A Level Physics 10.1 Practical Circuits MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Practise Cambridge AS & A Level Physics 9702 Section 10.1, Practical Circuits: electromotive force, internal resistance, lost volts and terminal potential difference. All 30 questions are original practice questions with fully worked answers – tap “Show Answer” under each question to check your reasoning.
Formula recap
E = W / Q | E = I(R + r) | V = E − Ir | lost volts = Ir | short-circuit current = E / r. Need the full list? See the A Level Physics Formula Sheet.
Electromotive Force
Q1. The electromotive force (EMF) of a cell is best defined as
- (A) the total energy transferred by the source per unit charge passing through it
- (B) the force exerted on the charges inside the cell
- (C) the potential difference across the cell when it delivers its maximum possible current
- (D) the energy transferred per unit charge in the external circuit only
Show Answer
Answer: A – EMF is the total energy the source transfers to each unit of charge, E = W / Q. It includes the energy later transferred both inside the source (across its internal resistance) and in the external circuit, so option D describes only part of it, and EMF is an energy per charge, not a force.
Q2. A source of EMF does 48 J of work in driving a charge of 3.0 C around a complete circuit. What is the EMF of the source?
- (A) 12 V
- (B) 16 V
- (C) 20 V
- (D) 144 V
Show Answer
Answer: B – E = W / Q = 48 / 3.0 = 16 V.
Q3. A cell is connected to an external resistor and a current flows. The terminal potential difference of the cell is less than its EMF because
- (A) some of the charge is used up inside the cell
- (B) the connecting wires have a large resistance
- (C) some of the energy transferred by the source per unit charge is dissipated across the internal resistance of the cell
- (D) the EMF of the cell falls as soon as any current flows
Show Answer
Answer: C – Inside the cell there is an internal resistance r. When a current I flows, a potential difference Ir is dropped across r inside the cell, so the terminal PD is V = E − Ir. Charge itself is never used up; the same current flows everywhere in a series circuit.
Q4. Four identical cells, each of EMF 1.5 V, are connected in series in a torch so that their EMFs act in the same direction. The total EMF is
- (A) 1.5 V
- (B) 3.0 V
- (C) 4.5 V
- (D) 6.0 V
Show Answer
Answer: D – EMFs in series in the same direction add: 4 × 1.5 = 6.0 V.
Q5. A 6.0 V battery and a 2.0 V battery are connected in series, but the 2.0 V battery is inserted the wrong way round so that its EMF opposes the other. The net EMF of the combination is
- (A) 4.0 V
- (B) 8.0 V
- (C) 3.0 V
- (D) 2.0 V
Show Answer
Answer: A – A reversed cell subtracts its EMF: 6.0 − 2.0 = 4.0 V.
Q6. In the equation E = V + Ir, the term Ir is often called the “lost volts”. It represents the
- (A) potential difference across the external resistor
- (B) potential difference across the internal resistance of the source
- (C) EMF of the source
- (D) potential difference measured on open circuit
Show Answer
Answer: B – The current must pass through the source’s own internal resistance r, dropping a PD of Ir there. This part of the EMF is “lost” to the external circuit, leaving the terminal PD V = E − Ir for the external components.
Q7. The EMF of a cell is measured by connecting a high-resistance voltmeter directly across the cell, with nothing else connected. This reading is (very nearly) the EMF because
- (A) the internal resistance of the cell becomes zero
- (B) the voltmeter adds its own EMF to that of the cell
- (C) almost no current flows, so the lost volts Ir are negligible and the terminal PD is almost equal to the EMF
- (D) a voltmeter always measures EMF rather than potential difference
Show Answer
Answer: C – A voltmeter actually measures terminal PD. With only a high-resistance voltmeter connected, the current is extremely small, so Ir is negligible and V = E − Ir is effectively E. This is the open-circuit measurement of EMF.
Q8. How much energy is supplied by a 9.0 V battery when a total charge of 5.0 C passes through it?
- (A) 1.8 J
- (B) 9.0 J
- (C) 14 J
- (D) 45 J
Show Answer
Answer: D – W = EQ = 9.0 × 5.0 = 45 J.
Q9. The volt, the unit of EMF, is equivalent to
- (A) 1 J C-1
- (B) 1 C s-1
- (C) 1 J s-1
- (D) 1 C J-1
Show Answer
Answer: A – Since E = W / Q, one volt is one joule per coulomb, 1 J C-1. (1 C s-1 is the ampere and 1 J s-1 is the watt.)
Q10. When a cell is on open circuit and supplies no current, the potential difference across its terminals is
- (A) zero
- (B) equal to its EMF
- (C) equal to the lost volts
- (D) greater than its EMF
Show Answer
Answer: B – With I = 0 there are no lost volts, so V = E − Ir = E. The terminal PD only falls below the EMF once a current is drawn.
Internal Resistance
Q11. A real cell is modelled as
- (A) an ideal source of EMF in parallel with a resistance
- (B) an ideal source of EMF in series with an internal resistance r
- (C) a resistor with no source of EMF
- (D) an ideal source of EMF whose resistance is always zero
Show Answer
Answer: B – The standard model is an ideal EMF E in series with a resistance r representing the opposition to current inside the cell itself. Every real source behaves this way, which is why E = I(R + r) for a load R.
Q12. A battery of EMF 12 V and internal resistance 0.50 Ω is connected to an external resistor of 5.5 Ω. The current in the circuit is
- (A) 0.50 A
- (B) 1.2 A
- (C) 2.0 A
- (D) 2.2 A
Show Answer
Answer: C – I = E / (R + r) = 12 / (5.5 + 0.50) = 12 / 6.0 = 2.0 A. (Ignoring the internal resistance would give 12 / 5.5 = 2.2 A, which is wrong.)
Q13. When a battery is connected to a 4.0 Ω resistor, the current is 2.0 A. When the same battery is connected instead to a 9.0 Ω resistor, the current is 1.0 A. The internal resistance of the battery is
- (A) 1.0 Ω
- (B) 0.50 Ω
- (C) 2.0 Ω
- (D) 5.0 Ω
Show Answer
Answer: A – Write E = I(R + r) for both cases: E = 2.0(4.0 + r) and E = 1.0(9.0 + r). Setting them equal: 8.0 + 2r = 9.0 + r, so r = 1.0 Ω. (This also gives E = 10 V.)
Q14. A graph of terminal potential difference V against current I for a cell is a straight line. The intercept on the V-axis, where I = 0, gives
- (A) the internal resistance
- (B) the short-circuit current
- (C) the maximum power output
- (D) the EMF of the cell
Show Answer
Answer: D – The graph follows V = E − Ir. At I = 0 the line meets the V-axis at V = E, so the V-intercept is the EMF.
Q15. For the same V against I graph of a cell, the gradient of the straight line is equal to
- (A) E
- (B) −r
- (C) r
- (D) 1 / r
Show Answer
Answer: B – Comparing V = E − Ir with y = mx + c, the gradient is −r. The line slopes downwards, and the magnitude of the gradient gives the internal resistance.
Q16. The terminal PD of a cell is 6.0 V when no current is drawn, and falls to 4.5 V when the cell supplies a current of 3.0 A. The internal resistance of the cell is
- (A) 0.20 Ω
- (B) 0.25 Ω
- (C) 0.50 Ω
- (D) 1.5 Ω
Show Answer
Answer: C – On open circuit V = E = 6.0 V. The lost volts are 6.0 − 4.5 = 1.5 V = Ir, so r = 1.5 / 3.0 = 0.50 Ω. This is also the magnitude of the gradient of the V-I graph between the two points.
Q17. A cell of EMF 1.5 V and internal resistance 0.30 Ω is short-circuited by connecting a wire of negligible resistance directly across its terminals. The current in the wire is
- (A) 5.0 A
- (B) 0.45 A
- (C) 2.0 A
- (D) 0.20 A
Show Answer
Answer: A – With external resistance R = 0, the only resistance is r, so I = E / r = 1.5 / 0.30 = 5.0 A. This is the maximum current the cell can supply.
Q18. A short circuit across a cell is dangerous mainly because
- (A) the EMF of the cell rises to a very high value
- (B) the terminal PD becomes greater than the EMF
- (C) the internal resistance of the cell drops to zero
- (D) the very large current dissipates power I2r inside the cell, heating it rapidly and possibly damaging it or starting a fire
Show Answer
Answer: D – In a short circuit the current is limited only by the small internal resistance, so it is very large. All the source’s energy is then dissipated inside the cell itself at a rate I2r, causing rapid heating.
Q19. As a larger current is drawn from a cell, for example by reducing the external resistance, the terminal potential difference
- (A) increases, because the EMF increases
- (B) decreases, because the lost volts Ir across the internal resistance increase
- (C) stays exactly equal to the EMF whatever the current
- (D) becomes zero
Show Answer
Answer: B – V = E − Ir. The EMF stays the same, but a bigger I means bigger lost volts, so the terminal PD falls. This is why car headlights dim when the starter motor draws a large current.
Q20. A battery has EMF 9.0 V and internal resistance 1.0 Ω. When it supplies a current of 2.0 A, its terminal potential difference is
- (A) 9.0 V
- (B) 11 V
- (C) 7.0 V
- (D) 4.5 V
Show Answer
Answer: C – V = E − Ir = 9.0 − (2.0 × 1.0) = 9.0 − 2.0 = 7.0 V.
Terminal PD & Circuit Calculations
Q21. A source of EMF 12 V and internal resistance 0.40 Ω is connected to a resistor of 5.6 Ω. The terminal potential difference of the source is
- (A) 11.2 V
- (B) 12 V
- (C) 0.80 V
- (D) 10.4 V
Show Answer
Answer: A – I = E / (R + r) = 12 / (5.6 + 0.40) = 12 / 6.0 = 2.0 A. Terminal PD = IR = 2.0 × 5.6 = 11.2 V. (Equivalently, E minus the lost volts: 12 − 2.0 × 0.40 = 12 − 0.80 = 11.2 V.)
Q22. A 12 V source of internal resistance 0.40 Ω drives a current of 2.0 A through an external resistor of 5.6 Ω. The power dissipated in the external resistor is
- (A) 24 W
- (B) 22.4 W
- (C) 12 W
- (D) 28 W
Show Answer
Answer: B – P = I2R = (2.0)2 × 5.6 = 4.0 × 5.6 = 22.4 W. (Using V = 12 V in VI would be wrong, because the PD across the resistor is only 11.2 V.)
Q23. A source of EMF 6.0 V and internal resistance 0.50 Ω is connected to a resistor of 2.5 Ω. The power wasted inside the source is
- (A) 0.50 W
- (B) 1.0 W
- (C) 2.0 W
- (D) 12 W
Show Answer
Answer: C – I = E / (R + r) = 6.0 / (2.5 + 0.50) = 6.0 / 3.0 = 2.0 A. Power wasted in r: P = I2r = (2.0)2 × 0.50 = 4.0 × 0.50 = 2.0 W.
Q24. A source of EMF 12 V and internal resistance 0.40 Ω supplies an external load of 5.6 Ω. The efficiency of the energy transfer to the load is closest to
- (A) 7%
- (B) 50%
- (C) 89%
- (D) 93%
Show Answer
Answer: D – The same current passes through R and r, so the fraction of the total power delivered to the load is R / (R + r) = 5.6 / (5.6 + 0.40) = 5.6 / 6.0 = 0.93, i.e. about 93%. The other 7% is wasted in the internal resistance.
Q25. A car battery must have a very low internal resistance so that
- (A) it can supply the very large current needed by the starter motor without most of its EMF being lost across the internal resistance
- (B) its EMF is raised from 12 V to 24 V when starting
- (C) no charge can ever leave the battery
- (D) the terminal PD is zero while the engine is starting
Show Answer
Answer: A – The starter motor needs a current of well over 100 A. The lost volts are Ir, so only a tiny r keeps Ir small and leaves most of the 12 V available at the terminals to drive that current.
Q26. A source of EMF 10 V and internal resistance 2.0 Ω delivers maximum power to an external load when the resistance of the load is
- (A) 0 Ω
- (B) 2.0 Ω
- (C) 10 Ω
- (D) as large as possible
Show Answer
Answer: B – Maximum power transfer occurs when the load resistance equals the internal resistance, R = r = 2.0 Ω. Note that at this point the efficiency is only 50%, since equal power is wasted in r – maximum power and maximum efficiency are not the same thing.
Q27. A rechargeable battery of EMF 12 V and internal resistance 0.50 Ω is being charged by driving a current of 4.0 A backwards through it. The potential difference across its terminals during charging is
- (A) 10 V
- (B) 12 V
- (C) 14 V
- (D) 8.0 V
Show Answer
Answer: C – When charging, the charger must overcome the EMF and also provide the PD across the internal resistance, so V = E + Ir = 12 + (4.0 × 0.50) = 12 + 2.0 = 14 V. During charging the terminal PD is greater than the EMF – the reverse of the discharging case.
Q28. Two identical cells, each of EMF 1.5 V and internal resistance 0.20 Ω, are connected in series to a resistor of 2.6 Ω. The current in the circuit is
- (A) 0.58 A
- (B) 1.2 A
- (C) 0.50 A
- (D) 1.0 A
Show Answer
Answer: D – In series the EMFs add and the internal resistances add: E = 3.0 V, r = 0.40 Ω. I = E / (R + r) = 3.0 / (2.6 + 0.40) = 3.0 / 3.0 = 1.0 A.
Q29. A source of EMF 9.0 V and internal resistance 0.80 Ω supplies a current of 1.5 A to a load. The efficiency of the energy transfer to the load is closest to
- (A) 13%
- (B) 67%
- (C) 80%
- (D) 87%
Show Answer
Answer: D – Terminal PD: V = E − Ir = 9.0 − (1.5 × 0.80) = 9.0 − 1.2 = 7.8 V. Useful power = VI = 7.8 × 1.5 = 11.7 W; total power supplied = EI = 9.0 × 1.5 = 13.5 W. Efficiency = 11.7 / 13.5 = 0.87, i.e. about 87%.
Q30. When a cell is being charged, its terminal potential difference is greater than its EMF because
- (A) the EMF of the cell increases during charging
- (B) the internal resistance becomes negative during charging
- (C) the charger drives current through the cell in the reverse direction, so it must supply the EMF plus the PD Ir dropped across the internal resistance: V = E + Ir
- (D) no current flows through the cell during charging
Show Answer
Answer: C – Charging pushes current into the positive terminal, backwards through the cell. The applied PD has to overcome the cell’s EMF and the drop across r, giving V = E + Ir, which is greater than E. On discharge the sign reverses and V = E − Ir.
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