ANSWERS AND MARK SCHEMES. (a) 3 A / 2 1 = 1.5 A 1. (b) 6 V 1. (c) resistance = V / I 1 = 6 / (b) I = V / R 1 = 3 / 15 1 = 0.

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1 QUESTIONSHEET (a) 3 A / 2 =.5 A (b) 6 V (c) resistance = V / I = 6 /.5 = 4 Ω QUESTIONSHEET 2 TOTAL / 6 (a) 5 Ω + 0 Ω = 5 Ω (b) I = V / R = 3 / 5 = 0.2 A Units are essential in calculations. Sometimes eamination questions give a blank space for the answer followed by the unit. Sometimes no unit is given and you are epected to supply it. It is a common mistake to leave the units out altogether don t get into this bad habit. TOTAL / 5

2 QUESTIONSHEET 3 (a) 6v 2 Ω 4 Ω symbols correct 2 Ω and 4 Ω resistors in parallel 5 Ω resistor in series (b) resistance of parallel resistors given by = + R t R R 2 = R t = Ω 3 4 so total resistance = Ω 3 5 Ω = 6 Ω 6 (c) 6 3 = 0.95 A For parallel resistors, it is sometimes easier to use the epression Product / Sum for the total resistance. i.e. here we get 2 4 / (2 + 4) = 8 / 6 = 4/3 Ω TOTAL / 9

3 QUESTIONSHEET 4 (a) (i) Power of lamp is 6 W when used on 2 V (ii) 6 W = 6 J/s 5 minutes = 300 s energy transferred = = 800 J (b) (i) I = P / V = 6 / 2 = 0.5 A (ii) R = V / I = 2 / 0.5 = 24 Ω (c) No difference 24 V shared between 2 lamps / pd across each lamp is 2 V Power is the rate of transferring electrical energy into other forms in this case, heat and light. A 6 W lamp transfers 6 Joules of electrical energy into heat & light each second, since W = J/s. The formula P = I / V is one you need to know. QUESTIONSHEET 5 TOTAL / 2 (a) (i) 0.6 0/20 = 0.3 A (ii) = 0.9 A (b) V = I R or = 6 V Current is inversely proportional to the resistance as long as the pd is constant. So half the current flows through the resistor that is twice as big. TOTAL / 7

4 QUESTIONSHEET 6 (a) = + = R R = 5 Ω (b) = 5 Ω (c) current through single resistor = V / R = 6 / 5 = 0.4 A current through parallel resistors = 0.2 A In many questions, the value of the parallel resistors is the same. In this case, the total resistance is half the value of one. QUESTIONSHEET 7 TOTAL / 7 (a) (i) = 7 Ω (ii) /R = /2 + /5 + /0 = 8 / 0 Ω R = 0 / 8 =.25 Ω (b) Current larger in (ii) current inversely proportional to resistance smaller resistance means larger current TOTAL / 8

5 QUESTIONSHEET 8 (a) = + = R = 4Ω R R 2 = + = R 2 = Total resistance = = 6Ω (b) I = V / R = 2 / 6 = 2 A (c) R = = 7 I = 2 / 7 or.7 A (d) R = = 9 Ω I = V / R = 2 / 9 =.33 A QUESTIONSHEET 9 TOTAL / (a) C (b) (i) a complete path for electricity to flow when connected to a voltage supply (ii) A & D (both) No electricity flows in the last circuit, because the cells are the opposite way round from each other, so they cancel out. TOTAL / 5

6 QUESTIONSHEET 0 (a) switch shown in either of the positions indicated A.5 A SWITCH (b) Parallel (c).5 / 3 = 0.5 A (d) 2 V QUESTIONSHEET V A2 SWITCH TOTAL / 5 (a) Resistance in circuit alters current ammeter resistance must be low so it doesn t affect current flow (b) (i) In parallel across component (ii) voltmeter must have high resistance little current flows through voltmeter most current flows through component For two parallel resistors, /R = /R + /R 2. So if one resistor (the voltmeter) R 2 has a high resistance, /R 2 is very small and the total resistance is almost the same as without the voltmeter. TOTAL / 6

7 QUESTIONSHEET 2 (a) pd/v I/A aes labelled + units correct plotting 2 smooth curve drawn (b) (i) no (ii) the graph is not a straight line or V / A R (iii) filament lamp TOTAL / 8

8 QUESTIONSHEET 3 (a) Voltage is proportional to current for a metal conductor at constant temperature (b) A Test component V circuit symbols correct ammeter in series voltmeter in parallel variable resistor in series QUESTIONSHEET 4 TOTAL / 6 (a) A = metal wire B = diode C = lamp filament (b) A (c) C An ohmic conductor obeys Ohm s law, so a graph of I against V must be a straight line through the origin. The gradient of graph C decreases so the resistance must be increasing. TOTAL / 5

9 QUESTIONSHEET 5 (a) (i) MACHINE two switches in series with machine (ii) each switch must be able to turn off machine (b) switches not in series each must be able to switch light on or off needs two parallel parts of circuit TOTAL / 7

10 QUESTIONSHEET 6 A (a) Set up Ohm s law circuit measure pd across wire and current through it measure length of wire Test component change length of wire and repeat eperiment V (b) (i) Length (cm) (ii) as length increases resistance increases directly proportional (iii) as length varies, resistance varies current varies and light dims or brightens The first two or three marks of this question could be obtained by drawing a labelled diagram. QUESTIONSHEET Resistance (Ω) TOTAL / 2 (a) A = battery B = lamp/light bulb C = switch D = variable resistor (b) varies current in a circuit / acts as dimmer switch You must be able to recognise circuit components. Look at the syllabus or ask your teacher which ones you must know. Note that a battery is made up of two or more cells joined in series TOTAL / 5

11 QUESTIONSHEET 8 (a) (i).5 V (ii) 0 V (iii) 3.0 V (iv).5 V (v).5 V (b) car bulb needs 2 v (c) increases QUESTIONSHEET 9 TOTAL / 7 (a) silver too epensive (b) copper is too heavy (c) rubber and PVC 2 (d) water will conduct electricity would get shock with wet hands QUESTIONSHEET 20 TOTAL / 7 (a) One connected after another in line 2 (b) 240 / 20 = 2 V (c) bulb from 40-bulb set uses 6 v higher voltage would cause it to blow (d) (i) fuse would blow repeatedly (ii) fault would not blow fuse could be dangerous to operator TOTAL / 9