A Level Physics 9.3 Resistance and Resistivity MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice questions cover Cambridge AS & A Level Physics 9702 Section 9.3, Resistance and Resistivity: Ohm’s law, the resistivity equation ρ = RA / L, I-V characteristics and the effect of temperature on resistance. Work through each question, then open the Show Answer toggle to check your answer against a fully worked explanation.
Formula recap: R = V / I | ρ = RA / L, so R = ρL / A | A = πr2 for a wire of radius r | resistivity of copper ρ = 1.7 × 10−8 Ω m. See the full A Level Physics Formula Sheet for every equation in one place.
Resistance & Ohm’s Law
Q1. A component has a potential difference of 12 V across it and a current of 3.0 A through it. What is its resistance?
- (A) 0.25 Ω
- (B) 4.0 Ω
- (C) 36 Ω
- (D) 0.40 Ω
Show Answer
Answer: B – Resistance is defined by R = V / I = 12 / 3.0 = 4.0 Ω.
Q2. Which statement correctly states Ohm’s law?
- (A) The current through any conductor is proportional to the potential difference for all temperatures.
- (B) The potential difference is proportional to the current only when the resistance is changing.
- (C) The current through a conductor is proportional to the potential difference across it, provided its temperature stays constant.
- (D) The resistance of a conductor is proportional to the current through it.
Show Answer
Answer: C – Ohm’s law states I ∝ V at constant temperature; a conductor obeying it has a constant resistance.
Q3. A 15 Ω resistor is connected across a supply. The current through it is 0.40 A. What is the potential difference across the resistor?
- (A) 6.0 V
- (B) 37.5 V
- (C) 0.027 V
- (D) 15 V
Show Answer
Answer: A – V = IR = 0.40 × 15 = 6.0 V.
Q4. Which component is a non-ohmic conductor under normal operating conditions?
- (A) A copper wire kept at constant temperature
- (B) A fixed resistor kept at constant temperature
- (C) A metal wire kept at constant temperature
- (D) A semiconductor diode
Show Answer
Answer: D – A diode does not obey Ohm’s law: its resistance changes greatly with the applied p.d. and it conducts in one direction only. The others are ohmic while their temperature is constant.
Q5. On an I-V graph (current on the y-axis, p.d. on the x-axis) a resistor gives a straight line through the origin passing through the point (6.0 V, 3.0 A). What is the resistance?
- (A) 0.50 Ω
- (B) 2.0 Ω
- (C) 3.0 Ω
- (D) 18 Ω
Show Answer
Answer: B – For I plotted against V, the gradient = I / V = 1 / R. Gradient = 3.0 / 6.0 = 0.50 A V−1, so R = 1 / 0.50 = 2.0 Ω (equivalently R = V / I = 6.0 / 3.0).
Q6. The ohm, the unit of resistance, can be expressed as
- (A) V A−1
- (B) A V−1
- (C) V C−1
- (D) J C−1
Show Answer
Answer: A – From R = V / I, 1 Ω = 1 V A−1: a conductor has a resistance of 1 Ω when a p.d. of 1 V drives a current of 1 A.
Q7. A current of 0.25 A flows through a 20 Ω resistor. What is the potential difference across it?
- (A) 0.0125 V
- (B) 80 V
- (C) 2.0 V
- (D) 5.0 V
Show Answer
Answer: D – V = IR = 0.25 × 20 = 5.0 V.
Q8. Resistors of 4.0 Ω and 6.0 Ω are connected in series. What is their total resistance?
- (A) 2.4 Ω
- (B) 5.0 Ω
- (C) 10 Ω
- (D) 24 Ω
Show Answer
Answer: C – In series, resistances add: R = 4.0 + 6.0 = 10 Ω.
Q9. Resistors of 6.0 Ω and 3.0 Ω are connected in parallel. What is their total resistance?
- (A) 2.0 Ω
- (B) 4.5 Ω
- (C) 9.0 Ω
- (D) 18 Ω
Show Answer
Answer: A – In parallel, 1 / R = 1 / 6.0 + 1 / 3.0 = 0.50, so R = 2.0 Ω (always less than the smallest resistor).
Q10. A conductor is described as ohmic. This means that, at constant temperature,
- (A) the current through it is the same in both directions only
- (B) the ratio V / I is constant
- (C) its resistance increases as the current increases
- (D) its I-V graph is a curve through the origin
Show Answer
Answer: B – An ohmic conductor has constant resistance at constant temperature, so V / I is constant and the I-V graph is a straight line through the origin.
Resistivity
Q11. The resistivity ρ of a wire of resistance R, length L and cross-sectional area A is given by
- (A) ρ = RL / A
- (B) ρ = A / (RL)
- (C) ρ = RA / L
- (D) ρ = L / (RA)
Show Answer
Answer: C – Starting from R = ρL / A and rearranging gives ρ = RA / L.
Q12. A copper wire (resistivity 1.7 × 10−8 Ω m) has length 2.0 m and diameter 1.0 mm. What is its resistance?
- (A) 0.043 Ω
- (B) 0.43 Ω
- (C) 4.3 Ω
- (D) 0.0034 Ω
Show Answer
Answer: A – Radius r = 0.50 mm = 5.0 × 10−4 m, so A = πr2 = π × (5.0 × 10−4)2 = 7.9 × 10−7 m2. Then R = ρL / A = (1.7 × 10−8 × 2.0) / (7.9 × 10−7) = 0.043 Ω.
Q13. A uniform wire has resistance 8.0 Ω. An identical wire of the same material and diameter but twice the length has resistance
- (A) 2.0 Ω
- (B) 4.0 Ω
- (C) 8.0 Ω
- (D) 16 Ω
Show Answer
Answer: D – R = ρL / A, so resistance is proportional to length: doubling L doubles R to 16 Ω.
Q14. A wire of resistance 12 Ω is replaced by a wire of the same material and length but twice the diameter. The new resistance is
- (A) 24 Ω
- (B) 3.0 Ω
- (C) 6.0 Ω
- (D) 48 Ω
Show Answer
Answer: B – Doubling the diameter doubles the radius, so the area A = πr2 increases by a factor of 4. Since R = ρL / A, the resistance falls to one quarter: 12 / 4 = 3.0 Ω.
Q15. A wire of length 1.5 m and cross-sectional area 3.0 × 10−7 m2 has resistance 0.50 Ω. What is the resistivity of the material?
- (A) 1.0 × 10−8 Ω m
- (B) 1.0 × 10−7 Ω m
- (C) 2.3 × 10−7 Ω m
- (D) 2.5 × 106 Ω m
Show Answer
Answer: B – ρ = RA / L = (0.50 × 3.0 × 10−7) / 1.5 = 1.0 × 10−7 Ω m.
Q16. A wire has a circular cross-section of diameter 2.0 mm. What is its cross-sectional area?
- (A) 3.1 × 10−6 m2
- (B) 1.3 × 10−5 m2
- (C) 3.1 × 10−3 m2
- (D) 1.6 × 10−6 m2
Show Answer
Answer: A – Use the radius, not the diameter: r = 1.0 mm = 1.0 × 10−3 m, so A = πr2 = π × (1.0 × 10−3)2 = 3.1 × 10−6 m2. Option B is the trap of using the diameter as the radius.
Q17. Wire X has twice the length and half the diameter of wire Y. Both wires are made of the same material. The ratio resistance of X / resistance of Y is
- (A) 2
- (B) 4
- (C) 8
- (D) 16
Show Answer
Answer: C – Doubling L doubles R. Halving the diameter quarters the area, which multiplies R by 4. Total effect: 2 × 4 = 8.
Q18. The unit of resistivity is
- (A) Ω m−1
- (B) Ω m2
- (C) Ω−1 m
- (D) Ω m
Show Answer
Answer: D – From ρ = RA / L, the unit is Ω × m2 / m = Ω m.
Q19. Resistivity, rather than resistance, is quoted as a property of a material because resistivity
- (A) depends only on the material (and its temperature), not on the size or shape of the sample
- (B) is always smaller than the resistance
- (C) does not change with temperature
- (D) is the same for all metals
Show Answer
Answer: A – Resistance depends on length and cross-sectional area as well as the material; resistivity removes the geometry, so it characterises the material itself at a given temperature.
Q20. A wire made of a material of resistivity 1.7 × 10−8 Ω m has cross-sectional area 1.0 × 10−7 m2 and resistance 1.7 Ω. What is its length?
- (A) 1.0 m
- (B) 5.0 m
- (C) 10 m
- (D) 100 m
Show Answer
Answer: C – L = RA / ρ = (1.7 × 1.0 × 10−7) / (1.7 × 10−8) = 10 m.
I-V Characteristics & Temperature
Q21. The I-V characteristic of an ohmic resistor at constant temperature is
- (A) a curve whose gradient increases with V
- (B) a curve whose gradient decreases with V
- (C) a horizontal line
- (D) a straight line through the origin
Show Answer
Answer: D – I ∝ V for an ohmic component, so the graph is a straight line through the origin and its gradient (1 / R) is constant.
Q22. As the p.d. across a filament lamp is increased from zero, its I-V graph curves so that the gradient decreases. This is because
- (A) the filament’s resistance increases as it heats up
- (B) the filament’s resistance decreases as it heats up
- (C) the current stops increasing
- (D) the filament becomes a superconductor
Show Answer
Answer: A – The current heats the filament, raising its temperature and resistance. Since gradient = 1 / R, the gradient falls as V increases.
Q23. A filament lamp is operated at increasing voltages. Its resistance
- (A) stays constant, because it obeys Ohm’s law
- (B) decreases, because more charge carriers are released
- (C) increases, because the filament gets hotter
- (D) becomes zero at high voltage
Show Answer
Answer: C – A filament lamp is non-ohmic: the larger current raises the filament temperature, and the resistance of the metal increases with temperature.
Q24. The resistance of a metal wire increases when its temperature rises because
- (A) the number of free electrons decreases
- (B) the lattice ions vibrate with greater amplitude, so the electrons collide with them more frequently
- (C) the electrons drift more slowly because they get heavier
- (D) the wire expands and becomes thinner
Show Answer
Answer: B – Heating makes the positive lattice ions vibrate more, increasing the collision rate for the drifting electrons, which reduces the current for a given p.d. – i.e. the resistance rises.
Q25. For a semiconductor diode, the I-V characteristic shows that
- (A) current is proportional to p.d. in both directions
- (B) a large current flows for a small reverse p.d.
- (C) current flows equally well in both directions above 0.6 V
- (D) almost no current flows in reverse bias, and forward current rises sharply only above about 0.6 V
Show Answer
Answer: D – A diode conducts in the forward direction only, with a sharp turn-on at about 0.6 V; in reverse bias the current is approximately zero.
Q26. Which description best matches the I-V graph of a diode in the forward direction?
- (A) A straight line through the origin
- (B) Almost zero current until about 0.6 V, then a steeply rising curve
- (C) A curve of decreasing gradient starting at the origin
- (D) A constant current for all voltages
Show Answer
Answer: B – Below the turn-on p.d. (about 0.6 V for silicon) the diode passes negligible current; above it the current rises very steeply with p.d.
Q27. A negative temperature coefficient (NTC) thermistor is warmed. Its resistance decreases mainly because
- (A) the lattice ions vibrate less
- (B) the thermistor contracts
- (C) heating releases more charge carriers, increasing the number density of electrons available to conduct
- (D) its length increases
Show Answer
Answer: C – In a semiconductor thermistor, a higher temperature frees more charge carriers. This effect outweighs the increased lattice vibration, so the resistance falls as temperature rises.
Q28. A thermistor is most suitable for use in
- (A) a temperature sensor in a potential divider circuit, such as a thermostat or fire alarm
- (B) a component that must keep a constant resistance at all temperatures
- (C) rectifying an alternating current
- (D) measuring the current in a circuit
Show Answer
Answer: A – Because its resistance changes strongly and predictably with temperature, an NTC thermistor in a potential divider produces an output voltage that varies with temperature – ideal for sensing and switching circuits.
Q29. The resistance of a light-dependent resistor (LDR) changes with light intensity so that
- (A) it increases as the light gets brighter
- (B) it is unaffected by light
- (C) it becomes zero in daylight
- (D) it decreases as the light gets brighter, because more charge carriers are released
Show Answer
Answer: D – Brighter light releases more charge carriers in the semiconductor of the LDR, so its resistance falls (from around 106 Ω in darkness to a few hundred ohms in bright light).
Q30. At one point on the I-V curve of a filament lamp, the p.d. is 6.0 V and the current is 0.40 A. The resistance of the filament at this operating point is
- (A) 2.4 Ω
- (B) 15 Ω
- (C) 0.067 Ω
- (D) 6.4 Ω
Show Answer
Answer: B – Even for a non-ohmic device, the resistance at an operating point is R = V / I = 6.0 / 0.40 = 15 Ω. Note this is not the gradient of the curve, and the value changes at other operating points.
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