Electromotive force and internal resistance: AQA A‑Level Physics questions

AQA A-Level Physics past-paper questions on 3.5.1.6 Electromotive force and internal resistance, 2008–2025, with mark schemes and examiner’s reports where available.

All 99 questions on Electromotive force and internal resistance, as a list

June 2025, Paper 1, Q05.1 (1 mark): Calculate the terminal potential difference of the cell at its maximum current. [1 mark] terminal potential difference = V
June 2025, Paper 1, Q05.2 (2 marks): Figure 6 shows an array of 20 of these cells connected to form a module. Figure 6 Calculate the internal resistance of the module shown in Figure 6. [2 marks] i
June 2025, Paper 1, Q05.3 (3 marks): Figure 7 shows the battery for an electric vehicle. This battery has an emf of 352 V and gives a maximum current of 500 A. The battery consists of a series of i
June 2025, AS Paper 2, Q03.1 (2 marks): For this circuit, it can be shown that: 12I = 0.037I 2 + 130 Explain, using conservation of energy, why this equation is valid. [2 marks]
June 2025, AS Paper 2, Q03.2 (2 marks): Show that I is approximately 11 A. [2 marks]
June 2025, AS Paper 2, Q03.3 (2 marks): Determine the resistance of X when it operates at a power of 65 W. [2 marks] resistance of X = Ω
June 2024, Paper 1, Q27 (1 mark): A 12 Ω resistor is connected across the terminals of a battery of emf 2.0 V and internal resistance 4.0 Ω. What is the pd across the resistor? [1 mark] A 0.25 V
June 2024, AS Paper 2, Q33 (1 mark): A student investigates the characteristics of a power supply. The experimental data are plotted on a graph with: • pd (potential difference) across the power su
June 2023, Paper 1, Q03.1 (2 marks): In Figure 2 the cell has emf ε and internal resistance r. Figure 2 The current in the circuit is I. The potential difference (pd) across R1 is V1 and the pd acr
June 2023, Paper 1, Q03.2 (3 marks): The sliding contact C is moved to end B of the variable resistor. The switch is closed. The digital voltmeter reads 2.89 V. Show that r is approximately 4.8 Ω.
June 2023, AS Paper 1, Q06.1 (1 mark): State what is meant by the emf (electromotive force) of a battery. [1 mark]
June 2023, AS Paper 1, Q06.4 (4 marks): Determine the resistance of R in Figure 10. [4 marks] resistance = Ω
June 2023, AS Paper 1, Q06.5 (3 marks): Another appliance is connected to the battery as shown in Figure 12. The current in the battery is 3.5 A when switch S is closed. Figure 12 Each LED requires a
June 2022, Paper 1, Q04.2 (3 marks): The terminal pd across the battery is 6.2 V. Calculate the emf of the battery. [3 marks] emf = V
June 2022, Paper 1, Q04.5 (2 marks): The student now makes a different circuit by connecting the variable resistor in parallel with the lamp. The contact is returned to its original position on the
June 2022, AS Paper 2, Q31 (1 mark): A 12 Ω resistor is connected across the terminals of a cell that has an emf of 2.0 V and an internal resistance of 4.0 Ω. What is the terminal pd? [1 mark] A 0.
June 2022, AS Paper 2, Q34 (1 mark): A practical power supply provides a steady current I for a time t to an external circuit. The emf of the power supply during t is equivalent to [1 mark] A the e
November 2021, Paper 1, Q28 (1 mark): Two identical batteries each of emf 1.5 V and internal resistance 1.6 Ω are connected in parallel. A 2.4 Ω resistor is connected in parallel with this combinati
November 2021, AS Paper 1, Q06.1 (2 marks): State what is meant by an emf of 5.30 V for this battery. [2 marks]
November 2021, AS Paper 2, Q20 (1 mark): In the circuit shown, the cell has an emf of 12 V and an internal resistance which is not negligible. When the resistance of the variable resistor is 10 Ω the v
November 2020, Paper 1, Q28 (1 mark): A resistor of resistance R and three identical cells of emf E and internal resistance r are connected as shown. What is the current in the resistor? [1 mark] A
November 2020, Paper 1, Q29 (1 mark): In the circuit, the reading of the voltmeter is V. When the switch is closed the reading becomes 3 V . What is the internal resistance of the cell? [1 mark] A 0
November 2020, Paper 3A, Q03.1 (1 mark): Complete Figure 10 to show a circuit including a voltmeter and an ammeter that is suitable for the investigation. [1 mark]
November 2020, Paper 3A, Q03.2 (4 marks): Describe • a procedure to obtain valid experimental data using your circuit • how these data are processed to obtain ε and r by a graphical method. [4 marks]
November 2020, Paper 3A, Q03.3 (2 marks): Show that ε is about 1.6 V. [2 marks]
November 2020, Paper 3A, Q03.4 (1 mark): Figure 13 shows the circuit when four resistors are connected. Figure 13 Show, using Figure 12, that the current in the power supply is about 0.25 A. [1 mark]
November 2020, Paper 3A, Q03.5 (4 marks): Deduce, for the circuit shown in Figure 13, • the potential difference (pd) across the power supply • r. [4 marks] pd = V r = Ω
November 2020, Paper 3A, Q03.6 (3 marks): Figure 14 shows the plots for n = 1 and n = 14 Figure 14 Three additional data sets for values of n between n = 1 and n = 14 are needed to complete the graph in
November 2020, Paper 3A, Q03.7 (2 marks): The experiment is repeated using a set of resistors of resistance 27 Ω. The relationship between n and I is now 27 r n I = − Show on Figure 14 the effect on the
November 2020, AS Paper 1, Q05.1 (2 marks): State what is meant by an emf of 1.5 V. [2 marks]
November 2020, AS Paper 1, Q05.2 (1 mark): 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]
November 2020, AS Paper 1, Q05.3 (2 marks): Calculate the energy dissipated per second in resistor R. [2 marks] energy dissipated per second = J s−1
November 2020, AS Paper 1, Q05.4 (2 marks): 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 t
November 2020, AS Paper 1, Q05.5 (4 marks): 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
November 2020, AS Paper 1, Q05.6 (3 marks): 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 la
November 2020, AS Paper 2, Q02.1 (2 marks): Explain why the student included the fixed resistor in this circuit. [2 marks]
November 2020, AS Paper 2, Q02.2 (3 marks): Determine the magnitude of the minimum gradient Gmin of a line that passes through all the error bars in Figure 5. [3 marks] magnitude of Gmin =
November 2020, AS Paper 2, Q02.3 (2 marks): The maximum gradient Gmax / V A−1 of a line passing through all the error bars in Figure 5 is –1.3 Determine, using Gmax and Gmin, the internal resistance of th
November 2020, AS Paper 2, Q02.4 (3 marks): The line of best fit passes through the data point (0.94, 0.37). Determine the emf of the cell. [3 marks] emf = V END OF SECTION A Section B Answer all question
June 2019, Paper 1, Q07.1 (2 marks): Explain why the reading on the voltmeter decreases as the brightness of the lamp increases. [2 marks]
June 2019, Paper 1, Q07.2 (2 marks): The variable resistor is adjusted so that the lamp is at its brightest. The reading V1 on the voltmeter is noted. A second identical cell is then connected in p
June 2019, AS Paper 2, Q30 (1 mark): Three identical cells, each of emf 1.5 V and internal resistance 6.0 Ω, are connected to resistor R. The resistance of R is 6.0 Ω. What is the current in R? [1
June 2019, AS Paper 2, Q33 (1 mark): The reading on the voltmeter halves when switch S is closed. What is the internal resistance of the cell? [1 mark] A 0.50 Ω B 1.0 Ω C 2.0 Ω D 4.0 Ω
June 2017, Paper 1, Q04.1 (2 marks): Explain how the engineer uses the graph to obtain the values for the emf and internal resistance of the solar cell. [2 marks]
June 2017, Paper 1, Q04.2 (4 marks): Deduce, using calculations, whether the circuits in Figure 5 and Figure 6 are suitable for this application. [4 marks]
June 2017, AS Paper 2, Q01.3 (1 mark): With switch S closed (in the on position) and no resistors connected between X and Y the voltmeter reading V is 1.62 V. The student concludes that this voltmete
June 2017, AS Paper 2, Q01.4 (1 mark): Figure 2 shows one particular combination and arrangement of two resistors that the student could use. Figure 2 When S is closed the voltmeter reading V is 1.14
June 2017, AS Paper 2, Q01.5 (1 mark): It can be shown that R V r V × = − where r is the internal resistance of the power supply. Determine (ε – V ) and R V for this circuit using the data given in q
June 2017, AS Paper 2, Q01.6 (1 mark): The student obtains values of V for five further different values of R. These data were used to produce the graph of (ε – V ) against R V in Figure 3. Plot the
June 2017, AS Paper 2, Q01.7 (2 marks): Use Figure 3 to determine r. [2 marks] r = Ω Figure 3
June 2017, AS Paper 2, Q07 (1 mark): Three cells each have an emf ε = 1.5 V and an internal resistance r = 0.6 Ω. Which combination of these cells will deliver a total emf of 1.5 V and a maximum cu
June 2016, AS Paper 2, Q33 (1 mark): A student investigates how the potential difference V across the terminals of a cell varies with the current I in the cell. Which graph correctly shows how V va
June 2016, AS Paper 2, Q34 (1 mark): A battery is connected to a 10 Ω resistor and a switch in series. A voltmeter is connected across the battery. When the switch is open (off) the voltmeter reads
Specimen, Paper 1, Q16 (1 mark): In the circuit shown, V is a voltmeter with a very high resistance. The internal resistance of the cell, r, is equal to the external resistance in the circuit.
June 2016, PHYA1 (old spec), Q7(b)(i) (1 mark): State the electrical property of the battery represented by the deflection PR. [1 mark] ________________________________________________________________________
June 2016, PHYA1 (old spec), Q7(b)(ii) (1 mark): Determine the value of the electrical quantity represented by the deflection PR. [1 mark] electrical quantity = _____________________________
June 2016, PHYA1 (old spec), Q7(c) (3 marks): With switch S1 kept closed, switch S2 is also closed. The spot moves to Q. Explain why the spot moves from R to Q. [3 marks] ___________________________________
June 2016, PHYA1 (old spec), Q7(d) (2 marks): Calculate the current in the battery when both switches are closed. [2 marks] current = ___________________________ A
June 2016, PHYA1 (old spec), Q7(e) (2 marks): Calculate the internal resistance of the battery. [2 marks] internal resistance = ___________________________ Ω
June 2015, PHYA1 (old spec), Q6(a) (2 marks): State what is meant by electromotive force. [2 marks]
June 2015, PHYA1 (old spec), Q6(b)(i) (1 mark): Suggest what was done to obtain the data for the plotted points. [1 mark]
June 2015, PHYA1 (old spec), Q6(b)(ii) (3 marks): The electromotive force (emf) of the potato cell is 0.89 V. Explain why the voltages plotted on Figure 4 are always less than this and why the difference betwee
June 2015, PHYA1 (old spec), Q6(b)(iii) (3 marks): Use Figure 4 to determine the internal resistance of the potato cell. [3 marks] internal resistance = Ω
June 2015, PHYA1 (old spec), Q6(c) (2 marks): A student decides to use two potato cells in series as a power supply for a light emitting diode (LED). In order for the LED to work as required, it needs a vol
June 2014, PHYA1 (old spec), Q5(a) (2 marks): Describe how the power dissipated in Y varies as its resistance is increased from 0.5 Ω to 6.5 Ω. [2 marks]
June 2014, PHYA1 (old spec), Q5(b)(i) (3 marks): Use data from Figure 2 to calculate the current through the battery. [3 marks] current A
June 2014, PHYA1 (old spec), Q5(b)(ii) (2 marks): Calculate the voltage across Y. [2 marks] voltage V
June 2014, PHYA1 (old spec), Q5(b)(iii) (2 marks): Calculate the internal resistance of the battery. [2 marks] internal resistance Ω
June 2014, PHYA1 (old spec), Q5(c) (3 marks): The student repeats the experiment with a battery of the same emf but negligible internal resistance. State and explain how you would now expect the power dissi
June 2013, PHYA1 (old spec), Q6(a)(i) (1 mark): Show that the potential difference (pd) across the internal resistance is 6.3V. (1 mark)
June 2013, PHYA1 (old spec), Q6(a)(vi) (2 marks): Calculate the total resistance of the circuit. circuit resistance Ω (2 marks)
June 2013, PHYA1 (old spec), Q6(b)(i) (3 marks): Using appropriate data values that you have calculated, complete the following table by calculating the rate of energy dissipation in each resistor. (3 marks) r
June 2013, PHYA1 (old spec), Q6(b)(ii) (2 marks): Hence show that energy is conserved in the circuit. (2 marks)
January 2013, PHYA1 (old spec), Q7(c) (2 marks): The lamp used in part (b) is connected by the same two wires to a power supply of the same emf but whose internal resistance is not negligible. State and explai
June 2012, PHYA1 (old spec), Q6(b) (3 marks): State and explain the effect on the brightness of the lamps in the circuit shown in Figure 2 if the battery has a significant internal resistance.
January 2012, PHYA1 (old spec), Q7(a)(i) (2 marks): Calculate the pd between points A and B in the circuit. answer = V (2 marks)
January 2012, PHYA1 (old spec), Q7(a)(ii) (2 marks): Calculate the internal resistance, r. answer = Ω (2 marks)
January 2012, PHYA1 (old spec), Q7(a)(iii) (2 marks): Calculate the total energy transformed by the battery in 5.0 minutes. answer = J (2 marks)
January 2012, PHYA1 (old spec), Q7(a)(iv) (2 marks): Calculate the percentage of the energy calculated in part (iii) that is dissipated in the battery in 5.0 minutes. answer % (2 marks)
January 2012, PHYA1 (old spec), Q7(b) (2 marks): State and explain one reason why it is an advantage for a rechargeable battery to have a low internal resistance. (2 marks) Copyright © 2012 AQA and its licenso
June 2011, PHYA1 (old spec), Q6(a) (2 marks): Explain why the terminal pd decreases as the current increases. (2 marks)
June 2011, PHYA1 (old spec), Q6(b)(i) (1 mark): Use the graph to find the emf, ε, of the cell. answer = V (1 mark)
June 2011, PHYA1 (old spec), Q6(b)(ii) (3 marks): Use the graph, on page 13, to find the internal resistance, r, of the cell. answer = Ω (3 marks)
June 2011, PHYA1 (old spec), Q6(c)(i) (2 marks): the same emf but double the internal resistance of the first cell labelling your graph A (2 marks)
June 2011, PHYA1 (old spec), Q6(c)(ii) (1 mark): the same emf but negligible internal resistance labelling your graph B. (1 mark)
June 2011, PHYA1 (old spec), Q6(d)(ii) (2 marks): Calculate the energy dissipated in the internal resistance of the cell per second. answer = W (2 marks)
June 2010, PHYA1 (old spec), Q5(c) (2 marks): State and explain the effect on the voltmeter reading if the internal resistance of the battery in the circuit in part (b) was not negligible.
June 2010, PHYA1 (old spec), Q7(a)(i) (1 mark): Define the emf of a battery. (1 mark)
June 2010, PHYA1 (old spec), Q7(a)(ii) (2 marks): When the switch is open the voltmeter reads 6.0V and when it is closed it reads 5.8V. Explain why the readings are different. (2 marks)
June 2010, PHYA1 (old spec), Q7(b) (3 marks): Calculate the internal resistance of the battery. answer = Ω (3 marks)
June 2010, PHYA1 (old spec), Q7(c) (2 marks): State and explain why it is important for car batteries to have a very low internal resistance.
June 2009, PHYA1 (old spec), Q7(a) (2 marks): Calculate the voltage across the terminals of the battery, when the current through the battery is 420 A. answer V (2 marks)
January 2009, PHYA1 (old spec), Q7(a)(i) (1 mark): Explain what is meant by the emf of the battery. (1 mark)
January 2009, PHYA1 (old spec), Q7(a)(ii) (1 mark): Explain what is meant by the internal resistance of the battery. (1 mark)
January 2009, PHYA1 (old spec), Q7(b)(i) (2 marks): Calculate the potential difference across the terminals of the battery. answer = V (2 marks)
January 2009, PHYA1 (old spec), Q7(b)(ii) (3 marks): Calculate the rate of dissipation of energy due to its internal resistance stating an appropriate unit. answer = (3 marks)
January 2009, PHYA1 (old spec), Q7(c) (2 marks): State and explain the effect of attempting to use a battery with a much higher internal resistance to start the car. (2 marks) Copyright © 2009 AQA and its lice
June 2008, PHA3W (2008 spec), Q5(a) (3 marks): 5 (a) In the circuit shown in Figure 8, the battery has an emf of 6.0 V. With the switch closed, the reading of the voltmeter is 5.5 V. Figure 8 Explain, withou
June 2008, PHA3W (2008 spec), Q5(b) (5 marks): 5 (b) In the circuit shown in Figure 9 the cell has an emf ∈ and internal resistance r. The voltage V across the cell is read on the voltmeter, which has infini

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