Edexcel IAL Physics: Electric Circuits — 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 — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Electric Circuits revision notes

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Questions

1. State the equation relating current to drift velocity, defining each symbol. [3]

2. Explain why the drift velocity of electrons in a metal is only millimetres per second, yet a lamp lights almost instantly when switched on. [3]

3. A copper wire has a thin section joined to a thick section.

(a) State which section has the greater drift velocity, with a reason. [2] (b) Explain why the thin section becomes hotter. [3]

4. A cell of e.m.f. 1.55 V and internal resistance 0.65 Ω is connected to a 3.20 Ω resistor.

(a) Calculate the current. [2] (b) Calculate the terminal p.d. [2] (c) Calculate the power wasted inside the cell. [2] (d) State and explain the effect on terminal p.d. of connecting a second 3.20 Ω resistor in parallel. [3]

5. Sketch or describe and explain the I–V characteristic of a thermistor (NTC), and explain how it differs from a filament lamp. [4]

6. A 9.0 V supply is connected across a 1.5 kΩ resistor in series with an LDR. The output is taken across the LDR.

(a) In darkness the LDR has resistance 12 kΩ. Calculate the output p.d. [2] (b) State and explain what happens to the output when light shines on the LDR. [3]

7. State the equation relating resistance to resistivity, defining each term, and name the CORE PRACTICAL that investigates it. [3]

8. A wire of resistivity 1.7 × 10⁻⁸ Ω m has length 2.5 m and cross-sectional area 0.50 × 10⁻⁶ m². Calculate its resistance. [2]

9. Using the circuit in Question 4, calculate the power dissipated in the external 3.20 Ω resistor. [2]

10. Explain, in terms of lattice vibrations and conduction electrons, why the resistance of a metallic conductor rises with temperature while the resistance of an NTC thermistor falls. [3]

11. State the two power equations obtained by combining P = VI with V = IR, and state what CORE PRACTICAL 8 determines. [3]

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Answers

1. I = nAvq [1], where n is the number density of charge carriers, A is the cross-sectional area [1], v is the drift velocity and q is the charge on each carrier [1].

2. Metals have a very high number density of free electrons [1], so only a very small drift velocity is needed to carry a large current [1]. The electric field is established throughout the circuit almost instantly, so electrons everywhere begin to drift at once [1].

3. (a) The thin section [1], because I and n are the same throughout and A is smaller, so from I = nAvq, v must be greater [1]. (b) The electrons move faster in the thin section [1], so they collide with the lattice ions more frequently [1], transferring more energy per second to that part of the wire, which therefore heats up more [1].

4. (a) I = E ÷ (R + r) = 1.55 ÷ 3.85 [1] = 0.403 A [1]. (b) V = IR = 0.403 × 3.20 [1] = 1.29 V [1]. (c) P = I²r = 0.4026² × 0.65 [1] = 0.105 W [1]. (d) Terminal p.d. decreases [1]. The parallel combination gives a lower external resistance, so a larger current flows [1], and the lost volts Ir inside the cell increase [1].

5. The curve steepens as V increases [1], because the current heats the thermistor [1] and in a semiconductor heating releases more charge carriers, so resistance falls [1]. In a filament lamp the curve flattens instead, because heating increases lattice vibration and resistance rises — the opposite behaviour [1].

6. (a) V_out = 9.0 × 12 ÷ (1.5 + 12) [1] = 8.0 V [1]. (b) The output decreases [1]. The LDR’s resistance falls in light [1], so it takes a smaller share of the total resistance and therefore a smaller share of the supply p.d. [1].

7. R = ρl/A [1], where ρ is resistivity, l is length and A is cross-sectional area [1] — investigated in CORE PRACTICAL 7, determining the electrical resistivity of a material [1].

8. R = ρl/A = (1.7 × 10⁻⁸ × 2.5) ÷ (0.50 × 10⁻⁶) [1] = 0.085 Ω [1].

9. P = I²R = 0.4026² × 3.20 [1] = 0.519 W [1].

10. In a metal, rising temperature increases lattice vibrations, which impede electron flow more, so resistance rises [1]. In an NTC thermistor, rising temperature releases more conduction electrons, and this increase in charge carriers dominates over the (smaller) increase in lattice vibration, so resistance falls overall [1] [1].

11. P = I²R and P = V²/R [1] [1]. CORE PRACTICAL 8 determines the e.m.f. and internal resistance of a cell, typically from a graph of terminal p.d. against current [1].

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Where marks are usually lost

• Saying electrons travel at the speed of light.
• Explaining the thin-wire heating without I = nAvq.
• Confusing the thermistor and filament lamp characteristics.
• Using the wrong resistance in the potential divider ratio.
• Mixing up ρ, R and A in the resistivity equation, or forgetting to convert area to m².
• Forgetting that power dissipated in the external resistor and the power wasted internally must be calculated separately.