Physics (AS & A Level) · Oct / Nov 2016 · Paper 22 · Question 5
9702
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5(a)[2]
total/sum of electromotive forces or e.m.f.s = total/sum of potential differences or p.d.s around a loop/(closed) circuit [2]
5(b)(i)[2]
(current in battery =) current in A + current in B or IA + IB (I =) 0.14 + 0.26 = 0.40 A [2]
5(b)(ii)[2]
E = V + Ir 6.8 = 6.0 + 0.40r or 6.8 = 0.40 (15 + r) r = 2.0 Ω [2]
5(b)(iii)[2]
R = V / I ratio (= RA / RB) = (6.0 / 0.14) / (6.0 / 0.26) = 42.9 / 23.1 or 0.26 / 0.14 = 1.9 (1.86) [2]
5(b)(iv)[4]
1. P = EI or VI or P = I 2R or P = V 2 / R = 6.8 × 0.40 = 0.402 × 17 = 6.82 / 17 = 2.7 W (2.72 W) [2]
2. output power = VI = 6.0 × 0.40 (= 2.40 W) efficiency = (6.0 × 0.40) / (6.8 × 0.40) = 2.40 / 2.72 = 0.88 or 88% (allow 0.89 or 89%) [2]
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