Revision questions
Question 20S: Short Answer
1. energy = power × time
microwave oven energy = 1 kW × 1/6 h = 0.166 kWh
kettle energy = 3.3 kW × 1/6 h = 0.55 kWh
toaster energy = 0.5 kW × 1/6 h = 0.083 kWh
total energy = 0.8 kWh daily
2. energy = (0.1 kW × 3 h) = 0.3 kWh daily
3. total energy = 90 × (0.8 kWh / d + 0.3 kWh / d) = 90 × 1.1 kWh = 99 kWh
cost = 99 kWh × 10 p / kWh = 990 p = £9.90
4. Answer depends on students' values.
5. 9 kW × 1/12 h × 10 p /kWh = 7.5 p (However don't forget the landlord has to maintain the shower,
pay a quarterly meter charge and plan for replacing the shower.)
6. Electrons going down a potential difference pass energy to vibrations of atoms in the filament.
Energy leaves the filament carried by electromagnetic radiation (light and infrared).
7. P = I V; power = 0.5 V  3 A = 1.5 W
8. Energy = power time (in seconds) = 1.5 W 10 h 3600 s hr1 = 54 000 J
Kinds of light bulb
Question 30S: Short Answer
1. Power in watts – rate of energy use; operating potential difference in volts, p.d. at which it is
designed to work.
2.
1
Bulb
Power / W
p.d. / V
Headlamp
36
12
Torch bulb
0.09
3
0.03
Filament bulb
100
230
I
Flashlight bulb
4.5
9
0.5
Energy Saving
24
230
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Resistance / 
Current / A
36 W
3A
12 V
I
I
4.0
R
100 W
 0.43 A
230 V
530
R 
24 W
0 10 A
3V
 100 
0.03 A
9V
 18 
0.5 A
230 V 2
Bulb
Power / W
p.d. / V
Energy Saving
bulb
24
230
Current / A
I
Resistance / 
24 W
 0.10 A
230 V
R
230 V 2
24 W
 2. 2 k 
3. To prevent too much power passing through the appliance – the fuse melts.
4.
Appliances
power
rating
p.d. / V
Iron
1200 W
230
Vacuum cleaner
900 W
230
Head lamp
48 W
12
Jug kettle
2.4 kW
230
Radio
100 W
230
Travel kettle
340 W
120
Microwave
cooker
1.4 kW
230
operating current /
A
Suggested
fuse values
choosing
from 3 A;
5A; 13 A
I
1200 W
 5 .2 A
230 V
13 A
I 
900 W
 3 .9 A
230 V
13 A
I
48 W
4A
12 V
5A
I
2400 W
 10 .4 A
230 V
13 A
I 
100 W
 0 .43 A
230 V
3A
I 
340 W
 2 .8 A
120 V
3A
I
1400 W
 6 .1 A
230 V
13 A
5. Voltmeter
6. The answer could be 'the p.d. across all parallel branches is the same' or 'the potential at the
points A, B and C is the same (assuming the connecting leads have negligible resistance). The
potentials at the three points F, E and D are equal but differ from A, B and C. Therefore the
potential difference between the stated pairs must be the same.'
7. Following the argument above there is no potential difference between the specified pairs and the
reading will be 0 V.
8. The cells are in series so the total p.d. is 8  1.5 V = 12 V.
9. Step-down transformer.
10. Mains supply is an alternating supply; battery supplies direct current; mains supply is at 230 V,
batteries usually run at much lower potential differences.
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Ions in chemical cells:
Large and small numbers 1
Question 40S: Short Answer
1. Q = NA e = 6  1023  1.6  10-19 C = 96 kC (1 Faraday)
2. V = 300  103 J / ( 2  96  103 C ) = 1.6 J C–1 or volts
3. 0.2 A = 0.2 C s–1
Number of electrons per second is
0.2 C s –1
1.6  10 19 C per electron
 1.25  1018 electrons s –1
4. 1.3  1018 s–1 / 6  1023 = 2.1 mol s–1
5. 2.1 mol  65.4 g / 2 = 68 g s–1
6. About 20 hours (5 g / (68 g s–1) = 73.4  104 s)
Ions in x-ray machines:
Large and small numbers 2
Question 50S: Short Answer
1. Pelvis: charge = Q = It = 350 × 10–3 A × 0.8 s = 0.28 C
energy = QV = 0.28 C × 65 × 103 V = 18.2 kJ
Hand: charge = Q = It = 80 × 10–3 A × 0.1 s = 8.0 mC
energy = QV = 8 × 10–3 C × 40 × 103 V = 320 J
2. Thicker tissue and heavier bones in the pelvis compared to the hand.
3. n = I / e = 0.35 A / 1.6  10–19 C = 2.2  1018 electrons s–1
E = e V = 1.6  10–19 C  65  103 V = 10.4  10–15 J
4. In 0.8 s, 0.8 s  2.18  1018 s–1 = 1.76  1018 electrons arrive carrying a total energy of 1.76 
1018  10.4  10–15 J = 18.3 kJ
Electrons in copper:
Large and small numbers 3
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Question 60S: Short Answer
1. n = I / e = 10 A / 1.6  1019 C = 6.25  1019 electrons s–1
2.
volume =
volume 
length 
number of electrons
number per unit volume
6.25  1019
 0.625 mm 3
10 20
0.625 mm 3
1 mm 2
 0.625 mm
3. All the electrons in this length drift through the section in 1 s so the drift velocity is about 0.625
mm s–1.
4. Only 1/100 of current therefore the drift velocity will be 6.25  10–3 mm s–1. It would take an
electron
t
1 mm
6.25  10 – 3 mm s –1
 160 s
to drift 1 mm.
Some circuit problems
Question 100S: Short Answer
1. Use the formula power = I 2R
Resistor Value
/
Power Rating /
W
Working Current
7.5
1.3
0.42 A
47
0.5
100 mA
3.3 k
2
25 mA
27 k
2.2
9 mA
680 k
0.25
0.6 mA
1.1 k
1
30 mA
2. 1 W resistor
I
1W
 13 mA
5600 
2 W resistor
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2W
 19 mA
5600 
I
so use 2 W resistor to carry 15 mA
3. A larger surface area allows quicker cooling
4. Conductance of the 2  resistor = 0.5 S, 5 resistor = 0.2 S, 10  resistor = 0.1 S.
5. 2  resistor = 0.5 S
6. combined conductance = 0.5 S + 0.2 S + 0.1 S = 0.8 S
7. I = G V = 0.8 S  3 V = 2.4 A
8. A straight line graph from the origin to 12 V; 8 cm
9. 100  setting:
V
12 V

 0.06 A
R 200 
I
P  I 2 R  (0.06 A) 2  100  = 3.6 W
20  setting:
V
12 V

 0 .1 A
R 120 
I
P  I 2 R  (0.1 A) 2  100  = 1.0 W
50  setting
I
V
12 V

 0.08 A
R 150 
P  I 2 R  (0.08 A) 2  100  = 0.64 W
Power of appliances
Question 110S: Short Answer
1. 720 J of energy is delivered per second to the kettle to heat the water and surroundings
2. The water will take longer to heat up in New York since the energy is transferred more slowly.
3.
P 340 W

 2. 8 ( 3 ) A
V
120 V
P 720 W
 
 3.1 (3) A
V
230 V
I120 
I 230
4. Almost double the p.d. across the element will result in double the current, leading to 4 times the
power and serious overheating which will overheat the insulation etc.
5. 5 A fuse will stop too high a current flowing.
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Effect of an ammeter in a circuit
Question 120S: Short Answer
1. P = I V; I = P / V = 24 W / 12 V = 2.0 A
2. Resistance of each bulb when operating normally:- R = V / I = 12 V / 2 A = 6 . Resistance of
bulb and ammeter = 16 ;
I = 12 V / 16  = 0.75 A.
In fact the increased resistance will reduce the current through the lamp, so it will not be at
normal brightness.
3. Resistance of two bulbs in parallel = 3 ; Total circuit resistance = 13 ; I = 12 / 13 = 0.92 A
Combining conductances
Question 125S: Short Answer
1.
G = 1 /10 = 0.1 S
2. G = G1 + G2 = 0.1 + 0.1 = 0.2 S
3. G = 1 /100 = 0.01 S
4. G = G1 + G2 = 0.1 + 0.01 = 0.11 S
5. Knowing that R = 1 / G makes calculation easy:
R = 1 / G = 1 / 0.11 S = 9.1 
6. The total resistance, 9.1 , is less than 10  which is the smallest resistance of the two parallel
resistors.
Circuit resistance
Question 130S: Short Answer
1. Circuit current = 0.8 A (Equivalent resistor for parallel arrangement = 2.4 ; total circuit
resistance = 7.4 )
2. Circuit current = 2.5 A (Equivalent resistor for series arrangement = 8 ; Total circuit resistance =
4.8 ) or current = 12 V / 12  + 12 V / 8  = 2.5 A
3. Circuit current 2.0 A = (Equivalent resistor for the parallel arrangements:- 2.35  and 3.53 ;
Total circuit resistance = 5.88 )
4. I = 4 V / (3  + 6 ) = 0.44 A
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5. p.d. = 3.2 V ( I
12 
= 4 V / (12 + 3 ) = 0.27 A; V
12 
= 0.27 A  12  = 3.2 V)
Combining resistors
Question 140S: Short Answer
Possible combinations
1. Three 1.0 k in series.
2. 6.8 k and 2.2 k in series.
3. 1.0 kand 1.0 k in parallel.
4. Two 5.6 k in parallel then connected in series with a 2.2 k.
5. Parallel combination of two sets of 3.3 k in series with 4.7 k
Electrical characteristics of a resistor
Question 150S: Short Answer
1. The resistor is ohmic for p.d.s up to about 30 V. At larger applied p.d.s its resistance is gradually
increasing (possibly because there is a heating effect).
2. Slope of the graph where it is linear  4700 
3. R = 50 V / 9.41 mA = 5370 
4. The answer comes from the student's graph which should cover the range 40 V to 50 V. The
resistance = 5170  i.e. 4700 + 10% occurs at an applied p.d. of about 43 V.
5.
P
V2
R
V  PR  1 W  4700   68 V
The algebra of power
Question 160S: Short Answer
1.
P 500 W

2A
V
250 V
V 250 V
R 
 125 
I
2A
I 
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2.
P 24 W

2A
V
12 V
V 12 V
R

6
I
2A
I
3.
P = I 2 / G; P = V 2 G
Tapping off a potential difference
Question 170S: Short Answer
1. 100 / (50  + 100  )  6 V = 4 V
2. 100 / (50  + 100  + 100 )  6 V = 2.4 V
3. Resistance of parallel combination = 50 . So the 6 V supply potential difference splits equally
between the two 50  and p.d. across AB is 3 V.
4. resistance of whole potential divider is 10 k
p.d. across 4 k  =
4 k
 20 V = 8 V
10 k 
p.d. across 6 k =
6 k
 20 V = 12 V
10 k
p.d. across 10 k =
10 k
 20 V = 20 V
10 k
5. The resistance has been connected as a variable resistor. The multimeter has an extremely high
resistance so that wherever one moves the sliding contact the p.d. is set up across the voltmeter
which always reads 6 V! The redrawn diagram should show the ends of the resistance connected
across the battery
6. The resistance of whole potential divider is 400 . The supply p.d. 12 V splits between the fixed
300  resistor and the 100  potentiometer in the ratio of their resistances. So there will be 9 V
across 300  and 3 V across 100 .
When the slide contact from B is next to A the p.d. tapped between A and B is 0 V; when the slide
is at the other end of the potentiometer the full p.d. of 3 V is across AB.
Loading the potential divider
Question 180S: Short Answer
1. Digital voltmeter reading = 3 V. Digital voltmeter reading = 5 V.
2. Moving coil meter will read 2.67 V; Equivalent resistance of 500  voltmeter and 125  of
potentiometer = 100 ; p.d. tapped off = 100 / (100  + 125)  6 V = 2.67 V. Moving coil
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meter will read 4.74 V; Equivalent resistance of 500  voltmeter and 50  of potentiometer = 45
; p.d. tapped off = 45 / (45  + 50 )  10 V = 4.74 V.
3. The moving coil meter changes the effective resistance to 33 . The p.d. read by the moving coil
voltmeter = 33 / (33  + 50 )  6 V = 2.4 V.
Controlling a robot arm
Question 200S: Short Answer
1.
I
V
3V

 1 mA
R 3  10 3 
2. 1.5 V
3. 0.1 V
4.
V 
10 
360 
 3 V = 0.1 V
Using a measurement amplifier as a comparator
Question 210S: Short Answer
1.
p.d. across LDR =
300 
 9 V = 1.5 V
1800 
2. The resistances are in the same ratio (1:5) so the potential differences across them are in the
same ratio too.
3. zero
4. The resistance of the LDR will increase, so the potential difference across it will increase. The
potential of terminal A increases, but the potential of terminal B stays the same. There will be a
potential difference across AB, with A the more positive terminal.
5. When the light level is too low, terminal A is more positive than terminal B. Terminal A is
connected to the positive (non-inverting) input of the amplifier; terminal B to the negative
(inverting) input. Thus the output of the amplifier will go positive and turn the lights on.
6. This circuit is a possible solution:
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LDR (80
1 k
+
A
B
9V
400 
5 k
–
output
+
measurement
amplifier
Internal resistance of power supplies
Question 220S: Short Answer
1. The two 1.5 V cells provide an emf of 3 V in series. If the current flowing is 0.5 A as stated, then
for the potential difference across the internal resistance to be 0.5 V (that is, 3 V–2.5 V) the
internal resistance of the cells combined would need to be 1 . The cells are in series so the
resistance of each is 0.5 .
2. The battery must have a low internal resistance so as to be able to deliver a current of 80 A from
an emf of only 12 V. If the internal resistance is 0.05  then the potential difference across this
with a current of 80 A flowing is 4 V. Thus the potential difference across the 12 V battery drops
to 8 V. This is a big enough change to dim headlights rated at 12 V.
3. The maximum current is 60 A. A torch bulb has a resistance of only a few ohm, so connected
across such a supply the potential difference across it would be very near to zero, with a current
of only is 60 A through it.
4. The 1000  voltmeter draws a current from the cell, of 1 mA when it reads 1.0 V. If the cell has
internal resistance some of its emf will be used in driving the current through the cell. A voltmeter
with very high resistance draws very little current, and reads nearer to the emf of the cell. If the
emf is 1.2 V then 0.2 V is used in driving the current of 1 mA through the internal resistance,
which is therefore 200 .
Resistance and conductance of thermistors
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Question 250S: Short Answer
1. Largest value is 1760 k. Smallest value is 0.0307 k, or 30.7 . Ratio of largest to smallest
values in first column is 176 k / 0.0307 k = 5700 approximately. When the ratio of largest to
smallest values in a batch of data much exceeds 10, a simple linear plot will rarely be helpful. A
logarithmic plot is probably called for. It is good practice to look first of all at the largest and
smallest values, when looking at a new set of data.
2. Here is the linear plot of resistance against temperature for the type 198-927.
200.0
150.0
100.0
50.0
0.0
–50
0
50
100
150
200
250
temperature / C
The graph shows that the resistance falls with temperature, rapidly at first and then more slowly.
The graph is almost useless because, to fit in the values of about 100 k, the scale has to make
values less than about 5 k invisible. The variation above 100 C cannot be seen at all.
3. Here is what the logarithmically scaled graph looks like.
1000.0
100.0
10.0
1.0
0.1
0.0
–50
0
50
100
150
temperature / C
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200
250
Equal intervals on the resistance scale now represent equal multiples of resistance (10). The
graph now shows the variation of resistance over the full temperature range. The resistance
always decreases with temperature. The graph is much less curved than before. If it were straight
it would show that the resistance decreases by a constant factor for each equal increase in
temperature. Because the graph curves upwards, the ratio of resistances between pairs of
equally spaced temperatures must decrease somewhat as the temperature rises.
4. The ratio gradually reduces, from 1.8 between –30 C and –20 C to just over 1.1 between 240
C and 250 C. This fits with the logarithmic graph curving upwards: that is, decreasing more
gradually as the temperature rises.
5. Here is what the combined graph looks like. All the five graphs are pretty well parallel. This
means that the resistances of all five thermistors vary in the same fundamental way with
temperature. There is such a thing as 'thermistor behaviour', not just the behaviour of this or that
thermistor.
10000.0
1000.0
100.0
10.0
1.0
0.1
0.0
–100
0
100
200
300
temperature / C
6. The conductance of the type 198-927 at 250 C is 32.5 mS. The conductance should rise with
temperature, rather faster for low temperatures than for high temperatures.
7. Here is what the logarithmic graph of resistance against 1 / T looks like.
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1000.0
100.0
10.0
1.0
0
0.001
0.002
0.003
0.004
0.005
0.1
0.01
1 / T (K)
Response time of thermistors
Question 260S: Short Answer
1. C
2. Longest response time B
Shortest response time C
Greatest sensitivity A
Least sensitivity C
Response time of A = 5.0 s
Response time of B = 8.0 s
Response time of C = 2.7 s
Sensitivity of A = 0.015 V °C-1
Sensitivity of B = 0.010 V °C-1
Sensitivity of C = 0.005 V °C-1
3. The output must be across the fixed resistor, since it rises when the thermistor resistance drops.
The fixed resistor takes a greater fraction of the supply voltage for both circuits.
4. The smaller thermistor warms up much more rapidly, having a much smaller response time than
the larger one.
5. The system is logging at about 1 sample s-1. The blue response time is about 3 ± 1 s and the red
response time about 100 ± 10 s.
6
There is no significant difference in response time between warming and cooling for either
thermistor.
7. The blue thermistor shows a decrease in output after it has reached its maximum reading. It
drops from just under 5 V peak to around 4.8 V after 130 s from the start. The slow response time
of the red thermistor means that it misses this variation.
8. The two average sensitivities are about equal, since the p.d. changes are about equal for
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eventually equal temperature rises. The red circuit is slightly more sensitive.
Response time of light sensors
Question 270S: Short Answer
1. C
2. The output must be across the fixed resistor, since it rises when the LDR resistance drops. The
fixed resistor takes a greater fraction of the supply voltage.
3. The first graph is best for the LDR (blue), since slow switching shows its slow response and the
second graph is best for the photodiode (red) since it seems to respond almost instantaneously at
the slower frequency. Response times: LDR (blue) about 0.5 ± 0.1 s and photodiode (red) about
5 ± 1 ms.
4. There is no significant evidence of difference in response time between brightening and
darkening for the photodiode. There is some evidence that the LDR responds a little quicker to
increases in intensity than to decreases. The LDR response has not finished decreasing before
the illumination increases again even at the lower frequency.
5. The LDR no longer follows the rapid fluctuations of intensity; it responds to some mean intensity
level similar to the human eye, which cannot see the flickering of the LED at 100 Hz.
6. At 1 Hz nominal the signal generator is 9% high in frequency; % error = (1.1 – 1.0) × 100% / 1.1 =
9% high. At 100 Hz nominal it is 7% high in frequency: 11 / 0.102 s = 108 Hz; % error = (108 –
100) × 100% / 108 = 7% high.
Both signals show a small amount of noise, but it is not random. It has a regular periodicity, and is
correlated for both sensors; in fact it is in anti-phase – blue peaks at the same time as red
troughs. It also has a frequency of about 50 Hz, so it cannot be due to fluctuating light intensity
from a nearby mains lamp (which occurs at 100 Hz). In any case, the LDR does not respond fast
enough to 'see' changes at this frequency. Thus the noise must be due to mains interference in
the sensing circuits. They must be affected by the 50 Hz mains signal with opposite senses of
induced voltage.
7. For example: As a light sensor on an automatic street light on / off switching circuit, there would
be no serious problem with the light turning on 0.5 s ‘late’.
In trying to measure the level of the rapid fluctuations in intensity from a mains lamp, an LDR
would be too slow and unsatisfactory. The photodiode would follow the fluctuations more quickly.
Baby, it's cold outside:
The uses of sensors in the care of newborn babies
Question 10C: Comprehension
1. Temperature, oxygen level, humidity. They are in a box in the Perspex hood. They measure
temperature, humidity and oxygen level of the air in the incubator.
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2. Skin temperature; heart rate, breathing rate, blood pressure
3. Miniaturisation of sensors means that the baby is not any longer 'scarcely visible under the
technology'. Parents can play a bigger role in care. High technology 'stick-on' sensors to measure
oxygen levels through the skin mean less stress on the baby. Less blood is taken for tests.
4. The sensor detects a breath beginning, and triggers a system to assist the breathing.
Electron beams
Question 70C: Comprehension
1. See A–Z section
2. The energy of the electrons passes to atoms in the screen, which then emit light carrying the
energy away.
3. Travel as rays, in a straight line, casting a shadow.
4. Rays of light are not deflected by an electric field.
Rays of light are not deflected by a magnetic field.
The cathode rays were deflected by a magnetic field.
5. A potential difference is applied to the deflecting plates, which sets up an electric field between
these plates.
6. The slits will narrow the beam making a smaller glowing spot on the end of the tube.
7. The beam moves upwards to the positive deflection plate; negative charge is attracted to positive.
8. Perpendicular to the plane of the paper
9. SI unit for mass is the kilogram; SI unit for charge is the Coulomb. e / m = 1 C / 1.3  10–11 kg =
7.7  1010 C kg–1
10. Kinetic energy given to each electron as it moves through potential V; eV = 1.6  10–19 C  5000 J
C-1 = 8.0  10–16 J
11. Power dissipated = IV = 2 10–3 A  5000 V = 10 W
12. Number of electrons arriving per second = 10 W / 8.0  10–16 J s–1 = 1.3  1016 s–1
13.
eV 
1
2 mv
2
ev
e
 2   v  2  1.76  1011 C kg –1  5000 V
m
m
2
11
v  1.76  10 m 2 s 2
v2  2
v  4.2  10 7 m s 1
14 Assume the electrons emitted from the cathode have no initial speed. All the potential energy is
converted to kinetic energy – this would not be the case if there were any collisions of electrons
with residual gas molecules in the tube.
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15 As a fraction, 4.2 107 m s–1 / 3 108 m s–1 = 0.14
Sensors and our senses
Question 80C: Comprehension
1. Nervous system and electrical sensors both detect change and react to the change; however
nervous system is connected to a much more sophisticated processor which makes decisions
about how to react based on far more data and previous experience.
2. touch – pressure pads
sight – light sensors
sound – microphone
taste and smell – chemical sensors
3. Detecting gas leaks, pollution, illnesses, drugs; quality control in food production, perfumery.
4. Both are sensitive to changes in intensity, the eye is better at detecting colour, the eye sees in
black and white at low intensities.
Sensors and Formula 1 racing
Question 90C: Comprehension
1. 4375 m. The throttle graph rises from near zero to about 85 % (we don't know if the negative
values on that scale mean anything – probably not). About 10 m.
2. The speed increased from 130 kilometres per hour to 180 kilometres per hour, an increase of 50
kilometres per hour. The average speed during this period is about 155 kilometres per hour. Point
A is at about 4375 m. Point B is at about 4475 m. As the average speed 155 kilometres per hour
is 43 m s–1, so time taken to travel 100 m is 2.3 s. The speed was 50 kilometres per hour, or
nearly 14 m s–1, in 2.3 s, giving an acceleration of 6.0 m s–2, more than half the acceleration of
free fall.
3. The braking force cause a deceleration, slowing the car progressively. Forces affect only the rate
of change of velocity.
4. The student book and this CD should help here – try searching on sensor.
Lamp and resistor in series
Question 190D: Data Handling
1.
RA 
16
3V
 12 
0.25 A
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RB 
3V
 7 .5 
0.4 A
2. A is an ohmic resistor, perhaps a metal wire. The resistance of B increases with current. It could
be a filament lamp.
3. From the graph when p.d. across A is 3 V the current is 0.25 A.
4. A and B are in series so 0.25 A also passes through B. So from the graph, the p.d. across B is
1.5 V.
Therefore the p.d. of the supply is 3.0 V + 1.5 V = 4.5 V.
5. At a p.d. of 6 V both have resistance 12 , and carry current 0.5 A. They each dissipate the same
power, 3 W.
Using non-ohmic behaviour
Question 270D: Data Handling
1. Current proportional to p.d. up to about 7 V – metals obey Ohm's law. Above 8 V the resistance is
increasing because the current is heating up the coil.
2. The average resistance below 7 V can be found from the gradient of the straight line section = 24

3. When the applied p.d. exceeds 8 V, the coil starts to get hot. Power at this pd is P = I V = 2.5 W
approx.
4. The diode does not conduct until the applied p.d. exceeds 0.6 V (the threshold). It is not an ohmic
device. The current increases faster than the applied p.d. (because the number of charge carriers
increases).
5.
Applied p.d. / V
0.50
0.55
0.60
0.65
0.70
0.75
0.80
Resistance / 
>1k
550
200
65
23
7.5
2.8
6. You will need to calculate or estimate the equivalent resistance of the parallel combination for
some applied p.d.s.
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3. Currents in the meter and the diode
300
250
200
150
100
50
0
0
0.2
0.4
0.6
0.8
1.0
p.d. / V
diode
meter
both
7. When the unprotected milliammeter reads 20 mA, the pd across it (V = I R) must be 0.5 V. When
the milliammeter is protected, the resistance of the diode is too large in comparison with the coil's
resistance to affect the circuit and carry any significant current.
If the circuit current rises to about 25 mA, the p.d. across the meter exceeds 0.6 V. The diode in
parallel will become conducting at that applied p.d. and divert some of the current away from the
coil. The student's graph shows that if the current through the combination rises much higher,
most of the current flows through the diode; the meter will not be damaged by high current.
However the current cannot rise indefinitely since the 68  resistor limits the circuit current to 175
mA.
Heating coils
Question 230X: Explanation–Exposition
1. In series the same current I passes through both resistors. Power dissipation P = I 2R; i.e. P  R.
If coil A has half the resistance of B we can predict that A heats up half as slowly as B.
If the coils are in parallel, the potential difference is common to both. Use P = V 2 / R; i.e. P  1 /
R or P  G. The prediction is that A heats up twice as fast as B.
2. The current through the coil will be larger when it is in parallel so we will expect the final
temperature to be higher in the second experiment.In more detail:- Consider the series circuit as
a potential divider: –1 / 3 of the potential difference is set up across A. Let the resistance of A be
'R'. The power dissipation = ((1 / 3)V)2 / R = V 2 / 9R. In the parallel circuit the power dissipation =
V 2 / R. Theoretically the temperature rise of the oil in the second experiment should be nine times
larger than in the first experiment.
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3. The rate of heat transfer to the oil will not be the same. Coil B has a bigger surface area and will
transfer heat faster. Although the same power source is used, its output can vary with the load
resistance.
The rate of heat transfer from a hot object to its surroundings depends on the temperature
difference, so the hotter oil will lose more energy.
Brightness of bulbs
Question 240X: Explanation–Exposition
1. When the resistance of the variable resistor is very low, it reduces the resistance of the parallel
resistance combination of itself and L2, which in turn lowers the p.d. across L2 and raises the p.d.
across L1. The total circuit resistance is also lower and the current through L1 is higher. Result: L1
is brighter than normal and may even blow; L2 is dimmer or may not even light at all.
2
If the resistance of R is made far higher than the resistance of the lamps, the resistance of the
parallel combination will always be slightly less than the resistance of L2 alone and the total circuit
resistance hardly changed from its original value. L2 now gets brighter. Result:- At its maximum
setting R has no effect on the circuit; it draws a negligible current.
Using a sensor in a potential divider
Question 260X: Explanation–Exposition
1.
+6V
R
2 k  3 k
Vout
10k
load circuit
equialent
0V
NB: A lower value resistor would
require a very bright light to change
over the p.d.
2. Lowering the value of the load resistor will affect the division of potential difference over the
potential divider. The change over will not be so noticeable and may not be a high enough
threshold voltage to trigger the counter.
3.
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+6V
o
hot >100 C
to switching
circuit
0V
NB: Using a variable resistor allows the
switch to be set in the real situation to
switch off at 100o C
4. 900 
Filament lamp and thermistor in series
Question 280X: Explanation–Exposition
1. With current on the y-axis and p.d. on the x-axis, the graph should slope steeply for low currents
(low resistance, high conductance), and then curve to become flatter at high currents (higher
resistance, lower conductance).
As the filament wire gets hot the conductance decreases and the resistance increases.
2. With current on the y-axis and pd on the x-axis, the curve should have a small slope to begin with
(high resistance, low conductance at small currents), and then start to rise more steeply (lower
resistance, higher conductance, as the thermistor becomes hotter).
As the thermistor gets hotter its conductance increases and the resistance decreases.
3.
When the switch is closed both components are cold. The thermistor has a high resistance which
limits the current. Most of the potential difference is across the thermistor and so most of the
power is dissipated in the thermistor. This heats up the thermistor so its conductance increases
and more of the potential difference is applied across the filament lamp. More power is dissipated
by the lamp which now glows brighter. The bulb glows more and more strongly as the thermistor's
conductance increases. As the filament gets hotter its resistance increases. Eventually the
situation stabilises when the resistance of the filament is far greater than that of the thermistor, so
that nearly all the power is dissipated by the lamp.
4. Surge currents often occur when circuits are turned on and this large initial current, although only
lasting for a fraction of a second, can still overload a component. The high resistance of the
thermistor (when cold) limits the size of the surge current.
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