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A-LevelPhysicsD.C. circuits9702/23 · May/June 2019 · Q610 Marks

A battery of electromotive force (e.m.f.) E and internal resistance 1.5 Ω is connected to a network of resistors, as shown in Fig. 6.1. [Figure 6.1] Resistor X has a resistance of 8.0 Ω. Resistor Y has a resistance of 2.0 Ω. Resistor Z has a resistance of R₂. The current in X is 0.60 A and the current in Y is 1.8 A.

  1. (a)Define the ohm. ......................................................................................................................................... [1][1]
  2. (b)A battery of electromotive force (e.m.f.) E and internal resistance 1.5 Ω is connected to a network of resistors, as shown in Fig. 6.1. [Figure 6.1] Resistor X has a resistance of 8.0 Ω. Resistor Y has a resistance of 2.0 Ω. Resistor Z has a resistance of R₂. The current in X is 0.60 A and the current in Y is 1.8 A. (i) Calculate:
  3. (b)(i)1.the current I in the battery I = ................................................................................. A [1][1]
  4. (b)(i)2.resistance R₂ R₂ = ................................................................................. Ω [2][2]
  5. (b)(i)3.e.m.f. E. E = ................................................................................. V [2][2]
  6. (b)(ii)Resistors X and Y are each made of wire. The two wires have the same length and are made of the same metal. Determine the ratio: 1. cross-sectional area of wire X cross-sectional area of wire Y ratio = .............................................................................. [2][2]
  7. (b)(ii)2.average drift speed of free electrons in X average drift speed of free electrons in Y ratio = .............................................................................. [2][2]

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Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

  1. [B1] volt / ampere
  2. [A1] I = 1.8 + 0.60 = 2.4 A
  3. [C1] (8.0 × 0.60) = 1.8 × (2.0 + Rz)
  4. [A1] Rz = 0.67 Ω
  5. [C1] E − (2.4 × 1.5) = (0.60 × 8.0) or E − (2.4 × 1.5) = 1.8 × (2.0 + 0.67) or E = 2.4 × [1.5 + (8.0 × 2.67)/ (8.0 + 2.67)]
  6. [A1] E = 8.4 V
  7. [C1] R = ρL/A or R ∝ 1/A
  8. [A1] ratio = Ry/Rx = 2.0/8.0 = 0.25
  9. [C1] I ∝ Av or Ix / Iү = Axvx / Ayvy
  10. [A1] ratio = (0.60 / 1.8) × (1 / 0.25) = 1.3

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About this question

Topic
D.C. circuits, A-Level Physics (syllabus 9702). Structured question, 10 marks.
Source
A-Level Physics, May/June 2019 Paper 2 Variant 3, question 6.

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