Edexcel International A Level Physics · Unit 2 · Waves and Electricity
E.m.f. and internal resistance
40 free, high-quality past-paper-style questions on E.m.f. and internal resistance for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 135 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U2-9-A-01Unit 2 · ASEasy1 mark
Which of the following is a correct description of the e.m.f. of a cell?
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Answer
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Notes
E.m.f. describes a source: it is the energy transferred to electrical energy per unit charge passing through the cell. Option B describes a p.d. in the external circuit, and option C is the terminal p.d., which is less than the e.m.f. whenever a current flows.
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Answer D
Answer: D
Exam tip: E.m.f. and p.d. are both “energy per unit charge”. The difference is the direction: to electrical energy in a source, from electrical energy in a component.
A battery of e.m.f. 9.0 V delivers a constant current of 0.12 A to a smoke alarm for 20 minutes. How much energy does the battery transfer to electrical energy in this time?
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Answer
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Notes
Charge = It = 0.12 × (20 × 60) = 144 C. Energy = εQ = 9.0 × 144 = 1300 J (1296 J).
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Answer C
Answer: C
Common error: Leaving the time in minutes. Convert to seconds before using Q = It.
A cell has an e.m.f. of 3.0 V and an internal resistance of 0.50 Ω. It delivers a current of 0.80 A. What is the potential difference across its terminals?
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Answer
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Notes
Lost volts = Ir = 0.80 × 0.50 = 0.40 V. Terminal p.d. V = ε − Ir = 3.0 − 0.40 = 2.6 V.
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Answer B
Answer: B
Exam tip: Terminal p.d. = e.m.f. − lost volts. With current flowing it is always less than the e.m.f. for a cell that is supplying energy.
A mobile phone battery has an e.m.f. of 3.85 V. During a video call, 2.4 × 103 C of charge passes through the battery and the potential difference across its terminals is 3.70 V throughout.
(a)
Define e.m.f.
(1)
(b)
Calculate the energy transferred to electrical energy by the battery during the call.
(2)
(c)
Calculate the energy dissipated inside the battery during the call.
(1)
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Answer
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Part (a)
the energy transferred to electrical energy per unit charge passing through the source / battery
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Part (b)
W = εQ = 3.85 × 2.4 × 103
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= 9.2 × 103 J (9240 J)
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Part (c)
(3.85 − 3.70) × 2.4 × 103 = 360 J
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Exam tip: The e.m.f. gives the energy supplied per coulomb; the terminal p.d. gives the energy delivered to the external circuit per coulomb. The difference is dissipated inside.
A torch bulb is lit by a single cell. While the bulb is lit its resistance is 3.5 Ω. The cell has an e.m.f. of 1.60 V and its internal resistance is 0.50 Ω.
Figure: torch bulb connected to the cell (dashed box: e.m.f. and internal resistance)
(a)
Calculate the current in the bulb.
(2)
(b)
Calculate the potential difference across the terminals of the cell.
(2)
(c)
State the value of the lost volts.
(1)
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Answer
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Notes
Part (a)
I = ε/(R + r) = 1.60 ÷ (3.5 + 0.50)
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= 0.40 A
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Part (b)
V = IR = 0.40 × 3.5
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= 1.4 V (allow error carried forward)
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Part (c)
1.60 − 1.4 = 0.20 V / 0.40 × 0.50 = 0.20 V
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Common error: Using the e.m.f. as the p.d. across the bulb. With current flowing, the bulb gets the terminal p.d., which is less than the e.m.f.
A digital voltmeter of very high resistance is connected across the battery of a bicycle light, with the lamp switched off. It reads 4.50 V. When the lamp is switched on, the voltmeter reads 4.05 V and an ammeter in series with the lamp reads 0.300 A.
Figure: the bicycle light circuit, with the high-resistance voltmeter across the battery
(a)
Explain why the first reading is equal to the e.m.f. of the battery.
(1)
(b)
Calculate the internal resistance of the battery.
(2)
(c)
Calculate the resistance of the lamp when it is lit.
(1)
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Answer
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Notes
Part (a)
the voltmeter draws (almost) no current, so there are no lost volts / Ir = 0, and the terminal p.d. equals the e.m.f.
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Part (b)
lost volts = 4.50 − 4.05 = 0.45 V
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r = 0.45 ÷ 0.300 = 1.5 Ω (1.50 Ω)
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Part (c)
R = 4.05 ÷ 0.300 = 13.5 Ω
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Exam tip: Use the lost volts, ε − V, with the current to find r; use the terminal p.d. with the current to find the external resistance.
The figure shows a battery of e.m.f. 4.5 V and internal resistance 1.0 Ω connected to two resistors in series. A voltmeter of very high resistance is connected as shown. What is the reading on the voltmeter?
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Answer
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Total resistance = 2.0 + 6.0 + 1.0 = 9.0 Ω, so I = 4.5 ÷ 9.0 = 0.50 A. The voltmeter is across the 6.0 Ω resistor only: V = 0.50 × 6.0 = 3.0 V.
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Answer B
Answer: B
Exam tip: Check what the voltmeter is connected across. Here it is one resistor, not the battery terminals, so neither the e.m.f. nor the terminal p.d. is the answer.
A cell is connected to an external resistor. The graph shows how the electric potential varies with position around the circuit. The current in the circuit is 0.40 A. What is the internal resistance of the cell?
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Answer
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Notes
The rise of 6.0 V is the e.m.f. The fall inside the cell, 6.0 − 5.2 = 0.8 V, is the lost volts. r = 0.8 ÷ 0.40 = 2.0 Ω.
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Answer A
Answer: A
Exam tip: On a potential diagram the e.m.f. is the rise inside the source, the lost volts are the fall inside the source, and the terminal p.d. is the fall across the external circuit.
The graph shows how the terminal potential difference varies with current for two different cells, P and Q. Which statement about the cells is correct?
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The intercepts on the V-axis give the e.m.f.s: P 1.6 V, Q 1.2 V. The gradients give the internal resistances: P 1.6 ÷ 2.0 = 0.80 Ω, Q 1.2 ÷ 3.0 = 0.40 Ω. The intercepts on the I-axis are the short-circuit currents: P 2.0 A, Q 3.0 A. So Q has the smaller e.m.f. but the larger short-circuit current.
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Answer D
Answer: D
Exam tip: On a V–I graph for a source: V-intercept = e.m.f., steepness = internal resistance, I-intercept = short-circuit current ε/r.
A battery supplies a steady current of 2.0 A to a small motor. The diagram shows the energy transfers that take place each second.
(a)
Calculate the e.m.f. of the battery.
(1)
(b)
Calculate the potential difference across the terminals of the battery.
(1)
(c)
Calculate the internal resistance of the battery.
(2)
(d)
Calculate the efficiency of the battery in this situation.
(1)
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Answer
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Part (a)
ε = P/I = 24 ÷ 2.0 = 12 V
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Part (b)
V = 18 ÷ 2.0 = 9.0 V
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Part (c)
I2r = 6.0 W, so r = 6.0 ÷ 2.02
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= 1.5 Ω (or (12 − 9.0) ÷ 2.0)
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Part (d)
18 ÷ 24 = 0.75 (75%)
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Exam tip: Energy per second supplied by the source = εI; energy per second delivered to the external circuit = VI; the difference, I2r, is wasted inside.
A laboratory cell has a large internal resistance. A voltmeter is connected across its terminals. The figure models the voltmeter as an ideal meter with a resistance of 10 kΩ.
(a)
Calculate the current in the circuit.
(1)
(b)
Calculate the reading on the voltmeter.
(1)
(c)
Explain why the voltmeter reading is less than the e.m.f. of the cell.
(1)
(d)
The voltmeter is replaced by a digital voltmeter of resistance 10 MΩ. Calculate the new reading.
(1)
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Answer
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Part (a)
I = 1.50 ÷ (2.0 × 103 + 10 × 103) = 1.25 × 10−4 A
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Part (b)
V = 1.25 × 10−4 × 10 × 103 = 1.25 V (allow error carried forward)
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Part (c)
the voltmeter draws a current, so there are lost volts across the internal resistance (Ir = 0.25 V)
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Part (d)
V = 1.50 × 10 × 106 ÷ (10 × 106 + 2.0 × 103) = 1.50 V (1.4997 V)
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Exam tip: A voltmeter reads the e.m.f. only if it draws no current. Its resistance must be much larger than the internal resistance of the source.
A power supply of e.m.f. 12 V and internal resistance 1.5 Ω is connected to a resistor of resistance 6.5 Ω. What is the current in the resistor?
Figure: power supply (dashed box: e.m.f. and internal resistance) connected to the resistor
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Answer
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Notes
ε = I(R + r), so I = 12 ÷ (6.5 + 1.5) = 12 ÷ 8.0 = 1.5 A.
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Answer A
Answer: A
Common error: Dividing the e.m.f. by the external resistance only. The e.m.f. drives the current through the external resistance and the internal resistance.
A lithium cell has an e.m.f. of 3.6 V. When it delivers a current of 0.80 A, the potential difference across its terminals is 3.2 V. What is the maximum current the cell could deliver if its terminals were short-circuited?
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Answer
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Lost volts = 3.6 − 3.2 = 0.4 V, so r = 0.4 ÷ 0.80 = 0.50 Ω. Short circuit: Imax = ε/r = 3.6 ÷ 0.50 = 7.2 A.
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Answer C
Answer: C
Exam tip: Find r from lost volts ÷ current first. In a short circuit the external resistance is zero, so the whole e.m.f. is across r.
A battery of e.m.f. 9.0 V and internal resistance 3.0 Ω is connected to a 6.0 Ω resistor. At what rate is energy dissipated inside the battery?
Figure: battery (dashed box: e.m.f. and internal resistance) connected to the resistor
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Answer
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I = 9.0 ÷ (6.0 + 3.0) = 1.0 A. Power dissipated inside = I2r = 1.02 × 3.0 = 3.0 W.
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Answer A
Answer: A
Exam tip: Three powers appear in these circuits: εI supplied by the source, I2R delivered to the load and I2r wasted inside. Decide which one is asked for.
Three cells, each of e.m.f. 1.5 V and internal resistance 0.40 Ω, are connected in series, but one of them has been inserted the wrong way round. The combination is connected to a 4.8 Ω resistor. What is the current in the resistor?
Figure: three cells in series, with the middle one reversed, connected to the resistor
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Answer
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Total e.m.f. = 1.5 + 1.5 − 1.5 = 1.5 V. All three internal resistances still add: r = 3 × 0.40 = 1.2 Ω. I = 1.5 ÷ (4.8 + 1.2) = 0.25 A.
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Answer A
Answer: A
Common error: Leaving out the internal resistance of the reversed cell. Its e.m.f. subtracts, but the current still flows through it, so its resistance adds.
A student plots a graph of the terminal potential difference V of a cell (vertical axis) against the current I (horizontal axis). The best-fit line meets the V-axis at 1.60 V and passes through the point (0.80 A, 1.20 V). What is the internal resistance of the cell?
Figure: terminal p.d. against current for the cell
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Answer
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V = ε − Ir, so the gradient is −r. Gradient = (1.20 − 1.60) ÷ (0.80 − 0) = −0.50 V A−1, so r = 0.50 Ω.
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Answer B
Answer: B
Exam tip: Internal resistance is minus the gradient of the V–I graph. Use a change in V divided by a change in I, never a single V/I ratio.
the internal resistance is in series with anything connected to the terminals, including a person / a short circuit
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so the current can never exceed ε/r, which is small / about 0.1 mA
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Part (c)
I = 5000 ÷ ((10 + 50) × 106) = 8.3 × 10−5 A
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V = IR = 8.33 × 10−5 × 10 × 106 = 830 V (833 V)
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Exam tip: Convert prefixes before substituting: 5.0 kV = 5000 V and 50 MΩ = 5.0 × 107 Ω. Most of the e.m.f. of this supply is lost inside it unless the load resistance is very large.
A battery of e.m.f. 9.0 V and internal resistance 1.8 Ω is used to power a heating element of resistance 7.2 Ω inside a hand warmer.
Figure: battery (dashed box: e.m.f. and internal resistance) connected to the heating element
(a)
Calculate the current in the heating element.
(1)
(b)
Calculate the power delivered to the heating element.
(1)
(c)
Calculate the percentage of the power supplied by the battery that is delivered to the heating element.
(2)
(d)
A second, identical heating element is connected in parallel with the first. Calculate the total power now delivered to the two elements.
(2)
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Answer
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Part (a)
I = 9.0 ÷ (7.2 + 1.8) = 1.0 A
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Part (b)
P = I2R = 1.02 × 7.2 = 7.2 W
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Part (c)
power supplied = εI = 9.0 × 1.0 = 9.0 W
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7.2 ÷ 9.0 = 0.80, so 80%
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Part (d)
external resistance = 3.6 Ω, so I = 9.0 ÷ (3.6 + 1.8) = 1.67 A
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P = 1.672 × 3.6 = 10 W (10.0 W)
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Exam tip: Adding a second element in parallel lowers the external resistance, so the current rises — but the lost volts rise too, so the total power does not double.
A small drone is powered by a pack of six rechargeable cells connected in series. Each cell has an e.m.f. of 1.2 V and an internal resistance of 0.15 Ω. In flight, the current from the pack is 2.0 A.
(a)
Calculate the e.m.f. and the internal resistance of the pack.
(2)
(b)
Calculate the terminal potential difference of the pack in flight.
(2)
(c)
Calculate the rate at which energy is dissipated inside the pack in flight.
(1)
(d)
When the pack is assembled, one cell is accidentally inserted the wrong way round. State the e.m.f. of the pack.
(1)
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Answer
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Part (a)
ε = 6 × 1.2 = 7.2 V
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r = 6 × 0.15 = 0.90 Ω
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Part (b)
lost volts = 2.0 × 0.90 = 1.8 V
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V = 7.2 − 1.8 = 5.4 V (allow error carried forward)
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Part (c)
I2r = 2.02 × 0.90 = 3.6 W
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Part (d)
5 × 1.2 − 1.2 = 4.8 V
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Common error: Treating a reversed cell as missing (giving 6.0 V). Its e.m.f. opposes the others, so the total drops by two cell e.m.f.s, not one.
A student determines the e.m.f. and internal resistance of a single cell. She uses a variable resistor to change the current and records the terminal potential difference V for each current I. Results: I = 0.10 A, V = 1.43 V; 0.20 A, 1.32 V; 0.30 A, 1.22 V; 0.40 A, 1.11 V; 0.50 A, 1.00 V.
Figure: the results with the line of best fit through (0.05 A, 1.48 V) and (0.55 A, 0.95 V)
(a)
Describe how the ammeter and the voltmeter should be connected in the circuit.
(2)
(b)
Her line of best fit on a graph of V against I passes through (0.05 A, 1.48 V) and (0.55 A, 0.95 V). Use these points to determine the internal resistance and the e.m.f. of the cell.
(3)
(c)
Explain why the student should open the switch between readings.
(2)
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Part (a)
ammeter in series with the cell and the variable resistor
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voltmeter in parallel with / directly across the terminals of the cell
The graph shows how the terminal potential difference V of a battery varies with the current I it delivers. What current would the battery deliver if its terminals were connected together by a wire of negligible resistance?
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Answer
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From the graph, ε = 4.5 V (the intercept) and the gradient is (3.0 − 4.5) ÷ 1.0 = −1.5 V A−1, so r = 1.5 Ω. Short circuit: I = ε/r = 4.5 ÷ 1.5 = 3.0 A. This is where the line, extended, would meet the I-axis.
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Answer C
Answer: C
Exam tip: The short-circuit current is the intercept on the current axis. If the graph stops early, either extend the line or calculate ε/r from the intercept and gradient.
A student measures the terminal potential difference V of a battery for different currents I and draws the line of best fit shown. What is the internal resistance of the battery?
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Use two points on the line far apart: (0.40 A, 4.00 V) and (1.20 A, 3.00 V). Gradient = (3.00 − 4.00) ÷ (1.20 − 0.40) = −1.25 V A−1, so r = 1.25 Ω.
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Answer B
Answer: B
Common error: Reading the bottom of the V-axis as zero. This axis starts at 2.8 V; a false origin does not change the gradient, but it does matter if you read p.d. values from the axis edge.
The figure shows a battery connected to a 15 Ω resistor through a switch S. The table in the figure gives the reading on the high-resistance voltmeter with S open and with S closed. When S is closed, at what rate is energy dissipated inside the battery?
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Answer
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S open: no current, so ε = 9.0 V. S closed: I = 7.5 ÷ 15 = 0.50 A and lost volts = 9.0 − 7.5 = 1.5 V. Power dissipated inside = lost volts × current = 1.5 × 0.50 = 0.75 W (or I2r with r = 3.0 Ω).
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Answer A
Answer: A
Exam tip: With the switch open the voltmeter reads the e.m.f.; with it closed it reads the terminal p.d. The difference is the lost volts, and lost volts × current is the power wasted inside the source.
In an experiment to find the e.m.f. and internal resistance of a cell, a student obtains the results shown. The current was left on for a long time while the larger currents were measured. Which is the most likely explanation for the shape of the graph at larger currents?
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Answer
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Beyond 0.6 A the terminal p.d. is lower than the straight line predicts, so the lost volts are larger than I × (the original r). The cell's internal resistance has changed (and its e.m.f. may have dropped) because a large current flowed for a long time. A zero error would shift every point equally, and the resistance of the variable resistor does not matter because V and I are both measured.
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Answer C
Answer: C
Exam tip: Keep currents modest and open the switch between readings, so the cell does not warm up and its internal resistance stays constant during the experiment.
A student uses a variable resistor to change the current I from a battery and measures the terminal potential difference V for each current. The graph shows her results and her line of best fit.
(a)
State the current at which the anomalous reading was taken.
(1)
(b)
Use the graph to state the e.m.f. of the battery.
(1)
(c)
Use the graph to determine the internal resistance of the battery.
(2)
(d)
Determine the resistance of the variable resistor when the current is 0.50 A.
(1)
(e)
State one precaution the student should take so that the internal resistance stays constant during the experiment.
(1)
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Part (a)
0.40 A (the point well above the line)
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Part (b)
intercept on the V-axis: 3.00 V (allow 2.98–3.02 V)
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Part (c)
gradient from a large triangle, e.g. (1.60 − 3.00) ÷ (0.70 − 0) = −2.0 V A−1
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r = −gradient = 2.0 Ω (allow 1.9–2.1 Ω)
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Part (d)
from the line, V = 2.00 V at 0.50 A, so R = 2.00 ÷ 0.50 = 4.0 Ω (allow 3.9–4.1 Ω)
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Part (e)
open the switch between readings / take each reading quickly / keep the currents small, so that the battery does not heat up
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Exam tip: Ignore an anomalous point when drawing the line of best fit, and use a gradient triangle that covers most of the line, with corners you can read accurately.
A battery is made by connecting a number of identical cells in series. The graph shows how the terminal potential difference varies with current for one of the cells and for the battery.
(a)
Use the graph to state the e.m.f. of one cell and of the battery, and deduce the number of cells in the battery.
(2)
(b)
Use the graph to determine the internal resistance of the battery.
(2)
(c)
Explain why both lines meet the current axis at the same point.
(1)
(d)
The battery is connected to a 3.0 Ω resistor. Calculate the terminal potential difference of the battery.
(2)
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Answer
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Part (a)
one cell 1.5 V, battery 4.5 V
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4.5 ÷ 1.5 = 3 cells
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Part (b)
gradient = −4.5 ÷ 3.0 = −1.5 V A−1
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r = 1.5 Ω
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Part (c)
short-circuit current = ε/r; for n cells in series both ε and r are n times larger, so the ratio is unchanged
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Part (d)
I = 4.5 ÷ (3.0 + 1.5) = 1.0 A
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V = 1.0 × 3.0 = 3.0 V (allow error carried forward; this is the point (1.0 A, 3.0 V) on the battery line)
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Exam tip: For identical cells in series, the e.m.f.s add and the internal resistances add. The short-circuit current of the battery is the same as that of one cell.
A data logger of very high resistance records the terminal potential difference of a battery. Switch S1 connects a 12 Ω resistor R1 across the battery. Switch S2 connects a second resistor R2 in parallel with R1. The graph shows the record as the switches are operated.
(a)
State the e.m.f. of the battery and explain how the graph shows this.
(2)
(b)
Calculate the current in R1 between 2 s and 6 s.
(1)
(c)
Determine the internal resistance of the battery.
(2)
(d)
Determine the resistance of R2.
(2)
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Answer
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Part (a)
9.0 V
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before 2 s (and after 9 s) no current flows, so there are no lost volts and the terminal p.d. equals the e.m.f.
current in R1 = 6.0 ÷ 12 = 0.50 A, so current in R2 = 1.5 A and R2 = 6.0 ÷ 1.5 = 4.0 Ω
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Exam tip: Each step down in the terminal p.d. shows that more current is being drawn. Find the total current from the lost volts, I = (ε − V)/r, then share it between the parallel branches.
A student sets up the circuit shown at the top of the figure to find the e.m.f. and internal resistance of a cell. By mistake, the voltmeter is connected across both the cell and the 2.0 Ω resistor R0. Her results are plotted on the graph below the circuit.
(a)
Write an equation for the voltmeter reading V in terms of the current I, the e.m.f. ε, the internal resistance r and R0.
(1)
(b)
Use the graph to determine the e.m.f. and the internal resistance of the cell.
(3)
(c)
State how the circuit should be changed so that the gradient gives the internal resistance directly.
(1)
(d)
Calculate the current the cell would deliver if its terminals were short-circuited.
(1)
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Answer
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Part (a)
V = ε − I(r + R0)
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Part (b)
ε = intercept = 1.50 V (allow 1.48–1.52 V)
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gradient = (0.50 − 1.50) ÷ 0.40 = −2.5 V A−1 (allow −2.4 to −2.6)
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r = 2.5 − 2.0 = 0.50 Ω (allow 0.4–0.6 Ω)
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Part (c)
connect the voltmeter directly across the terminals of the cell only
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Part (d)
I = ε/r = 1.50 ÷ 0.50 = 3.0 A (allow error carried forward)
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Common error: Taking minus the gradient as r without checking what the voltmeter is across. Any resistance inside the voltmeter connection adds to the gradient.
In an experiment to find the e.m.f. ε and internal resistance r of a cell, the voltmeter has a zero error: every reading is 0.05 V higher than the true value. The ammeter is accurate. The student plots terminal p.d. against current and uses the intercept and gradient. What is the effect of the zero error on the values obtained?
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Answer
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Notes
Adding the same 0.05 V to every reading moves the whole line up by 0.05 V without changing its slope. The intercept, ε, is 0.05 V too large; the gradient, and so r, is unchanged.
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Answer D
Answer: D
Exam tip: A zero error shifts every point by the same amount, so it changes an intercept but not a gradient. This is one reason for using a graph rather than a single pair of readings.
A source of e.m.f. ε and internal resistance r is connected to a load resistor of resistance 3r. What percentage of the power supplied by the source is delivered to the load?
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Answer
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The same current flows through both resistances, so the power divides in the ratio of the resistances. Fraction to the load = R/(R + r) = 3r/4r = 75%.
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Answer C
Answer: C
Exam tip: Fraction of energy reaching the load = I2R ÷ I2(R + r) = R/(R + r) = V/ε. The current cancels.
A student makes a cell by pushing a copper strip and a zinc strip into a lemon. She connects a resistor of resistance R across the cell and measures the p.d. V across it with a digital voltmeter. There is no ammeter. Results: R = 500 Ω, V = 0.41 V; R = 1000 Ω, V = 0.56 V; R = 2000 Ω, V = 0.69 V; R = 4000 Ω, V = 0.78 V.
Figure: the lemon cell connected to resistor R, with the digital voltmeter across R
(a)
Calculate the current when R = 500 Ω.
(1)
(b)
Use the readings for R = 500 Ω and R = 4000 Ω to determine the internal resistance and the e.m.f. of the lemon cell.
(3)
(c)
The student connects the lemon cell to a torch bulb of resistance 10 Ω. Explain, with a calculation, why the bulb does not light.
(2)
(d)
Suggest why a voltmeter with a very high resistance is needed for this experiment.
An electric scooter has a battery that powers both the motor and the lights, which are connected in parallel across the battery terminals. When the rider climbs a steep hill, the motor draws a much larger current than on level ground. The rider notices that the lights become dimmer and that the battery becomes warm.
Figure: the motor and the lights connected in parallel across the battery (dashed box: e.m.f. and internal resistance)
(a)
State what is meant by the internal resistance of a battery.
(1)
(b)
Explain why the lights become dimmer and the battery becomes warm while the scooter climbs the hill. You may use equations in your answer.
(6)
Show mark scheme
Answer
Mark
Notes
Part (a)
the resistance of the materials inside the battery (chemicals and electrodes), which dissipates energy when current flows through it
✓ 1
Part (b)
Indicative content:
—
the battery has an e.m.f. ε, the energy supplied per unit charge, which stays (nearly) constant
✓ 1
the battery has an internal resistance r in series with the external circuit
✓ 1
climbing the hill increases the total current I drawn from the battery
✓ 1
so the lost volts Ir increase and the terminal p.d. V = ε − Ir decreases
✓ 1
the lights are across the terminals, so their p.d., current and power decrease and they are dimmer
✓ 1
the power dissipated inside the battery, I2r, increases (with the square of the current), so the battery warms up
✓ 1
Exam tip: In an extended explanation, link each step to the next: more current → more lost volts → lower terminal p.d. → less power in the lamps. A bare list of equations earns little.
A battery of unknown e.m.f. ε and internal resistance r is connected to a 4.0 Ω resistor and the current is 0.50 A. When the 4.0 Ω resistor is replaced by an 8.0 Ω resistor, the current is 0.30 A.
Figure: the battery connected to the 4.0 Ω resistor (dashed box: e.m.f. and internal resistance)
(a)
Determine the internal resistance and the e.m.f. of the battery.
(3)
(b)
Calculate the power delivered to the 4.0 Ω resistor.
(1)
(c)
Calculate the power that would be delivered to a 2.0 Ω resistor connected to the same battery, and comment on your answer.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
ε = 0.50(4.0 + r) and ε = 0.30(8.0 + r)
✓ 1
2.0 + 0.50r = 2.4 + 0.30r, so r = 2.0 Ω
✓ 1
ε = 0.50 × 6.0 = 3.0 V
✓ 1
Part (b)
P = I2R = 0.502 × 4.0 = 1.0 W
✓ 1
Part (c)
I = 3.0 ÷ (2.0 + 2.0) = 0.75 A, P = 0.752 × 2.0 = 1.1 W (1.125 W)
✓ 1
this is more than with 4.0 Ω (1.0 W) or 8.0 Ω (0.72 W), so here the power rises as the load resistance falls towards r / the power to the load is greatest when the load resistance equals the internal resistance
✓ 1
Exam tip: Two unknowns need two equations. Write ε = I(R + r) for each situation, then eliminate ε by setting the two expressions equal.
A battery is connected to a variable load resistor of resistance R. The graph shows how the power delivered to the load varies with R. What is the e.m.f. of the battery?
Show mark scheme
Answer
Mark
Notes
The power in the load is greatest when R equals the internal resistance, so r = 4.0 Ω. At the peak, V across the load = √(PR) = √(2.25 × 4.0) = 3.0 V and I = 3.0 ÷ 4.0 = 0.75 A. ε = I(R + r) = 0.75 × 8.0 = 6.0 V. At maximum power, half the e.m.f. is across the load.
✓ 1
Answer B
Answer: B
Exam tip: Maximum power is delivered to the load when R = r. Then the terminal p.d. is half the e.m.f., and only half of the power supplied reaches the load.
A torch contains two new cells and one old cell connected in series with a lamp, as shown in the figure. The lamp has a resistance of 3.1 Ω while it is lit. What is the potential difference across the terminals of the old cell?
Show mark scheme
Answer
Mark
Notes
Total e.m.f. = 1.5 + 1.5 + 1.2 = 4.2 V; total resistance = 3.1 + 0.10 + 0.10 + 0.90 = 4.2 Ω, so I = 1.0 A. Old cell: V = ε − Ir = 1.2 − 1.0 × 0.90 = 0.30 V.
✓ 1
Answer A
Answer: A
Exam tip: Apply V = ε − Ir to one cell using the current in the whole series circuit. A cell with a large internal resistance can have a very small terminal p.d., so it adds little to the lamp.
A cell is connected to resistors of different resistance R in turn. The graph shows how the terminal potential difference V of the cell varies with R. The dashed line shows the value that V approaches when R is very large. What is the internal resistance of the cell?
Show mark scheme
Answer
Mark
Notes
For very large R the current is negligible, so ε = 3.0 V. At the marked point, I = V/R = 1.2 ÷ 2.0 = 0.60 A and the lost volts are 3.0 − 1.2 = 1.8 V. r = 1.8 ÷ 0.60 = 3.0 Ω.
✓ 1
Answer D
Answer: D
Exam tip: When the load resistance is very large, the current is tiny and the terminal p.d. approaches the e.m.f. Use the terminal p.d. and the load resistance to find the current.
A battery is connected to a variable external resistance. The graph shows how the power delivered to the external circuit varies with the current from the battery. The e.m.f. ε and internal resistance r of the battery are constant.
(a)
Explain why the power delivered to the external circuit is zero when the current is 6.0 A.
(1)
(b)
Use the graph to determine the internal resistance and the e.m.f. of the battery.
(3)
(c)
Calculate the external resistance when the power delivered to it is greatest, and comment on your answer.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
6.0 A is the short-circuit current: the external resistance is zero, so the terminal p.d. (and P = IV) is zero / all the e.m.f. is across r
✓ 1
Part (b)
short circuit: ε = 6.0r
✓ 1
at the peak, P = εI − I2r: 18 = 3.0ε − 9.0r
✓ 1
18 = 18r − 9.0r, so r = 2.0 Ω and ε = 12 V
✓ 1
allow equivalent methods, e.g. the peak is at R = r, so I = ε/2r and Pmax = ε2/4r
—
Part (c)
R = P/I2 = 18 ÷ 3.02 = 2.0 Ω
✓ 1
this equals the internal resistance: the power delivered to the load is greatest when R = r
✓ 1
Exam tip: Power delivered to the external circuit = power supplied − power wasted inside = εI − I2r. Two points on the graph give two equations for the two unknowns.
The figure shows the circuit a student uses to determine the e.m.f. and internal resistance of a cell. R0 is a fixed resistor of a few ohms.
(a)
Suggest why the fixed resistor R0 is included in the circuit.
(1)
(b)
Describe how the student should use this circuit to determine the e.m.f. and the internal resistance of the cell. Your answer should include how the results are processed.
(6)
Show mark scheme
Answer
Mark
Notes
Part (a)
it limits the largest current (when the variable resistor is set to zero), so the cell does not overheat / its internal resistance does not change / the ammeter is protected
✓ 1
Part (b)
Indicative content:
—
close the switch and record the voltmeter reading V and the ammeter reading I
✓ 1
adjust the variable resistor to obtain at least six pairs of readings over as wide a range of current as possible
✓ 1
open the switch between readings / take readings quickly, so the cell does not warm up and r stays constant
✓ 1
plot a graph of V (vertical axis) against I (horizontal axis) and draw a straight line of best fit
✓ 1
the intercept on the V-axis is the e.m.f., since V = ε − Ir
✓ 1
the internal resistance is minus the gradient of the line, found from a large triangle
✓ 1
Exam tip: Link the graph to the equation: compare V = −rI + ε with y = mx + c. That tells you which axis each quantity goes on and what the gradient and intercept mean.
The graph shows how the potential difference across a filament lamp varies with the current in it. The lamp is connected to a battery of e.m.f. 6.0 V and internal resistance 2.0 Ω.
(a)
Explain why the current in the circuit is given by the point where the line V = 6.0 − 2.0I crosses the curve.
(1)
(b)
Determine the current in the lamp and the potential difference across it.
(2)
(c)
Calculate the power dissipated inside the battery and the fraction of the power supplied by the battery that is delivered to the lamp.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the same current passes through the battery and the lamp, and the p.d. across the lamp equals the terminal p.d. of the battery, so both relationships must be satisfied at once
✓ 1
Part (b)
line drawn from (0, 6.0 V) through (1.0 A, 4.0 V) / to (1.4 A, 3.2 V)
✓ 1
crosses the curve at I = 1.0 A (allow 0.95–1.05 A), V = 4.0 V (allow 3.8–4.2 V)
Exam tip: When a component is not ohmic, you cannot use I = ε/(R + r) with a fixed R. Draw the source line V = ε − Ir on the component's graph and read off where they cross.
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