I × V
- Substitute values: P = 12² / 4.84 = 144 / 4.838... = 29.76 W
- Round appropriately to 30 W (penalise 1 s.f. like 30 W if derived poorly, but 30 is accepted from 2.97×10¹ style formatting).
Question 08.6
✅ Correct Answer
Figure 14 circuit does have a wider voltage range (0–12 V vs restricted range in Figure 12), but lower efficiency because power is wasted across the fixed parallel network paths/resistors.
💡 Key Knowledge
Figure 12 acts as a potential divider where voltage across the lamp cannot reach 0 V because part of the resistance wire is always in series with it. Figure 14 connects components in parallel across the 12 V supply, allowing the full 0–12 V range across the lamp branch, but draws higher total current from the battery, reducing overall system efficiency.
❌ Common Errors
Vague statements like "Figure 14 is better" without referencing specific voltage limits or energy dissipation/efficiency principles fail to gain credit.
Topics
Physics · 3.5 Electricity
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.