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AQA AS Physics P12 Electricity Kerboodle Answers

This page contains the AQA AS Physics P12 Electricity  Kerboodle Answers for revision and understanding of  Electricity. This page also contains the link to the notes and video for the revision of this topic. 

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C12.1 Current and charge AQA AS Physics C12 Electricity Kerboodle Answers: Page No. 203

1.Ans- a i) 3.5 c ii in 10 min. 210 c b Calculate the average current in a wire through which a charge of 15 C passes in i ss, 3.0A ii 100 s. 0.15 A 2.Ans- a HA 3.8x 1015 b S.OA. 1.9 x 1022 3.Ans- a the charge flowing along the beam each minute 72 mC b the number of electrons that pass along the beam each minute. 4.5 x 1017 4 .ANS- 1600 s b 0.1 A. 8000 s

12.2 Potential difference and power AQA AS Physics C12 Electricity Kerboodle Answers : Page No. 205

1. 29kJ b O.OSX 720J 2.a )the current through the light bulb 2.0A b the energy transfer to the light bulb in 1800 s. 22kJ 3. i) 48kJ ii 3.5A 4.a) 12 kJ b. 4.5W c 2700s

12.3 Resistance AQA AS Physics C12 Electricity Kerboodle Answers: Page No. 208

1. 6.0.0, I OV, 0.125 mA, 160.0, 2.5 mA b 7.5.0 2.Ans- 310 3.Ans- o.11mo 4.a 1.8 x10-6 nm b33mm

12.4 Components and their characteristicsAQA AS Physics C12 Electricity Kerboodle Answers : Page No. 211

1.a.0.25A, 12D b The filament would become brighter and hotter until it melts and breaks as a result. 2.Ans- a at20’C 0.03mA 3.a. b The diode would then be ‘reverse-biased’ so the current in the circuit would be negligible. 4.a. the resistance at SO ‘C 30.4 n b. the temperature when the resistance is 30.00. 46°C Baneer 6

Practice questions: Page No. 212-213

(b) (c) 2 .(a) ΔQ = I Δt = 40 × 10-3 × 3 × 60 = 7.2 C (b) (c) (d) 3 (a) (b) Current is zero for all negative voltages For positive voltages, curve showing a steep rise in current starting at about +0.6 V (with voltage not greatly to exceed 1 V) 4. . (a) (i) Readings need to be taken more quickly than could be achieved by manual timing (ii) Rate: more than 40 samples per second Reason: the current rises very rapidly over the first 0.1 s, and you should have about 4 results on this first section (b) Initial resistance is low so initial current is high • Temperature of filament increases (or filament heats up) • Resistance increases as temperature rises • Increase in resistance causes current to fall • Current is steady when energy supplied = energy lost from filament (or when temperature is constant) • Maximum heating is produced at start when current is highest • Melting of filament causes it to fail (could be mechanical failure caused by temperature rise) • When switched on energy is supplied more rapidly than it is lost so filament melts 5. (a) (b). 6. (a) (i)Circuit to show: • Battery, switch, wire, variable resistor and ammeter in series • Voltmeter in parallel with wire (ii) Switch on, measure I and V Change length of wire Measure new I and V Measure length of wire each time (iii) (b) 7 (a) (i) A component (or substance) which has no electrical resistance (ii)• Correct high temperature graph with abrupt discontinuous vertical line indicating that R has become zero at a certain temperature • Temperature axis labelled (e.g. Transition) at the corresponding temperature (b) Reason: When resistance is zero there is no energy (heat, or power) lost Applications: Power cables, electromagnets, generators, motors, transformers, MRI scanners, monorail trains, particle accelerators, fusion reactors. Banner 7

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