Q1. The graph below shows how a sinusoidal alternating voltage varies with time when connected across a resistor, R.

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1 Q1. The graph below shows how a sinusoidal alternating voltage varies with time when connected across a resistor, R. (a) (i) State the peak-to-peak voltage. peak-to-peak voltage...v (1) (ii) State the peak voltage. peak voltage...v (1) (iii) Calculate the root mean square (rms) value of the alternating voltage. rms voltage...v (2) (iv) Calculate the frequency of the alternating voltage. State an appropriate unit. frequency...unit... (3) Page 1 of 31

2 (b) (c) On the graph above draw a line to show the dc voltage that gives the same rate of energy dissipation in R as produced by the alternating waveform. An oscilloscope has a screen of eight vertical and ten horizontal divisions. Describe how you would use the oscilloscope to display the alternating waveform in the graph above so that two complete cycles are visible (2)... (3) (Total 12 marks) Page 2 of 31

3 Q2. (a) An alternating current supply provides an output voltage of 12 V rms at a frequency of 50 Hz. Describe how you would use an oscilloscope to check the accuracy of the rms output voltage and the frequency of the supply. The quality of your written communication will be assessed in your answer (6) (b) The power supply in part (a) is connected to a 12 V 24 W lamp. (i) Calculate the rms current in the lamp. answer =... A (1) Page 3 of 31

4 (ii) Calculate the peak current in the lamp. answer =... A (1) (iii) Calculate the peak power of the lamp. answer =... W (2) (Total 10 marks) Q3. An oscilloscope is used to investigate various voltage sources. In order to do this a voltage source is connected to the y-input and the time base is switched off. Figure 1 below shows the screen of the oscilloscope when the y-input is not connected to a voltage source. Figure 1 Page 4 of 31

5 Figure 2 shows the screen when a 1.5V cell is connected to the y-input. Figure 2 (a) On the grid below show the appearance of the screen if the y-input is connected to a 2.5V dc supply. (1) Page 5 of 31

6 (b) The y-input is now connected to a sinusoidal ac voltage supply and the screen is shown in Figure 3. Figure 3 (i) Explain why a vertical line is now seen on the screen. (2) (ii) Calculate the peak-to-peak voltage of the ac supply. answer =... V (2) Page 6 of 31

7 (iii) Calculate the root mean square voltage of the supply. answer =... V (2) (Total 7 marks) Q4. Domestic users in the United Kingdom are supplied with mains electricity at a root mean square voltage of 230V. (a) State what is meant by root mean square voltage (1) (b) (i) Calculate the peak value of the supply voltage. answer =... V (2) (ii) Calculate the average power dissipated in a lamp connected to the mains supply when the rms current is 0.26 A. answer =... W (1) Page 7 of 31

8 (c) The frequency of the voltage supply is 50 Hz. On the axes below draw the waveform of the supplied voltage labelling the axes with appropriate values. (4) (Total 8 marks) Q5. An alternating current (ac) source is connected to a resistor to form a complete circuit. The trace obtained on an oscilloscope connected across the resistor is shown in the diagram below. The oscilloscope settings are: Y gain 5.0 V per division time base 2.0 ms per division. (i) Calculate the peak voltage of the ac source. answer =... V (1) Page 8 of 31

9 (ii) Calculate the rms voltage. answer =... V (1) (iii) Calculate the time period of the ac signal. answer =... ms (1) (iv) Calculate the frequency of the ac signal. answer =... Hz (2) (Total 5 marks) Page 9 of 31

10 Q6. The diagram below shows an ac waveform that is displayed on an oscilloscope screen. The time base of the oscilloscope is set at 1.5 ms per division and the y-gain at 1.5 V per division. (a) For the ac waveform shown, (i) Calculate the frequency answer... Hz (3) (ii) Calculate the peak voltage answer... V (2) (iii) the rms voltage answer... V (2) Page 10 of 31

11 (b) State and explain the effect on the oscilloscope trace if the time base is switched off (2) (Total 9 marks) Q7. An oscilloscope is connected to an alternating voltage source of rms value 4.2 V at a frequency of 2.5 khz. (a) Calculate the peak-to-peak alternating voltage. peak-to-peak voltage =... (2) (b) Figure 1 represents the screen of the oscilloscope. Figure 1 Determine (i) the voltage sensitivity of the oscilloscope, voltage sensitivity =... Page 11 of 31

12 (ii) the time base setting of the oscilloscope. time base setting =... (3) (c) The time base of the oscilloscope is switched off and the voltage sensitivity is set to 0.5 V div 1. The oscilloscope is connected across a 1.75 V battery of internal resistance 3.5 Ω which is connected to a 10 Ω resistor as shown in Figure 2. Figure 3 represents the screen of the oscilloscope which shows the spot when registering zero volts. Figure 2 Figure 3 (i) Draw a spot on Figure 3 showing the appearance on the screen when the switch is open. Label this spot O. (ii) When the switch is closed determine the current flowing through the 10 Ω resistor. current =... Page 12 of 31

13 (iii) Draw a spot on Figure 3 showing the appearance on the screen when the switch is closed. Label this spot C. (5) (Total 10 marks) Q8. The circuit in Figure 1 shows a sinusoidal ac source connected to two resistors, R 1 and R 2, which form a potential divider. Oscilloscope 1 is connected across the source and oscilloscope 2 is connected across R 2. Figure 1 (a) Figure 2 shows the trace obtained on the screen of oscilloscope 1. The time base of the oscilloscope is set at 10 m/s per division and the voltage sensitivity at 15 V per division. Figure 2 For the ac source, calculate (i) the frequency, Page 13 of 31

14 (ii) the rms voltage. (4) (b) The resistors have the following values: R 1 = 450 Ω and R 2 = 90 Ω. Calculate (i) the rms current in the circuit, (ii) the rms voltage across R 2. (2) (c) Oscilloscope 2 is used to check the calculated value of the voltage across R 2. The screen of oscilloscope 2 is identical to that of oscilloscope 1 and both are set to the same time base. Oscilloscope 2 has the following range for voltage sensitivity: 1 V per div., 5 V per div., 10 V per div. and 15 V per div. State which voltage sensitivity would give the most suitable trace. Explain the reasons for your choice (3) (Total 9 marks) Page 14 of 31

15 Q9. A sinusoidal alternating voltage source of frequency 500 Hz is connected to a resistor of resistance 2.0 kω and an oscilloscope, as shown in Figure 1. Figure 1 (a) The rms current through the resistor is 5.3 ma. Calculate the peak voltage across the resistor (2) (b) The settings on the oscilloscope are timebase: 250 µs per division, voltage sensitivity: 5.0 V per division. Draw on the grid, which represents the screen of the oscilloscope, the trace that would be seen (4) (Total 6 marks) Page 15 of 31

16 Q10. An oscilloscope is connected to a sinusoidal ac source as shown in Figure 1. The frequency and the voltage output of the ac source can be varied. Figure 1 At a certain frequency the ac signal has an rms output of 7.1 V. Figure 2 shows the trace obtained on the screen of the oscilloscope when one horizontal division corresponded to a time of 5.0 ms. Figure 2 (a) Calculate, for the signal shown in Figure 2, (i) the peak voltage, Page 16 of 31

17 (ii) the frequency. (3) (b) The voltage output and frequency of the signal are now changed so that the peak voltage is 80 V and the frequency is 200 Hz. State which two controls on the oscilloscope have to be altered so that four full cycles again appear on the screen but the peak to peak distance occupies the full screen. Determine the values at which these two controls have to be set. control 1:... value of the setting: control 2:... value of setting: (5) (Total 8 marks) Page 17 of 31

18 Q11. (a) The circuit shown in Figure 1 may be used to determine the internal resistance of a battery. An oscilloscope is connected across the battery as shown. Figure 2 represents the screen of the oscilloscope. Figure 1 Figure 2 The time base of the oscilloscope is switched off throughout the experiment. Initially the switches S 1 and S 2 are both open. Under these conditions the spot on the oscilloscope screen is at A. (i) Switch S 1 is now closed, with S 2 remaining open. The spot moves to B. State what the deflection AB represents. (ii) Switch S 1 is kept closed and S 2 is also closed. The spot moves to C. State what the deflection AC represents. (iii) The vertical sensitivity of the oscilloscope is 0.50 V div 1. Calculate the current through the 14 Ω resistor with both switches closed. (iv) Hence, calculate the internal resistance of the battery. (6) Page 18 of 31

19 (b) The oscilloscope is now connected to an alternating voltage source of rms value 3.5 V. (i) Calculate the peak value of the alternating voltage.. (ii) Draw on Figure 3 what you would expect to see on the oscilloscope screen, if the time base is still switched off and the voltage sensitivity is altered to 2.0 V div 1. Figure 3 (3) (Total 9 marks) Page 19 of 31

20 M1. (a) (i) 128 V 1 (ii) 64 V CE from (i) 1 (iii) V rms = 64 / 2 =45.3 V CE from (ii) 2 (iv) frequency = 1 / 0.01 = 100 Hz do not accept khz for unit mark unless correct for candidate value if use 10 s instead of 10 ms then can score second two marks 3 (b) horizontal line through y = 45 (44 48) x =0 CE from (a)(iii)+ / - half square straight line must extend to at least to 6.0 ms 2 (c) connect to y-input adjust / change time base so that each division is 2.0 ms OR 20 ms across screen reference to y-gain / sensitivity if inappropriate numbers quoted for y gain then lose last mark 3max [12] M2. (a) The candidate s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear. The candidate s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria. High Level (Good to excellent): 5 or 6 marks The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question. The candidate states that the power supply is connected to the input of the oscilloscope. The time base is switched off and the y gain adjusted until a complete vertical line is seen on the screen. The length of the line is measured and this is converted to peak to peak voltage using the calibration. The peak voltage is divided by root two to get the rms voltage and this is compared with the stated value. The time base is now switched on and adjusted until a minimum of one cycle is seen on the screen. The length of one cycle is measured and this is converted to time using the time base setting. Frequency is the reciprocal of this time. Page 20 of 31

21 Intermediate Level (Modest to adequate): 3 or 4 marks The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate. The candidate states that the power supply is connected to the input of the oscilloscope. The y gain adjusted. The length of the line/height of peak is measured. The peak voltage is divided by root two to get the rms voltage. The time base is now switched on and adjusted until a minimum of one cycle is seen on the screen. The length of one cycle is measured and this is converted to time using the time base setting. Low Level (Poor to limited): 1 or 2 marks The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate. The candidate states that the power supply is connected to the input of the oscilloscope. The length of the line/height of peak is measured. The time base is now switched on and adjusted until a minimum of one cycle is seen on the screen. The length of one cycle is measured and this is converted to time. The explanation expected in a competent answer should include a coherent selection of the following points concerning the physical principles involved and their consequences in this case. power supply connected to oscilloscope input time base initially switched off y gain adjusted to get as long a line as possible length of line used to find peak to peak voltage rms voltage found time base switched on and adjusted to get several cycles on the screen use the time base setting to find period use period to find frequency compare vales with stated values 6 (b) (i) (use of P = IV) (ii) I = 24/12 = 2.0 (A) peak current = = 2.8 (A) 1 1 (iii) peak power = = 48 (W) 2 [10] Page 21 of 31

22 M3. (a) 1 (b) (i) the voltage reverse/changes direction/sign this makes the spot move up and down or correct explanation of lack of horizontal movement 2 (ii) length of line = 8 divisions peak to peak = = 4.0 V 2 (iii) (peak = 2.0 V) rms = 2.0/ 2 = 1.4 V 2 [7] M4. (a) the square root of the mean of the squares of all the values of the voltage in one cycle (1) or the equivalent dc/steady/constant voltage that produces the same heating effect/power (1) 1 (b) (i) peak voltage = (1) peak voltage = 325 V (or 324 V) (1) (ii) average power = = 60 W (1) 2 1 Page 22 of 31

23 (c) shape and symmetrical with consistent values of x at y = 0 and consistent y max (must be at least one cycle) (1) appropriate scale y-axis (1) correct peak values (to within one 2 mm square) (1) correct period (accept 0.02 or 20) (1) 4 [8] M5. (i) 10.0 (V) (1) (ii) V rms = 10.0/ 2 = 7.1 (V) (1) (iii) time period = 3 2 = 6 (ms) (1) (iv) frequency = 1/0.006 or 1/6 (1) frequency = 167 (1) (Hz) [5] Page 23 of 31

24 M6. (a) (i) use of 1.5 cycles (1) conversion to time eg time for 1.5 cycles = = 15ms (1) calculation of frequency eg frequency = 1 / = 100 ± 3Hz (1) (ii) peak voltage = (1) = 3.0V (1) (iii) rms voltage = 3.0/ 2 (1) (ce from (a) (i)) rms voltage = 2.12V (1) 7 (b) vertical line is formed (1) of length equal to twice the peak voltage (1) because trace no longer moves horizontally or spot moves just up and down (1) max 2 [9] M7. (a) V 0 = 2 V rms = V (1) (5.94 V) V p-p (= 2 V 0 ) = = 11.8 V (1) 2 (b) (i) voltage sensitivity = 11.8/5.9 = 2.0 V div 1 (1) (ii) T (= 1/f = 1/2500) = s (1) time base = /8 = s div 1 (1) 3 (c) (i) spot at (1.75/0.5) = 3.5 div (1) (ii) (use of sum of emf = sum of pd) 1.75 = I ( ) (1) I = 0.13 A (1) (iii) V (= RI = ) = 1.3 V (1) [or V = ε Ir = = 1.3 V] spot at (1.3/0.5) = 2.6 div (1) (accept 2.5 to 2.75 div) 5 [10] Page 24 of 31

25 M8. (a) (i) T = 40(ms) (1) Hz (1) (allow C.E. for value of T) (ii) peak voltage (= 3 15) = 45 (V) (1) rms voltage =32 V (1) (31.8 V) 4 (b) (i) Irms = = 59mA (1) (58.9mA) (use of 32 V gives 59(.2) ma) (allow C.E. for value of V rms from (a)) (ii) V rms = = 5.3(1) V (1) (allow C.E. for value of I rms from (i)) [or V 2 =V 1 ] 2 (c) V peak = 5.31 =7.5(1) (V) (1) best choice: 5 V per division (1) (allow C.E. for incorrect V rms and for suitable reason) reason: others would give too large or too small a trace (1) 3 [9] M9. (a) (V = IR gives) V rms = ( ) = 10.6 (V) (1) V 0 = V rms 2 = = 15 V (1) (14.99 V) [or calculate I 0 (= 7.5 ma) and then V 0 ] 2 Page 25 of 31

26 (b) (use of T = gives) T = = = 2 0(ms) (1) trace to show: correct wave shape (sinusoidal) (1) correct amplitude (3 divisions) (1) correct period (8 divisions) (1) 4 [6] M10. (a) (i) (use of Vrms = gives) V 0 = = 10 V (1) (ii) T = 10 (ms) (1) (use of f = gives) f = = 100 Hz (1) 3 (b) control 1: time base (1) (or time period) (use of T = gives) T = = (s) (1) setting = 2.5 ms (div 1 ) (1) control 2: voltage sensitivity or Y-plate setting (or Y-gain) (1) setting = 20 V (div 1 ) (1) 5 [8] M11. (a) (i) the emf (of the battery) (1) (ii) the voltage across the battery when current flows [or terminal voltage or pd supplied to the circuit] (iii) V = (3 0.5) = 1.5 (V) (1) current = (1.5/14) = 0.11 A (1) (0.107 A) Page 26 of 31

27 (iv) ( = V + Ir and emf = = 1.75 (V) gives) 1.75 = r (1) r = 2.3 Ω [or use of = I (R + r) with I = gives r = 2.4 Ω and I = gives r = 1.9 Ω] (allow C.E. for value of I from (iii)) 6 (b) (i) peak value = = 4.9 V (1) (ii) oscilloscope screen to show vertical line of height 2.5 divisions above central axis (1) and below central axis (1) 3 [9] Page 27 of 31

28 E1. This question on alternating currents was generally very well done. There were few major problems with part (a) although a minority did leave the rms voltage in surd form thus not completing the calculation. Part (b) was also well answered and most candidates drew their line with care, using a ruler. Part (c) however, was answered poorly and it is apparent that a significant proportion of candidates were not clear on how to use an oscilloscope. Reference to the time base was seen more often than reference to y gain or y sensitivity but frequently neither of these was mentioned. It was also quite common for candidates to assume that the vertical scale only needed to cover the peak voltage and not the peak to peak voltage i.e. eight divisions covered 64 V rather than 128 V. On the other hand a far greater proportion was able to deduce that each horizontal division needed to represent 2.0 ms. E2. Part (a) required students to describe the use of an oscilloscope to measure peak voltage and frequency of an alternating current supply. This was answered well by a good proportion of students and many were confident in their description of the use of the time base to determine frequency and the y-gain to measure peak voltage. In a number of good quality answers students mentioned switching off the time base and measuring peak to peak voltage so as to find an accurate rms voltage. It was evident however, that a minority of students were unfamiliar with the use of an oscilloscope and consequently gave very vague answers which scored few marks. The calculation of rms and peak current were well done with the only common error occurring when students assumed that the 12 V quoted in the question referred to peak voltage. Those doing this were not heavily penalised as their answers were carried forward in the subsequent calculations. E3. Previous papers have suggested that the majority of candidates have a good understanding of the use of an oscilloscope as a voltmeter and this proved to be the case here too. The vast majority were able to successfully complete both parts of the question and only a few confused peak to peak with peak voltage. Most candidates provided evidence that they understood why a vertical line is produced if an alternating voltage is applied when the time base is switched off. However, some answers were spoilt by a lack of precision in explanations of how the voltage was varying references to alternating currents rather than voltages were common. E4. Part (a) of this question on the meaning of root mean square voltage was not answered well, with the majority responding by quoting a formula rather than by referring to an equivalent direct voltage. In contrast part (b) was answered extremely well, with full marks being a common outcome. In part (c), candidates were required to draw a graph of an alternating voltage and although some made a commendable attempt at this, there were a significant proportion of careless answers with waves of varying amplitude and time period being a common occurrence. There was also a tendency not to label the axes with appropriate values in spite of candidates being instructed to do this in the rubric. Page 28 of 31

29 E5. This was the most accessible question in the paper and candidates are clearly familiar with the use of an oscilloscope and the idea of peak and rms voltages. The only consistent error was the failure to convert the time period to seconds when calculating the frequency. E6. In part (a) (i) the majority of candidates were able to relate the time-base setting to time period and from this determine the frequency. Many however, did not use the whole trace and did not recognise that there were one and half cycles across the ten divisions. Instead, they tried to judge the number of divisions occupied by one cycle and consequently obtained a value for frequency of less than 100 Hz. Part (a) (ii) & (iii) were answered very well with only a minority of candidates confusing peak voltage with peak to peak voltage. Part (b) was less well done and it was rare for candidates to score full marks. It was not uncommon for candidates to state that two horizontal lines were produced when the time base is switched off. Some also confused this situation with what would occur if a source of direct current had been used and stated that the trace or spot is deflected upwards. E8. The calculation to determine frequency in part (a) surprisingly brought difficulties to the surface. Many candidates simply did not know how to obtain frequency from the period, but many others had trouble in converting the waveform to an actual value of T. Calculating the rms voltage in part (ii) was usually carried out correctly, although many candidates just used the voltage sensitivity of 15 V per division as the peak voltage. Although the majority of candidates had part (b) correct, a significant number became completely confused and converted back to peak values, under the impression that they were rms values. Part (c) introduced a new type of problem for candidates, where they had to select a particular voltage sensitivity to suit the output voltage across the resistor. The general failing, even for good candidates, was failing to convert the answer of part (b) back to peak voltage, thereby losing a mark. Given that the answer to part (b) was treated as a consequential error in part (c), most candidates then chose the correct voltage sensitivity, but failed to support their choice by stating that the other sensitivities would either give a trace that was off the screen or else a trace that was too small for any meaningful measurements to be made. E9. This question gave the most consistently correct answers in the whole paper but there were some poor efforts in part (a), e.g. leaving the voltage at V rms, not converting to V 0, or just calculating the value of the peak current. The drawing of the oscilloscope trace in part (b) was usually correct, although many candidates, having worked out the value of the period T correctly, failed to translate this to the correct number of squares in the trace. Page 29 of 31

30 E10. High marks were usually obtained in this question. Very few candidates failed to gain full marks in part (a) although it was sad to see some giving the peak voltage as V rms / 2. In part (b) a large number of candidates were obviously familiar with the correct terminology of the oscilloscope controls, but other terms were also accepted. The calculations of the new settings were generally well performed. E11. In part (a), although many candidates gave the emf as the answer to (i), they struggled with part (ii) and failed to put into words what they obviously knew. Very few referred to the voltage across the battery when a current flowed, but rather used a vague description of the voltage across the circuit. Terminal pd was a phrase which occurred infrequently but which the examiners were pleased to see. The calculation in part (iii) was straightforward but many candidates incurred a significant figure penalty here by giving the answer as 0.1 A, i.e. not reducing correctly from the answer of A. Part (iv) realised many correct answers and it was pleasant to see that candidates were using correctly the equation involving the emf and the internal resistance. Part (b) involved a use of the oscilloscope which has not been tested before, and it was encouraging to find a large number of correct answers. The initial calculation was carried out correctly by the large majority of candidates and they also drew the vertical line on the screen correctly. The common error in this section was simply showing two points, without the line joining them. This did incur a penalty. Page 30 of 31

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