A-LevelPhysicsD.C. circuits9702/21 · May/June 2021 · Q511 Marks

A battery has electromotive force (e.m.f.) 4.0V and internal resistance 0.35Ω. The battery is connected to a uniform resistance wire XY and a fixed resistor of resistance R, as shown in Fig. 5.1. [Figure 5.1] Wire XY has resistance 0.90 Ω. The potential difference across wire XY is 1.8V. A cell of e.m.f. 1.2V is connected to the circuit in (b), as shown in Fig. 5.2. [Figure 5.2] The connection P is moved along the wire XY. The galvanometer reading is zero when distance XP is 0.30 m.

(a)State Kirchhoff's second law.[2]
(b)(i)Calculate: the current in wire XY[1]
(b)(ii)the number of free electrons that pass a point in the battery in a time of 45 s[2]
(b)(iii)resistance R.[2]
(c)(i)Calculate the total length L of wire XY.[2]
(c)(ii)The fixed resistor is replaced by a different fixed resistor of resistance greater than R. State and explain the change, if any, that must be made to the position of P on wire XY so that the galvanometer reading is zero.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[M1] sum of e.m.f.(s) = sum of p.d.(s) or (algebraic) sum of e.m.f.(s) and p.d.(s) is zero
[A1] around a loop/around a closed circuit
[A1] I = 1.8/0.90
[A1] = 2.0 A
[C1] Q = It
[A1] number = (2.0 × 45) / 1.60 × 10⁻¹⁹
[A1] = 5.6 × 10²⁰
[C1] 4.0 = 1.8 + [2.0 × (0.35 + R)] or 4.0 = 2.0 × (0.90 + 0.35 + R)
[A1] R = 0.75 Ω
[C1] 1.2/1.8 = 0.30/L
[A1] L = 0.45 m
[B1] p.d. across XY decreases/p.d. across XP decreases
[B1] (so) P is moved towards Y/away from X/to the right

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