AQA AS Level Physics Paper 1, November 2020: Question 5
14 marks · Medium difficulty · Short Answer
Calculate power, energy, and analyze a circuit containing cells with internal resistance, a resistor, and a lamp.
Practise this questionQuestion
Question text
05 A cell has an emf of 1.5 V and an internal resistance of 0.65 Ω.
The cell is connected to a resistor R.
05.1 State what is meant by an emf of 1.5 V.
[2 marks]
05.2 The current in the circuit is 0.31 A.
Show that the total power output of the cell is approximately 0.47 W.
[1 mark]
05.3 Calculate the energy dissipated per second in resistor R.
[2 marks]
energy dissipated per second16 = J s−1
05.4 The cell stores 14 kJ of energy when it is fully charged. The cell’s emf and internal
resistance are constant as the cell is discharged.
Calculate the maximum time during which the fully-charged cell can deliver energy to
resistor R.
[2 marks]
17maximum time = s
05.5 A student uses two cells, each of emf 1.5 V and internal resistance 0.65 Ω, to operate
a lamp. The circuit is shown in Figure 7.
Figure 7
The lamp is rated at 1.3 V, 0.80 W.
Deduce whether this circuit provides the lamp with 0.80 W of power at a potential
difference (pd) of 1.3 V.
Assume that the resistance of the lamp is constant.
[4 marks]
05.6 The lamp operates at normal brightness across a pd range of 1.3 V to 1.5 V.
State and explain how more of these cells can be added to the circuit to make the
lamp light at normal brightness for a longer time.
No further calculations are required.
[3 marks]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark
details
2nd marking point obtains both marks
Work done in moving 1 C of charge through the cell
1.5 J of work is done in moving 1 C of charge through the cell
OR Max 1 mark available for the following:
Amount of energy converted from other forms to electrical AO1.1a
The emf is the terminal pd when there is no
energy per 1 C of charge
current in the cell (and this equals 1.5 V)
AO2.1f
05.1 2
1.5 J of energy converted from other forms to electrical energy
1.5 J of energy per 1 C of charge.
per unit charge (passing across the emf)
Allow a statement of Kirchhoff’s 2nd law for 1
OR mark. Where the law is in symbol form, the
meaning of the symbols must be stated.
Work done in moving 1 C of charge (whole way) round circuit Need a clear communication of internal and
external resistances.
1.5 J of work is done in moving 1 C of charge the (whole way)
round circuit
P= VI
05.2 And 1 AO2.1b
Seen to more than 2 sf with supporting
(P) = 0.465 (W) – SICS – – JUNE 2020
equation with subject seen in working
Alternative for 1 mark:
𝜀
Use of I = 𝑅+𝑟
Or
pd across R = 1.5 – 0.65 x 0.31
Use of appropriate power equation to determine wasted power or
or pd across R = 1.2985 (V)
or
power dissipated in R = total power – their wasted power total resistance = 1.5/ 0.31
or
total resistance = 4.839 (Ω)
AO1.1a
05.3 or R = 4.2 (Ω) 2
AO2.1b
or P = I2 x their R
or
𝑉2
P = 𝑅 using their V and R
(P =) 0.40 W
7407_1 Version post-stand
ID
details
Use of E = P t AO1.1a
Allow use of the equation with their values.
20 or E = VI t
An answer of 3.5 x 104 is worth 1 mark
Or
05.4 E= QV and Q=It 2
4 AO2.1b
(t =) 3.0(1) x 10 (s) – SICS – – JUNE 2020
ID
details
MAX 3 from (1 to 4) or (5 to 8)
It is suitable, because:
𝑃
(1) Current required in lamp = 0.62 A or use of I= seen
𝑉
𝑉2
(2) Resistance of lamp = 2.11 Ω or use of R= 𝑃 𝑠𝑒𝑒𝑛
(3) current in each cell = 0.31 A
(4) lost volts = 0.2 V
or Check Figure 7
lost volts = 0.65 x 0.31 Must have the correct conclusion to award 4
marks.
AO3.1a
05.5 Conclusion: yes, terminal pd = 1.5 – 0.2 seen 4 AO3.1a
or AO3.1a
terminal pd= 1.5 – 0.65 x 0.4 /1.3 AO3.1b21
Allow max 3 from a combination of two route
OR [(2) and (7) worth total of 1 mark]
(5) total internal resistance = 0.325 Ω
(6) total resistance in circuit = 2.44 Ω
(7) Resistance of lamp = 2.11 Ω
(8) pd splits in ratio of 0.325:2.11
𝟐.𝟏𝟏 ×𝟏.𝟓
Conclusion: yes, pd across lamp is 𝟐.𝟒𝟒 (= 1.3 V) seen
7407_1 Version post-stand
Question Answers Addi
Must link the cells being added in parallel to
one or both reason to gain three marks.
(Cells must be added) in parallel
Alternative:
Because: • In parallel
more energy stored in the bank of cells / less power from • Current shared by cells
each cell • Takes longer to convert the energy
AO3.1a
stored in each cell.
5.6 3 AO3.1a
Alternative:
without increasing the voltage across the bulb (above 1.5 V) AO2.1g
• In parallel
or • Less internal resistance
• Less power / energy wasted
without increasing the terminal pd (above 1.5V)
Cells in series statement means no marks can
be obtained.
Total 14
How to answer it
Electromotive Force and Internal Resistance Study Guide
What this question tests
This question assesses your understanding of electromotive force (emf), internal resistance, terminal potential difference, power calculations in DC circuits, energy conservation, and the effects of combining cells in series and parallel.
Defining Electromotive Force (EMF)
✅ Correct Answer
Work done in moving 1 C of charge through the cell (or round the whole circuit), stating that 1.5 J of energy is converted from other forms to electrical energy per 1 C of charge.
💡 Key Knowledge
EMF is defined as the energy converted from other forms (chemical, mechanical, etc.) into electrical energy per unit charge. Always mention both the energy conversion concept and the per-charge (per coulomb) aspect to secure full marks.
Show That: Total Power Output
📐 Calculation Steps
- Identify relevant formula: P = E × I (Power = emf × current)
- Substitute values: P = 1.5 V × 0.31 A
- Evaluate: 0.465 W , which rounds to approximately 0.47 W .
❌ Common Errors
Writing down 0.47 W directly without showing the unrounded intermediate value (0.465 W) or missing the supporting equation will lose you the mark on a "show that" question.
Energy Dissipated per Second in Resistor R
✅ Correct Answer
0.40 W (or J s⁻¹)
📐 Step-by-Step Calculation
- Find power wasted in the internal resistance: P = I² × r = 0.31² × 0.65 = 0.0624 W
- Subtract wasted power from total power: 0.465 W - 0.0624 W = 0.4026 W
- Alternatively, find terminal pd: V = E - Ir = 1.5 - (0.31 × 0.65) = 1.2985 V , then use P = V² / R or P = I²R .
Calculating Maximum Time
📐 Step-by-Step Calculation
- Recall energy-power relationship: Energy = Power × time ( E = P t )
- Rearrange for time: t = Energy / Power
- Convert energy to Joules: 14 kJ = 14,000 J
- Calculate: t = 14000 / 0.465 = 30107 s
- Give to appropriate sig figs: 3.0 × 10⁴ s (or 3.01 × 10⁴ s).
🧠 Exam Technique
Always use the total power output of the cell (0.465 W) rather than just the power delivered to resistor R, because the total stored energy depletes based on the total power generated by the cell.
Deducing Circuit Suitability for a Lamp
💡 Key Knowledge & Analysis
Figure 7 shows a parallel combination of two identical cells connected to a lamp. For parallel cells of emf E and internal resistance r:
- Combined emf remains 1.5 V .
- Combined internal resistance halves: r_total = 0.65 / 2 = 0.325 Ω .
- Lamp resistance: R = V² / P = 1.3² / 0.80 = 2.11 Ω .
- Circuit current: I = E / (R + r_total) = 1.5 / (2.11 + 0.325) = 0.615 A .
- Terminal pd across lamp: V = E - Ir_total = 1.5 - (0.615 × 0.325) = 1.30 V .
✅ Conclusion
Yes, it is suitable. The terminal pd provided matches the required 1.3 V specification of the lamp, delivering 0.80 W of power.
Adding Cells for Longer Operating Time
✅ Correct Answer
Add more cells in parallel.
💡 Explanation
- Connecting cells in parallel increases the total energy stored in the cell bank without increasing the overall emf (voltage).
- The current drawn is shared between the cells, reducing the load/power drained from each individual cell.
- This keeps the terminal pd within the acceptable operating range (1.3 V to 1.5 V) while extending battery life.
Topics
Physics · 3.5 Electricity
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.