A-LevelPhysicsD.C. circuits9702/22 · May/June 2019 · Q513 Marks

5 (b) A battery of electromotive force (e.m.f.) 5.6V and internal resistance r is connected to two external resistors, as shown in [Figure 5.1]. The reading on the voltmeter is 4.8 V. (c) The battery in (b) is now connected to a battery of e.m.f. 7.2V and internal resistance 3.5Ω. The new circuit is shown in [Figure 5.2].

(a)State Kirchhoff's second law.[2]
(b)(i) 1.the combined resistance of the two resistors connected in parallel[2]
(b)(i) 2.the current in the battery.[2]
(b)(ii)Show that the internal resistance r is 2.5Ω.[2]
(b)(iii)Determine the ratio power dissipated by internal resistance r / total power produced by battery[3]
(c)Determine the current in the circuit.[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.

[M1] sum of e.m.f.(s) = sum of p.d.(s)
[A1] around a loop/around a closed circuit
[C1] 1/R = 1/R1 + 1/R2 (or 1/R = 1/90 + 1/18)
[A1] R = 15 Ω
[C1] I = V/R (or I = 4.8/15 or I = 4.8/90 + 4.8/18)
[A1] I = 0.32 A
[C1] E = V + Ir (or E = I(R + r))
[A1] 5.6 = 4.8 + 0.32 r so r = 2.5 (Ω) (or 5.6 = 0.32 × (15 + r) so r = 2.5 (Ω))
[C1] P = EI or P = VI or P = I²R or P = V²/R
[C1] ratio = (0.32² × 2.5) / (5.6 × 0.32) or 0.256 / 1.792
[A1] = 0.14
[C1] 7.2 – 5.6 – 2.5I – 3.5I = 0
[A1] I = 0.27 A

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