Grade 11 Physical Sciences · Electricity and Magnetism

Use emf and internal resistance

Use emf = I(R + r) to find a battery's emf, internal resistance, current or lost volts.

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Question 1.8 · May/June 2025 · Paper 1 · National (DBE) · 2 marks

Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 E.

1.8 The circuit below was used in a practical investigation. The battery has emf

ε

and internal resistance r. Ignore the resistance of the connecting wires.

Graphs K and L below were obtained from the readings taken.

Which ONE of the combinations below CORRECTLY indicates the voltmeter readings taken to obtain graphs K and L?(2) Diagram for question 1.8

A Graph K: $V_{1}$ ; Graph L: $V_{1}$
B Graph K: $V_{1}$ ; Graph L: $V_{2}$
C Graph K: $V_{2}$ ; Graph L: $V_{1}$
D Graph K: $V_{2}$ ; Graph L: $V_{2}$

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: use emf and internal resistance

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3. Question 8.2.4 · November 2024 · Paper 1 · National (DBE) · 4 marks
$L_{1}$ and $L_{2}$ are two light bulbs that have the following ratings:

$L_{1}$ : $36$ W ; $20$ V and $L_{2}$ : $48$ W ; $32$ V

The two bulbs are connected as shown in the circuit diagram below. The battery has an internal resistance of $0, 6 Ω$ while the conducting wires and the ammeters have negligible resistance. $R_{1}$ and $R_{2}$ are resistors.
8.2 If both light bulbs operate as RATED, calculate the:
8.2.4 Emf of the battery(4)
Show the memo (4 ticks)Memo
1. ✓
  Formula $V_{2} = I_{TOT} R_{2} = 3, 3 (4) = 13, 2$ V
  
  $ε = V_{e x t} + I r$
  1 mark · Use emf and internal resistance
2. ✓
  External potential difference $= (13, 2 + 32) + \dots$
  1 mark · Use emf and internal resistance
3. ✓
  Lost volts $\dots + 3, 3 (0, 6)$
  1 mark · Use emf and internal resistance
4. ✓
  Answer $ε = 47, 18$ V
  1 mark · Use emf and internal resistance
4. Question 8.2.2 · November 2023 · Paper 1 · National (DBE) · 3 marks
The circuit diagram below shows a battery with an emf of $12$ V and an internal resistance of $0, 5 Ω$ connected to three resistors, a light bulb, a switch, an ammeter and connecting wires. The ammeter and connecting wires have negligible resistance.
8.2 Switch S is initially CLOSED.

Calculate the:
8.2.2 Reading on the ammeter(3)
Show the memo (3 ticks)Memo
1. ✓
  Formula $ε = I (R + r)$
  1 mark · Use emf and internal resistance
2. ✓
  Substitution $12 = I (7, 5 + 0, 5)$
  1 mark · Use emf and internal resistance
3. ✓
  Answer with unit $I = 1, 5$ A
  1 mark · Use emf and internal resistance
5. Question 8.2.2 · May/June 2022 · Paper 1 · National (DBE) · 4 marks
In the circuit below a battery of UNKNOWN emf and an internal resistance of $0, 5 Ω$ is connected to two resistors of $4 Ω$ and $8 Ω$ each, and a resistor R of unknown resistance.

Ignore the resistance of the connecting wires.
8.2 When switch S is OPEN, voltmeter V $_{1}$ reads $3, 2$ V.

Calculate the:
8.2.2 Emf of the battery(4)
Show the memo (4 ticks)Memo
1. ✓
  Formula $ε = I (R + r)$
  1 mark · Use emf and internal resistance
2. ✓
  External resistance $ε = 0, 8 (4 + 8 + \dots)$
  1 mark · Use emf and internal resistance
3. ✓
  Internal resistance $ε = 0, 8 (4 + 8 + 0, 5)$
  1 mark · Use emf and internal resistance
4. ✓
  Answer with unit $ε = 10$ V
  1 mark · Use emf and internal resistance
6. Question 8.4 · November 2025 · Paper 1 · National (DBE) · 6 marks
Two identical headlights, $L_{1}$ and $L_{2}$ , and a starter motor, M, of a car are connected to a battery, as shown in the circuit diagram below. The resistance of each headlight is $12 Ω$ , while the resistance of the starter motor is $0, 1 Ω$ . The emf ( $ε$ ) and internal resistance (r) of the battery are unknown.

The ammeters and the connecting wires have negligible resistance, while the voltmeter has a high resistance. Switches $S_{1}$ and $S_{2}$ are initially open.
8.4 Calculate the emf of the battery.(6)
Show the memo (6 ticks)Memo
1. ✓
  Parallel resistance of the headlights $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$
  $\frac{1}{R _{p}} = \frac{1}{12} + \frac{1}{12}$
  $R_{p} = 6 Ω$
  (OR $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{( 12 ) ( 12 )}{12 + 12} = 6 Ω$ )
  1 mark · Calculate equivalent resistance
2. ✓
  FormulaWhen $S_{1}$ closed and $S_{2}$ open:
  $ε_{1} = I (R + r)$
  1 mark · Use emf and internal resistance
3. ✓
  Substitution: headlights circuit $ε_{1} = 2, 24 (6 + r)$
  1 mark · Use emf and internal resistance
4. ✓
  Substitution: motor circuitWhen $S_{1}$ open and $S_{2}$ closed:
  $ε_{2} = I (R + r) = 120 (0, 1 + r)$
  1 mark · Use emf and internal resistance
5. ✓
  Equating $ε_{1} = ε_{2}$ (equating)
  $2, 24 (6 + r) = 120 (0, 1 + r)$
  $r = 0, 012 Ω$
  1 mark · Use emf and internal resistance
6. ✓
  Answer with unit $ε = I (R + r) = 2, 24 (6 + 0, 012) = 13, 47$ V
  OR $ε = I R + I r = 120 (0, 1 + 0, 012) = 13, 47$ V
  1 mark · Use emf and internal resistance

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