IB DP Physics: Electric current, resistance and circuits (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topic B.5, Current and circuits, from the IB Physics guide (first assessment 2025). The content is the same for SL and HL. Meters are ideal unless a question says otherwise. Use e = 1.60 × 10⁻¹⁹ C where needed. Read the study guide first if a topic is new, and use the revision notes for recall. See the course hub and the printable checklist. To find weak areas first, take the free 10-minute diagnostics.

Questions

1. A phone battery charges at a steady 3.2 mA for 5.0 hours, with a p.d. of 3.7 V across it. (a) Calculate the charge that flows. [2] (b) Calculate the energy transferred to the battery. [2]

2. Explain, in terms of charge carriers, why copper conducts while a plastic does not, and why the resistance of a copper wire rises as it warms. [4]

3. A constantan wire has length 0.60 m and diameter 1.2 mm. Its resistivity is 4.9 × 10⁻⁷ Ω m (given). (a) Calculate its resistance. [3] (b) A second wire of the same material is twice as long and has half the diameter. Determine its resistance. [2]

4. The table gives readings for a small filament lamp.

p.d. / V	1.0	2.0	4.0	6.0
current / A	0.10	0.16	0.24	0.30

(a) Calculate the resistance of the lamp at 1.0 V and at 6.0 V. [2] (b) State whether the lamp is ohmic, and explain the change in resistance in terms of the lattice. [4]

5. Two resistors of 6.0 Ω and 12 Ω are connected across an ideal 12 V supply, first in series and then in parallel. Calculate the total power dissipated in each arrangement and state the ratio of parallel power to series power. [5]

6. An ideal 12 V cell is connected to a 10 Ω resistor in series with a parallel pair of 15 Ω and 30 Ω resistors. Calculate: (a) the current from the cell, [3] (b) the current in the 30 Ω resistor. [3]

7. A student measures the terminal p.d. of a cell at two currents: 1.38 V at 0.20 A and 1.14 V at 0.60 A. (a) Determine the internal resistance and the emf of the cell. [4] (b) Calculate the current if the cell is short-circuited. [1] (c) Calculate the external resistance when the current is 0.50 A. [2]

8. A 12 V supply is connected across a 4.0 kΩ fixed resistor in series with an LDR. The LDR has resistance 1.0 kΩ in bright light and 20 kΩ in the dark. (a) Calculate the p.d. across the LDR in bright light and in the dark. [3] (b) The fixed resistor and the LDR are swapped. State and explain how the p.d. across the LDR in the dark changes. [2]

9. A 12 V cell of negligible internal resistance is connected to two 5.0 kΩ resistors in series. A voltmeter of constant resistance 5.0 kΩ is connected across one of the resistors. Calculate the voltmeter reading, and state what an ideal voltmeter would read. [4]

10. A solar panel of area 1.6 m² receives light of intensity 800 W m⁻² and is 18% efficient. (a) Calculate the electrical energy it delivers in 3.0 hours. [3] (b) Give one advantage and one disadvantage of this panel compared with a chemical cell as a source for a remote sensor. [2]

11. A heater coil is made from nichrome wire of diameter 0.30 mm and resistivity 1.1 × 10⁻⁶ Ω m (given). It is connected to a cell of emf 12 V and internal resistance 0.50 Ω. The coil resistance is 5.5 Ω. (a) Calculate the length of wire in the coil. [3] (b) Calculate the current and the power dissipated in the coil. [3] (c) Calculate the power dissipated inside the cell and the fraction of the total power that heats the coil. [3]

Answers

1. (a) Time = 5.0 × 3600 = 18 000 s [1]. q = IΔt = 3.2 × 10⁻³ × 18 000 = 58 C [1]. [2] (b) W = Vq [1] = 3.7 × 57.6 = 2.1 × 10² J (213 J) [1]. [2] Examiner insight: Converting hours to seconds before substituting is where most slips occur; a unit missing from the final line often costs the answer line.

2. Copper has many free (mobile) electrons [1]; plastic has almost no mobile charge carriers, so the same p.d. gives negligible current [1]. On warming, the lattice ions vibrate with larger amplitude [1], so electrons collide with them more often, which lowers their mobility and raises the resistance [1]. [4] Examiner insight: Answers earn credit for naming the particles involved (free electrons, lattice ions) and the collision link; “it gets harder for current to flow” on its own is too vague.

3. (a) r = 0.60 mm, so A = π(0.60 × 10⁻³)² = 1.13 × 10⁻⁶ m² [1]. R = ρL/A [1] = (4.9 × 10⁻⁷ × 0.60) ÷ 1.13 × 10⁻⁶ = 0.26 Ω [1]. [3] (b) R ∝ L/A, and A is a quarter, so R is 2 × 4 = 8 times larger [1]: 2.1 Ω [1]. [2] Examiner insight: Show the radius step explicitly; using the diameter in place of the radius in A is a common way to lose all the later credit.

4. (a) At 1.0 V: R = 1.0 ÷ 0.10 = 10 Ω [1]. At 6.0 V: R = 6.0 ÷ 0.30 = 20 Ω [1]. [2] (b) Not ohmic, because R is not constant (current is not proportional to p.d.) [1]. More current heats the filament [1]. The lattice ions vibrate with larger amplitude [1], so there are more collisions with electrons and R rises [1]. [4] Examiner insight: Quote values from the table to support “not ohmic”; a bare statement about the curve gains less than a comparison of two resistances.

5. Series: R = 18 Ω [1], P = V²/R = 144 ÷ 18 = 8.0 W [1]. Parallel: Rp = (6.0 × 12) ÷ 18 = 4.0 Ω [1], P = 144 ÷ 4.0 = 36 W [1]. Ratio = 36 ÷ 8.0 = 4.5 [1]. [5] Examiner insight: V²/R is the quickest form because the supply p.d. is fixed; choosing a form that needs an unknown current adds steps.

6. (a) 1/Rp = 1/15 + 1/30 = 3/30, so Rp = 10 Ω [1]. Total R = 20 Ω [1]. I = 12 ÷ 20 = 0.60 A [1]. [3] (b) p.d. across the pair = 0.60 × 10 = 6.0 V [1]. Current in 30 Ω = 6.0 ÷ 30 [1] = 0.20 A [1]. [3] Examiner insight: Use the p.d. across the parallel pair, not the supply p.d.; checking that the branch currents add to 0.60 A catches this.

7. (a) ε = V + Ir for both readings [1]. 1.38 + 0.20r = 1.14 + 0.60r [1], so r = 0.24 ÷ 0.40 = 0.60 Ω [1]. ε = 1.38 + 0.20 × 0.60 = 1.50 V [1]. [4] (b) I = ε/r = 1.50 ÷ 0.60 = 2.5 A [1]. [1] (c) ε = I(R + r), so R + r = 1.50 ÷ 0.50 = 3.0 Ω [1]; R = 3.0 − 0.60 = 2.4 Ω [1]. [2] Examiner insight: Stating the relation you use before substituting makes your method clear even if an arithmetic slip follows.

8. (a) Bright: V = 12 × 1.0 ÷ 5.0 = 2.4 V [1]. Dark: V = 12 × 20 ÷ 24 = 10 V [1]. The share of the p.d. is proportional to the resistance in a series pair [1]. [3] (b) After the swap the fixed resistor is 4.0 kΩ, the LDR is still 20 kΩ in the dark, so nothing changes in series: p.d. across the LDR stays 10 V [1] because the order of series components does not alter the current or the individual p.d.s [1]. [2] Examiner insight: A reasoned “no change” is a complete answer; saying the p.d. changes without checking the series relationship is a common trap.

9. The voltmeter is in parallel with 5.0 kΩ, so the pair has 1/R = 1/5.0 + 1/5.0, R = 2.5 kΩ [1]. Total = 2.5 + 5.0 = 7.5 kΩ [1]. Reading = 12 × 2.5 ÷ 7.5 = 4.0 V [1]. Ideal voltmeter: 6.0 V [1]. [4] Examiner insight: Compare the two readings: the difference shows why the guide’s ideal-meter assumption matters.

10. (a) Power in = 800 × 1.6 = 1280 W [1]. Electrical power = 0.18 × 1280 = 230 W [1]. Energy = 230.4 × 10 800 = 2.5 × 10⁶ J [1]. [3] (b) Advantage: no fuel is used up, so it can recharge from sunlight (or: no emissions in use) [1]. Disadvantage: output depends on light, so it fails at night or in heavy cloud [1]. [2] Examiner insight: Tie the pair to the remote-sensor context; generic points score less than ones that mention reliability or maintenance.

11. (a) A = π(0.15 × 10⁻³)² = 7.07 × 10⁻⁸ m² [1]. L = RA/ρ [1] = (5.5 × 7.07 × 10⁻⁸) ÷ 1.1 × 10⁻⁶ = 0.35 m [1]. [3] (b) I = ε/(R + r) = 12 ÷ 6.0 = 2.0 A [1]. P = I²R = 2.0² × 5.5 [1] = 22 W [1]. [3] (c) Power in cell = I²r = 4.0 × 0.50 = 2.0 W [1]. Total = 22 + 2.0 = 24 W (= εI) [1]. Fraction = 22 ÷ 24 = 0.92 [1]. [3] Examiner insight: Using I²R with the same current for both the coil and the cell keeps the working short; mixing in V²/R with the wrong p.d. is a frequent source of error.

Where marks are usually lost

• Not converting hours, minutes, millimetres or kilohms before substituting.
• Using the diameter rather than the radius in A = πr².
• Quoting “R rises” without explaining it using lattice vibrations and collisions.
• Using the supply p.d. across one part of a series-parallel circuit.
• Treating the terminal p.d. as the emf when current flows through a real cell.
• Forgetting to invert 1/Rp at the end.
• Giving a one-sided answer when asked to compare energy sources.
• Leaving units off final answers or over-rounding intermediate values.

Next steps

• Recap with the revision notes.
• Re-read the study guide for any topic that felt shaky.
• Visit the course hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Or book a free trial class.

Official syllabus

Source: Physics guide, first assessment 2025, International Baccalaureate Organization (Diploma Programme), sub-topic B.5 Current and circuits.