# Prelim Revision. Questions and Answers. Electricity

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1 Prelim Revision Questions and Answers Electricity

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4 SECTION A Answer questions on the answer sheet 8. Specimen Paper The diagram shows an 8V supply connected to two lamps. The supply has negligible internal resistance. 9. Specimen Paper The diagram shows a circuit used to determine the e.m.f. E and the internal resistance r of a cell. In 6 s, the total electrical energy converted in the two lamps is A B C D E J 4 J 3 J 48 J 64 J Which graph correctly shows how the potential difference V across the terminals of the cell varies with the current I in the circuit?

5 0. Specimen Paper The graph represents a sinusoidal alternating voltage Paper An electron is accelerated from rest through a potential difference of.0 kv. The kinetic energy gained by the electron is The r.m.s. (root mean square) voltage is A B C D E J J J J J A B C D E 5 V 0 V 0 V 0 V 0 V Paper The e.m.f. of a battery is A B C D E the total energy supplied by the battery the voltage lost due to the internal resistance of the battery the total charge which passes through the battery the number of coulombs of charge passing through the battery per second the energy supplied to each coulomb of charge passing through the battery.

6 Paper The diagram shows the trace on an oscilloscope when an alternating voltage is applied to its input Paper A potential difference, V, is applied between two metal plates. The plates are 0.5 m apart. A charge of +4.0 mc is released from rest at the positively charged plate as shown. The timebase is set at 5 ms/div and the Y-gain is set at 0 V/div. Which row in the table gives the peak voltage and the frequency of the signal? Peak voltage/v Frequency/Hz A 7. 0 B 4 50 C 0 0 D 0 50 E The kinetic energy of the charge just before it hits the negative plate is 8.0 J. The potential difference between the plates is A B C D E 3. 0 V. V.0 V V V Paper A battery of e.m.f. 4 V and negligible internal resistance is connected as shown. The reading on the ammeter is.0a. The resistance of R is A B C D E 3.0 Ω 4.0 Ω 0 Ω Ω 8 Ω

7 . 009 Paper A 5.0 μf capacitor is charged until the potential difference across it is 500V Paper Which of the following combinations of resistors has the greatest resistance between X and Y? The charge stored in the capacitor is A B C D E C C.5 0 C C C Paper The potential difference between two points is A B C D E the work done in moving one electron between the two points the voltage between the two points when there is a current of one ampere the work done in moving one coulomb of charge between the two points the kinetic energy gained by an electron as it moves between the two points the work done in moving any charge between the two points

8 . 00 Paper A circuit is set up as shown Paper One volt is equivalent to one A B C D E farad per coulomb ampere per ohm joule per ampere joule per ohm joule per coulomb The capacitor is initially uncharged. Switch S is now closed. Which graph shows how the potential difference, V, across R, varies with time, t?. 0 Paper The output of a 50 Hz a.c. supply is connected to the input of an oscilloscope. The trace produced on the screen of the oscilloscope is shown. The time-base control of the oscilloscope is set at A B C D E ms/div 0 ms/div 0 ms/div 00 ms/div 00 ms/div

9 . 0 Paper An a.c. supply with an output voltage of 6 0 V r.m.s. is connected to a 3 0 Ω resistor. Which row in the table shows the peak voltage across the resistor and the peak current in the circuit? Peak voltage/v Peak current/a A 6 B 6 C 6 D 6 E Paper In an experiment to find the capacitance of a capacitor, a student makes the following measurements. potential difference across capacitor = ( ) V charge stored by capacitor = ( ) μc Which row in the table gives the capacitance of the capacitor and the percentage uncertainty in the capacitance? Capacitance/ F Percentage uncertainty A 0.0 B C 50 D 50 5 E

10 SECTION B All answers must be written clearly and legibly in ink. Marks 4. Specimen Paper The diagram illustrates a cathode ray tube used in an oscilloscope. Electrons are released from the hot cathode and are accelerated uniformly by a p.d. of.0 kv between the cathode and anode. The distance between the cathode and anode is 0.0 m. (a) (b) Show that the amount of work done in accelerating an electron between the cathode and anode is J. Assuming that the electrons start from rest at the cathode, calculate the speed of an electron just as it reaches the anode. (c) (i) Show that the acceleration of an electron as it moves from the cathode to the anode is ms. (ii) Calculate the time taken by an electron to travel from the cathode to the anode. (d) The distance between the cathode and anode in the cathode ray tube is increased. The same p.d. is applied between the cathode and anode. Explain the effect this has on the speed of an electron just as it reaches the anode. (9)

11 Marks 6. Specimen Paper A 0 F capacitor is to be charged using the circuit shown below. The V battery has negligible internal resistance. The capacitor is initially uncharged. After switch S is closed, the charging current is kept constant at 0 A by adjusting the resistance of the variable resistor, R, while the capacitor is charging. (a) What is the initial resistance of R? (b) While the capacitor is charging, should the resistance of R be increased or decreased to keep the current constant? Justify your answer. (c) Calculate (i) (ii) the charge on the capacitor 45 s after the switch S is closed the potential difference across R at this time. 3 (9)

12 . 008 Paper Marks (c) The brake lights of the car consist of a number of very bright LEDs. An LED from the brake lights is forward biased by connecting it to a V car battery as shown. The battery has negligible internal resistance. (i) Explain, in terms of charge carriers, how the LED emits light. (ii) The LED is operating at its rated values of 5.0 V and. W. Calculate the value of resistor R. 3 (4)

13 Paper Marks Electrically heated gloves are used by skiers and climbers to provide extra warmth. (a) Each glove has a heating element of resistance 3.6 Ω. Two cells, each of e.m.f..5 V and internal resistance 0.0 Ω, are used to operate the heating element. Switch S is closed. (b) (i) Determine the value of the total circuit resistance. (ii) Calculate the current in the heating element. (iii) Calculate the power output of the heating element. When in use, the internal resistance of each cell gradually increases. What effect, if any, does this have on the power output of the heating element? Justify your answer. (7)

14 Paper Marks (a) (b) State what is meant by the term capacitance. An uncharged capacitor, C, is connected in a circuit as shown. The V battery has negligible internal resistance. Switch S is closed and the capacitor begins to charge. The interface measures the current in the circuit and the potential difference (p.d.) across the capacitor. These measurements are displayed as graphs on the computer. Graph shows the p.d. across the capacitor for the first 0.40 s of charging. Graph shows the current in the circuit for the first 0.40 s of charging. (i) Determine the p.d. across resistor R at 0.40 s. (ii) Calculate the resistance of R.

15 (iii) The capacitor takes. seconds to charge fully. At that time it stores 0.8 mj of energy. Calculate the capacitance of the capacitor. Marks 3 (c) The capacitor is now discharged. A second, identical resistor is connected in the circuit as shown. Switch S is closed. Is the time taken for the capacitor to fully charge less than, equal to, or greater than the time taken to fully charge in part (b)? Justify your answer. (9)

16 Paper Marks A battery of e.m.f. 6.0 V and internal resistance, r, is connected to a variable resistor R as shown. The graph shows how the current in the circuit changes as the resistance of R increases. (a) (b) Use information from the graph to calculate: (i) the lost volts in the circuit when the resistance of R is.5 ; (ii) the internal resistance, r, of the battery. The resistance of R is now increased. What effect, if any, does this have on the lost volts? You must justify your answer. (6)

17 Paper Marks A volt battery of negligible internal resistance is connected in a circuit as shown. The capacitor is initially uncharged. Switch S is then closed and the capacitor starts to charge. (a) (b) (c) (d) Sketch a graph of the current against time from the instant switch S is closed. Numerical values are not required. At one instant during the charging of the capacitor the reading on the ammeter is 5.0 ma. Calculate the reading on the voltmeter at this instant. Calculate the maximum energy stored in the capacitor in this circuit. The 500 Ω resistor is now replaced with a.0 kω resistor. What effect, if any, does this have on the maximum energy stored in the capacitor? Justify your answer. 3 (8)

18 4. 00 Paper Marks An experiment is carried out to measure the time taken for a steel ball to fall vertically through a fixed distance using an electronic timer. (a) The experiment is repeated and the following values for time recorded s, 0 53 s, 0 50 s, 0 50 s, 0 55 s, 0 5 s. Calculate: (i) (ii) the mean value of the time; the approximate random uncertainty in the mean value of the time. (b) Part of the circuit in the electronic timer consists of a 6 mf capacitor and an 8 k resistor connected to a switch and a 4 5 V supply. (i) (ii) Calculate the charge on the capacitor when it is fully charged. Sketch the graph of the current in the resistor against time as the capacitor charges. Numerical values are required on the current axis. (6)

19 6. 00 Paper Marks A signal generator is connected to a lamp, a resistor and an ammeter in series. An oscilloscope is connected across the output terminals of the signal generator. The oscilloscope control settings and the trace displayed on its screen are shown. (a) (b) (c) For this signal calculate: (i) the peak voltage; (ii) the frequency. The frequency is now doubled. The peak voltage of the signal is kept constant. State what happens to the reading on the ammeter. The resistor is now replaced by a capacitor. The procedure in part (b) is repeated. State what happens to the reading on the ammeter as the frequency is doubled. (d) The capacitor will be damaged if the potential difference across it exceeds 6 V. The capacitor is now removed from this circuit and connected to a different a.c. supply of output 5 Vr.m.s.. Explain whether or not the capacitor is damaged. (7)

20 4. 0 Paper Marks (a) A supply of e.m.f. 0 0 V and internal resistance r is connected in a circuit as shown in Figure. The meters display the following readings. Reading on ammeter = 5 A Reading on voltmeter = 7 50 V (i) What is meant by an e.m.f. of 0 0 V? (ii) Show that the internal resistance, r, of the supply is 0 Ω. (b) A resistor R is connected to the circuit as shown in Figure. The meters now display the following readings. Reading on ammeter = 0 A Reading on voltmeter = 6 0 V (i) Explain why the reading on the voltmeter has decreased. (ii) Calculate the resistance of resistor R. 3 (7)

21 5. 0 Paper Marks A student carries out an experiment using a circuit which includes a capacitor with a capacitance of 00 μf. (a) (b) Explain what is meant by a capacitance of 00 μf. The capacitor is used in the circuit shown to measure the time taken for a ball to fall vertically between two strips of metal foil. The ball is dropped from rest above foil A. It is travelling at 5 ms when it reaches foil A. It breaks foil A, then a short time later breaks foil B. These strips of foil are 0 80 m apart. The computer displays a graph of potential difference across the capacitor against time as shown. (i) (ii) Calculate the current in the 4 kω resistor at the moment foil A is broken. Calculate the decrease in the energy stored in the capacitor during the time taken for the ball to fall from foil A to foil B. 3 (6)

22 Physics Higher Marking Scheme Section A Question Paper Answer 8 Specimen E 9 Specimen C 0 Specimen B Specimen E Specimen B E E D 008 A 008 B D C B 009 C 009 A C 7 00 C 9 00 A 0 00 D 00 D 00 C 3 00 E 8 0 E 9 0 C 0 0 B 0 B 0 A 3 0 D 4 0 E

23 Physics Higher Marking scheme Section B Paper Question Sample Answer Mark Allocation Notes Marks Specimen 3 (a) (i) No reading on voltmeter R / R sensor = R 3 / R 4 Specimen 3 (a) (ii) Place sensor in melting ice adjust R 3 until V = 0 (or note reading on V) Place sensor in boiling water note reading on V Divide difference in the two readings by a hundred to give the number of volts per degree celsius Specimen 4 (a) Work done = QV Work done =.6 x 0 9 x 000 Work done = 3. x 0 6 J Specimen 4 (b) Change in E K = electrical energy transferred ½ m v 0 = 3. x 0 6 ½ x 9. x 0-3 x v = 3. x x 0-3 x v = 3. x 0 6 v = 7.03 x 0 4 v = 7.03 x 0 4 ms - Specimen 4 (c) (i) v = u + as (.65 x 0 7 ) = 0 + ( x a x 0.) 7.03 x 0 4 = 0.a a = 3.5 x 0 5 ms - Specimen 4 (c) (ii) v = u + at.65 x 0 7 = 0 + (3.5 x 0 5 x t) t = 7.6 x 0-9 s

24 Paper Question Sample Answer Mark Allocation Notes Marks Specimen 4 (d) No effect same work done in accelerating electron electron has same E K so same speed. Specimen 5 (a) V XY = 4.5 V Specimen 5 (b) R T = R + R t R T = {} R T = 000 Ω 3 I circuit = V T / R T I circuit = 9 / 000 I circuit = A V DE = I DE R DE V DE = x 000 V DE = 8.5 V V th / V T = R th / R T V th / 9 = 000 / ( ) V th = 8.5 V Specimen 5 (c) V + > V (V > V ) V 0 positive transistor switches on current in relay coil or relay switch closes Specimen 5 (d) n-channel enhancement MOSFET a transistor award ½ mark Specimen 6 (a) R = V R / I R R = / 0 x 0-6 R = Specimen 6 (b) Decreased as C charges up current decreases to compensate for this decrease R to increase I

25 Paper Question Sample Answer Mark Allocation Notes Marks Specimen 6 (c) (i) Q = I x t Q = 0 x 0-6 x 45 Q = 9 x 0-4 C Specimen 6 (c) (ii) V C = Q / C V C = 9 x 0-4 / 0 x 0-6 V C = 4. (V) 3 V R = V S V C V R = 4. V R = 7.9 V 008 (c) (i) electrons and holes recombine at/in the junction (and energy is released) not meet/come together/travel to can say +ve and ve charge carriers but not just charge carriers do not accept depletion layer for junction

26 Paper Question Sample Answer Mark Allocation Notes Marks 008 (c) (ii) V S = 5 ½ mark for this anywhere 3 V S = 7 V I = P / V I =. / 5 I = 0.44 (A) R = V / I R = 7 / 0.44 R = 6 R = V / P R = 5 /. R =.4 I = P / V I =. / 5 I = 0.44 (A) ½ for both equations accept marks for.4 on its own R = V / I R = / 0.44 R = 7.3 ½ mark for all 3 equations R = R = (a) (i) 4 deduct ½ mark for an incorrect or missing unit (a) (ii) I = E / R T V / R or consistent with (a) (i) I = ( x.5) / 4 I = 0.75 A

27 Paper Question Sample Answer Mark Allocation Notes Marks (a) (iii) P = I R or consistent with (a) (i) and P = 0.75 x 3.6 (a) (ii) P =.0 W V = IR V = 0.75 x 3.6 V =.7 V THEN P = IV P = 0.75 x.7 P =.0 W P = V / R P =.7 / 3.6 P =.0 W ½ mark for both equations

28 Paper Question Sample Answer Mark Allocation Notes Marks (b) Power output is less Current is less P = I R R (load) is constant Power output is less t.p.d./v is less P = V / R R (load) is constant Power is less Current is less P = I V t.p.d./v is also less (a) quantity of charge stored per volt if C = Q / V is written, Q and V must be defined to get the mark coulombs per volt Coulomb per Volt gets 0 marks ratio of charge to p.d/voltage (b) (i) 8.6 = 3.4 V deduct ½ mark for an incorrect or missing unit (b) (ii) R = V / I or consistent with (b) (i) R = 3.4 / R = 5

29 Paper Question Sample Answer Mark Allocation Notes Marks (b) (iii) V = V (from diagram) E = ½ C V x 0-3 = ½ x C x C = F (50 F) mark for this anywhere ½ mark for this anywhere E = ½ Q V 0.8 x 0-3 = ½ x Q x Q = C E = ½ Q / C 0.8 x 0-3 = ½ x x C C = F (50 F) (c) time is less circuit resistance is less current/rate of flow of charge is greater ½ mark for both equations MUST say time/it is less or 0 marks e.g. capacitor charges faster 0 marks deduct ½ mark for current/voltage through capacitor (a) R / R = R 3 / R 4 R LDR /. = 6 / 4 R LDR =.8 (k ) from graph irradiance = 0 48 Wm - points and 3, marks are independent if stop here the unit for R is needed if no equation/substitution and R =.8 k, award marks maximum 3 range accepted Wm - if only 0.48 Wm - is written, then award mark only

30 Paper Question Sample Answer Mark Allocation Notes Marks (b) (i) V = ( R / R + R ) x V S V = (.0 /.0 +. ) x V = 7.5 V Need final line or lose ½ mark I = V / R t I = ( / ) x (A) V = I R LDR V = x 000 V = 7.5 V Need final line or lose ½ mark V / V = R / R 7.5 / 4.5 = /. = (b) (ii) V O = ( V V ) x R f / R V O = ( ) x 40 / 0 V O = -. V (b) (iii) Yellow LED is lit because it is forward biased Need first statement otherwise 0 marks awarded (a) (i) V TPD = I R V TPD =.5 x 3 V TPD = 4.5 (V) MUST be consistent with (ii) i.e. if +. V in (ii) then must state blue LED lit if stop at 4 5 V, award mark 4.5 gets ½ mark awarded lost volts = E - V TPD lost volts = lost volts =.5 V all marks are independent

31 Paper Question Sample Answer Mark Allocation Notes Marks (a) (ii) r = lost volts / I r =.5 / 3.0 r = 0.5 or consistent with (a) (i) r = E / I r = 6.0 / r = 0.5 E = Ir + IR 6.0 = (3.0 x.5 ) + ( 3.0 x r ) r = (b) current decreases must attempt explanation answer on own gets zero marks so lost volts (V = Ir) decreases (a) (i) V P = = 5 mv if go further i.e. work out V RMS, award zero marks (a) (ii) (Period = 4 ms) f = / T f = / 4 x 0-3 f = 50 Hz (b) (i) inverting (mode) (b) (ii) V RMS = V peak / Can use gain formula first gives 3 V RMS = 6. x 0-3 / V O = 4 V etc. V RMS = 4.38 x 0-3 (V) V O / V = - R f / R V O / 4.38 x 0-3 = - 0 x 0 6 / 5 x 0 3 V O = (-) 8.8 V

32 Paper Question Sample Answer Mark Allocation Notes Marks (b) (iii) trace will be clipped /flattened at ( + 9 V ) or almost square wave max output voltage will be + 9 V / V S op-amp saturates saturation occurs (a) no origin label deduct ½ mark (b) V R = I R V R = 5 x 0-3 x 500 V R =.5 (V) 3 V C =.5 V C = 9.5 V (c) E = ½ C V E = 0.5 x 47 x 0-6 x E = 3.4 x 0-3 J Must use V otherwise maximum ½ mark for correct formula. Q = CV Q = 47 x 0-6 x Q = 5.64 x 0-4 C E = ½ Q V E = 0.5 x 5.64 x 0-4 x E = 3.4 x 0-3 J ½ mark for both equations ½ mark for both substitutions

33 Paper Question Sample Answer Mark Allocation Notes Marks (d) Maximum energy the same/ no effect Values of C and V are same as before 00 4 (a) (i) 0.5 s secs is a unit error 00 4 (a) (ii) Random uncertainty = {max min} / no. Random uncertainty = { } / 6 Random uncertainty = 0.0 s 00 4 (b) (i) Q = C V Q =.6 x 0-3 x 4.5 Q = 7. x 0-3 C 00 4 (b) (ii) I = V / R I = 4.5 / 8000 I = 0.5 ma No graph, no marks. Wrong or missing units for current deduct ½ mark. Must start at 0.5 ma on axis and must tend towards zero. Line must not cross axis. Deduct ½ mark for missing origin (a) (i) R / R = R 3 / R 4 R = / 4000 R = 00 Ω 00 5 (a) (ii) V P = 4 0 V V Q = 4 8 V Voltmeter reading = 0 8 V

34 Paper Question Sample Answer Mark Allocation Notes Marks 00 5 (b) V o = ( V V ) ( R f / R ) V o = ( ) ( / ) V o = 0 (V) V o = (3 0 3 ) ( / ), is wrong Physics maximum ½ mark awarded 3 (But, due to saturation, the actual output voltage is) 0 to V voltage saturated is wrong Physics (a) (i) V P = 0 V 00 6 (a) (ii) f = / T f = / 0.0 f = 00 Hz 00 6 (b) Stays the same / constant / no change / nothing 00 6 (c) Increases / doubles 00 6 (d) The capacitor will be damaged The peak voltage from this power supply is greater than 6 V. 0 4 (a) (i) 0 joules of energy are given to each coulomb (of charge) passing through the supply. Any mention of voltage flowing, travelling or moving is wrong Physics and gets zero marks. It is the t.p.d. on open circuit. It is the reading on voltmeter (across cell) on open circuit.

35 Paper Question Sample Answer Mark Allocation Notes Marks 0 4 (a) (ii) I = E / ( R + r ).5 = 0 / ( 6 + r ) r =.0 Deduct ½ mark if last line is missing. Deduct ½ mark if wrong unit. r = lost volts / I r = /.5 r =.0 R / R = V / V r / 6.0 =.5 / 7.5 r = (b) (i) (Total) resistance decreases (circuit) current increases lost volts increases All independent marks. The use of up and down arrows is not accepted.

36 Paper Question Sample Answer Mark Allocation Notes Marks 0 4 (b) (ii) Parallel resistance = R = V / I R = 6.0 /.0 R = 3.0 Note potential divider ratio could be used to get R P = 3 0 Ω 3 / R T = / R + / R / 3 = / 6 + / R R = 6.0 Total resistance = E / I R = 0.0 /.0 R = 5.0 Resistance of parallel network = 5 Resistance of parallel network = 3.0 / R T = / R + / R / 3 = / 6 + / R R = (a) 00 μc of charge increases voltage across plates by volt. 00 μc per volt. One volt across the plates of the capacitor causes 00 μc of charge to be stored. 0 5 (b) (i) I = E / R I = / 400 I = A (8.6 ma)

37 Paper Question Sample Answer Mark Allocation Notes Marks 0 5 (b) (ii) E = ½ C V initial stored enegy = 0.5 x ( 00 x 0-6 ) x initial stored enegy = J If this number is rounded off (e.g. to 0 04) deduct ½ mark. 3 final stored enegy = 0.5 x ( 00 x 0-6 ) x 4 final stored enegy = J Difference = decrease in stored energy = 0.08 J Deduct ½ mark if missing or wrong unit. 0 6 (a) (i) inverting Do not accept inverted or inverse. 0 6 (a) (ii) V O = - R f / R x V Formula is wrong if negative = - 80 / 0 x V sign is missing or using any V = -.5 V other correct points from graph. 0 6 (a) (iii) Output cannot be greater than (approximately 85% of) the supply voltage. Accuracy of calculation is determined from ± ½ scale division reading from graph. Output voltage saturates no It saturates no Op amp saturates - yes Saturation of the amplifier has been reached.

38 Paper Question Sample Answer Mark Allocation Notes Marks 0 6 (b) Either or both labels missing, 3 deduct ½ mark. Either or both units on axes is missing, deduct ½ mark. [END OF MARKING INSTRUCTIONS]

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