A-LevelPhysicsElectricity9702/21 · May/June 2020 · Q512 Marks

Metal wire is used to connect a power supply to a lamp. The wire has a total resistance of 3.4 Ω and the metal has a resistivity of 2.6 × 10⁻⁸ Ω m. The total length of the wire is 59 m. Show that the wire has a cross-sectional area of 4.5 × 10⁻⁷ m².

(a)(i)Metal wire is used to connect a power supply to a lamp. The wire has a total resistance of 3.4 Ω and the metal has a resistivity of 2.6 × 10⁻⁸ Ω m. The total length of the wire is 59 m. Show that the wire has a cross-sectional area of 4.5 × 10⁻⁷ m².[2]
(a)(ii)The potential difference across the total length of wire is 1.8 V. Calculate the current in the wire. current = ..................................................... A[1]
(a)(iii)The number density of the free electrons in the wire is 6.1 × 10²⁸ m⁻³. Calculate the average drift speed of the free electrons in the wire. average drift speed = ..................................................... m s⁻¹[2]
(b)(i)A different wire carries a current. This wire has a part that is thinner than the rest of the wire, as shown in Fig. 5.1. [Figure 5.1] State and explain qualitatively how the average drift speed of the free electrons in the thinner part compares with that in the rest of the wire.[2]
(b)(ii)State and explain whether the power dissipated in the thinner part is the same, less or more than the power dissipated in an equal length of the rest of the wire.[2]
(c)(i)Three resistors have resistances of 180 Ω, 90 Ω and 30 Ω. Sketch a diagram showing how two of these three resistors may be connected together to give a combined resistance of 60 Ω between the terminals shown. Ensure you label the values of the resistances in your diagram.[1]
(c)(ii)A potential divider circuit is produced by connecting the three resistors to a battery of electromotive force (e.m.f.) 12 V and negligible internal resistance. The potential divider circuit provides an output potential difference VOUT of 8.0 V. Fig. 5.2 shows the circuit diagram. [Figure 5.2] On Fig. 5.2, label the resistances of all three resistors and the potential difference VOUT.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[C1] R = ρL / A
[A1] A = (2.6 × 10⁻⁸ × 59) / 3.4 = 4.5 × 10⁻⁷ m²
[A1] I = 1.8 / 3.4 = 0.53 A
[C1] I = Anvq
[A1] v = 0.53 / (4.5 × 10⁻⁷ × 6.1 × 10²⁸ × 1.60 × 10⁻¹⁹) = 1.2 x 10⁻⁴ m s⁻¹
[M1] (cross-sectional) area/A is less
[A1] (I, n, e the same so) average drift speed is greater
[M1] (area is less so) more resistance/R
[A1] (I is the same, so) more power/P
[M1] (P = I²ρL / A so) P ∝ 1/A
[A1] (A is less so) more P
[B1] 180 Ω and 90 Ω resistors shown connected in parallel
[M1] resistors connected in parallel labelled as 180 Ω and 90 Ω and the other resistor labelled as 30 Ω
[A1] VOUT or 8.0 V labelled across the two resistors in parallel