A-LevelPhysicsD.C. circuits9702/21 · May/June 2019 · Q610 Marks

A battery of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 6.1. [Figure 6.1] The current in the circuit is I and the potential difference across the variable resistor is V.

(a)Explain, in terms of energy, why V is less than E.[1]
(b)State an equation relating E, I, r and V.[1]
(c)The resistance R of the variable resistor is varied. The variation with I of V is shown in Fig. 6.2. [Figure 6.2] Use Fig. 6.2 to:
(c)(i)explain how it may be deduced that the e.m.f. of the battery is 2.8 V[1]
(c)(ii)calculate the internal resistance r.[2]
(d)The battery stores 9.2 kJ of energy. The variable resistor is adjusted so that V = 2.1 V. Use Fig. 6.2 to:
(d)(i)calculate resistance R[1]
(d)(ii)calculate the number of conduction electrons moving through the battery in a time of 1.0 s[1]
(d)(iii)determine the time taken for the energy in the battery to become equal to 1.6 kJ. (Assume that the e.m.f. of the battery and the current in the battery remain constant.)[3]

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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] energy is dissipated in the internal resistance
[1] E = V + Ir
[1] (graph shows) maximum value of potential difference is 2.8 (V) or (graph shows) when current/I (from battery) is zero, V is 2.8 (V) / E
[1] r = (-)gradient or r = (E – V) / I or substituted values from the graph for E, V and I
[1] r = 1.4 Ω
[1] R = 2.1/0.50 = 4.2 Ω
[1] number = (0.50 / (1.60 × 10⁻¹⁹)) = 3.1 × 10¹⁸
[1] energy = EIt or P = EI and P = W/t
[1] (9.2 – 1.6) × 10³ = 2.8 × 0.50 × t
[1] t = 5.4 x 10³ s

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