Chapter 9 Test - Mark Scheme Shiblee College of Commerce, Koh-i-Noor Town, Faisalabad PHYS 1

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9 Energy, power, and resistance OCR Physics A Exam-style mark scheme This resource sheet may have been changed from the original 1 Question Answer Marks Guidance 1 a potential difference = ener... gy lost by charges charge flow B1 Do not allow equation copied from data booklet. 1 b One similarity from: both are energy transferred per unit charge both are measured in volt or joule/coulomb. and One difference from: p.d. is electrical energy transferred to other forms per unit charge e.m.f. is energy in other forms (chemical, mechanical, etc.) transferred to electrical energy per unit charge. B1 B1 The direction of transfer is the key factor. 1 c i Q = I Δt = 1.2 ×10−3 × 2 × 602 Q = 8.64C C1 A1 Must convert mA to A and hours to seconds. 1 c ii Energy = ƐQ = 2.8 × 8.64 = 24.2 J A1 Allow e.c.f. if error in part i. 2 a i R from graph = 30 Ω I = 5 30 = 0.167 A C1 A1 2 a ii P = I² R = 0.167² × 30 = 0.833 W A1 Allow e.c.f. if part i was incorrect. Allow P =V 2 R = 52 30 . 2 b i E k = 1 2 mv 2 = eV E k = 1.6 ×10–19 × 80 ×103 = 1.28 ×10–14 J A1 Must convert the p.d. to S.I. units first. 2 b ii eV = 1 2 mv 2 v = 2eV m = 2 ×1.28 ×10−14 1.7 ×10−27 = 3.88 ×106 ms−1 A1 Allow e.c.f. from part i provided method is clear and accurate. 3 a For a metallic wire at constant temperature, the current is (directly) proportional to the p.d. across its ends. B1 B1 No credit for R =V I in any form. 3 b i Straight line through the origin. B1 3 b ii Horizontal section for negative V followed by steep rising curve [Show More]

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