Download Q1. The graph below shows how a sinusoidal alternating voltage

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
(b)
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.
(2)
(c)
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.
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(3)
(Total 12 marks)
Page 2 of 31
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.
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(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
(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
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
(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.
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(2)
(ii)
Calculate the peak-to-peak voltage of the ac supply.
answer = ..................................... V
(2)
Page 6 of 31
(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.
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(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
(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
(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
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
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answer ............................................ Hz
(3)
(ii)
Calculate the peak voltage
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answer ........................................... V
(2)
(iii)
the rms voltage
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answer ............................................ V
(2)
Page 10 of 31
(b)
State and explain the effect on the oscilloscope trace if the time base is switched off.
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(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
(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)
(ii)
Draw a spot on Figure 3 showing the appearance on the screen when the switch is
open. Label this spot O.
When the switch is closed determine the current flowing through the 10 Ω resistor.
current = ....................................
Page 12 of 31
(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, R1 and R2,
which form a potential divider. Oscilloscope 1 is connected across the source and oscilloscope
2 is connected across R2.
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,
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Page 13 of 31
(ii)
the rms voltage.
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(4)
(b)
The resistors have the following values: R1 = 450 Ω and R2 = 90 Ω.
Calculate
(i)
the rms current in the circuit,
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(ii)
the rms voltage across R2.
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(2)
(c)
Oscilloscope 2 is used to check the calculated value of the voltage across R2. 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.
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(3)
(Total 9 marks)
Page 14 of 31
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.
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(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.
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(4)
(Total 6 marks)
Page 15 of 31
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,
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Page 16 of 31
(ii)
the frequency.
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(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: ......................................................................................
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control 2: …...................................................................................................
value of setting: ............................................................................................
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(5)
(Total 8 marks)
Page 17 of 31
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 S1 and S2 are both open. Under these conditions the spot on the oscilloscope
screen is at A.
(i)
Switch S1 is now closed, with S2 remaining open. The spot moves to B. State what
the deflection AB represents.
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(ii)
Switch S1 is kept closed and S2 is also closed. The spot moves to C. State what the
deflection AC represents.
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(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.
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(iv)
Hence, calculate the internal resistance of the battery.
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(6)
Page 18 of 31
(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.
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(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
M1.
(a)
(i)
128 V
1
(ii)
64 V
CE from (i)
1
(iii)
Vrms= 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
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)
I = 24/12 = 2.0 (A)
1
(ii)
peak current = √2 × 2.0 = 2.8 (A) 1
(iii)
peak power = √2 × 12 × √2 × 2.0 = 48 (W)
2
[10]
Page 21 of 31
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 = 8 × 0.5 = 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 = 230 × √2 (1)
peak voltage = 325 V (or 324 V) (1)
2
(ii)
average power = 230 × 0.26 = 60 W (1)
1
Page 22 of 31
(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)
1
(ii)
Vrms = 10.0/√2 = 7.1 (V) (1)
1
(iii)
time period = 3 × 2 = 6 (ms) (1)
1
(iv)
frequency = 1/0.006 or 1/6 (1)
frequency = 167 (1) (Hz)
2
[5]
Page 23 of 31
M6.
(a)
(i)
use of 1.5 cycles (1)
conversion to time eg time for 1.5 cycles = 10 × 1.5 = 15ms (1)
calculation of frequency eg frequency = 1 / 0.010 = 100 ± 3Hz (1)
(ii)
peak voltage = 1.5 × 2 (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)
V0 = √2 Vrms = √2 × 4.2 V (1) (5.94 V)
Vp-p (= 2 × V0) = 2 × 5.94 = 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) = 4.0 × 10–4 s (1)
time base = 4.0 × 10–4/8 = 5.0 10–5 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 (3.5 + 10) (1)
I = 0.13 A (1)
(iii)
V (= RI = 10 × 0.13) = 1.3 V (1)
[or V = ε – Ir = 1.75 – 0.13 × 3.5 = 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
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
(31.8 V)
=32 V (1)
4
(b)
(i)
Irms =
= 59mA (1)
(58.9mA)
(use of 32 V gives 59(.2) mA)
(allow C.E. for value of Vrms from (a))
(ii)
Vrms = 59 × 10–3 × 90 = 5.3(1) V (1)
(allow C.E. for value of Irms from (i)) [or V2 =V1
]
2
(c)
Vpeak = 5.31×
=7.5(1) (V) (1)
best choice: 5 V per division (1)
(allow C.E. for incorrect Vrms and for suitable reason)
reason: others would give too large or too small a trace (1)
3
[9]
M9.
(a)
(V = IR gives)
Vrms = (5.3 × 10–3 × 2 × 103) = 10.6 (V) (1)
V0 = Vrms √2 = 10.6√2 = 15 V (1)
(14.99 V)
[or calculate I0 (= 7.5 mA) and then V0]
2
Page 25 of 31
(b)
(use of T =
gives) T =
= 2 × 10–3 = 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 =
(ii)
T = 10 (ms) (1)
(use of f =
gives) V0 = 7.1√2 = 10 V (1)
gives) f =
= 100 Hz (1)
3
(b)
control 1: time base (1) (or time period)
(use of T =
gives)
T=
= 5 × 10–3 (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
(iv)
(
= V + Ir and emf = 3.5 × 0. 5 = 1.75 (V) gives)
1.75 = 1 .5 + 0.1 07r (1)
r = 2.3 Ω
[or use of
= I (R + r) with I = 0. 107 gives r = 2.4 Ω
and I = 0. 11 gives r = 1.9 Ω]
(allow C.E. for value of I from (iii))
6
(b)
(i)
peak value = 3.5√2 = 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
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
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 Vrms, not converting to V0, 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
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 Vrms/√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 0.107 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.
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