Year 10 Advanced Science Electricity Test 2013
Name: ___________________________________
Max marks = 116
% = ________ out of 110
Examination Report & Suggested answers – E. Hung @2010 Nov
Part A : Twenty four Multiple Choice Questions @ 1½ marks – Total = 36% marks
1. Which of the following statements does not represent ohm's law? (potential difference = voltage)
A. current / potential difference = constant
B. potential difference / current = constant
C. potential difference = current x resistance
D. current = resistance x potential difference (Only V=IR is the Ohm’s Law)
(This seems to be a difficult questions to many students)
2. The unit of charge is _____________.
A. ampere (current)
B. watt (power)
C. volt (voltage)
D. coulomb
3. The potential difference (=voltage) required to pass a current 0.2 A in a wire of resistance 20 is ____.
A.
100 V
B.
4V
V=IR = 0.2x20 = 4V
C.
.01 V
D.
40 V
4. The resistance of an electric bulb drawing 1.2 A current at 6.0 V is ___________.
A.
0 .5 
B.
5
R=V/I = 6/1.2 = 5 
C.
0.2 
D.
2
5. Three resistances of 100 , 35  and 30  ohm are connected in series. The overall resistance will be
A. 55 
B. 135 
C. 160 
D. 165  Being in series, just adding all resistance = 100+35+30 = 165 
6. Two resistors 20  , 20  are connected in parallel, what is the overall equivalent resistance?
A.
10 
B.
C.
D.
15 
20 
40 
7. The following shows a combination circuit, which is the correct
description of the circuit?
A. 6  , 3  and 9  are in series
B. 9  and 6  are in parallel and the combination is in series with 3 
C. 3  , 6  and 9  are in parallel
D. 3  , 6  are in parallel and 9  is in series
( A series circuit with a parallel part)
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Year 10 Advanced Science – Electricity – Page 1
8.
When a fuse is rated 8 A, it means _____________.
A. it will not work if current is less than 8 A.
B. it will work only if current is 8 A
C. it will melt if current exceeds 8 A
(A fuse is used to protect the circuit from an overloading of current; the circuit has
a current over 8A, it is be broken and prevent overheating of the circuit and possibly having a fire)
D. it has a resistance of 8 
9. The equivalent resistance across AB is
A. 4 
B. 2 
C. 1 
D. 0.5 
(This is a rather difficult question to many students. This is
actually a series circuit with 2 parallel parts. You need to find the
equivalent resistance of each parallel part first)
(Each parallel part has a Equivalent resistance of ½ , therefore
the Total Equivalent resistance = ½ + ½ =1 }
10. Which type of substance is easy to produce static electricity:
A. conductors
B. insulators
C. electrolytes
D. semiconductors
11. Kilowatt - hour is the unit of ______________.
A. potential difference
B. electric power
C. electrical energy (Unit of Household Electricity)
D. charge
12. The free electrons of a metal _____________, this allows metal to conduct electricity.
A. do not collide with each other
B. are free to escape through the surface
C. are free to fall into the nuclei
D. are free to move anywhere in the metal (related to Q10, making it easy to conduct electricity)
13. Heat produced in a current carrying wire in 5s is 60 J. The same current is passed through another wire of half the resistance.
The heat produced in 5 s will be _____________.
A.
15 J
B.
30 J (Less resistance in the heating coil, less heat is transformed from the same amount of current. This is a difficult
question – can use an advanced formula: P=VI  P=(IR)(I)  P=I2R)
C.
60 J
D.
120 J
14. Three equal resistances when combined in series are equivalent to 90  . Their equivalent resistance when
combined in parallel will be __________.
A.
270 
B.
30 
C.
810 
D.
10  (This may help students if they can draw diagrams)
(3 equal resistance in series give 90, therefore, each resistor must have 90/3=30)
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15. The resistivity of a wire depends on ____.
A. length (the longer, the higher the resistance)
B. material Some materials have high resistance, eg. Tungsten, nichrome; others like silver and
copper have very little resistance)
C. area of cross- section (the thinner, the higher the resistance)
D. all of the above
Year 10 Advanced Science – Electricity – Page 2
16. A charge of 30 coulomb passes through a wire in 3 sec. The current flow in the wire, in ampere, is:
A. 3.3
B. 10 Q=IT  I = Q/T = 30/3 = 10A
C. 90
D. 30
17. The filament of bulb is made of:
A. mercury
B. copper
C. tungsten
(High resistance, good to transform electricity to heat, then light)
D. argon
18. In parallel combination, total resistance:
A. decreases (that is why light bulbs in parallel have more current and become brighter)
B. increases
C. may decrease or increase according to situation
D. no particular observation
19. A dynamo develops 0.5 ampere at 6 volt. The power produced is:
A. 15 W
B. 17 W
C. 3 W
P=VI = 6 x 0.5 = 3 Watts
D. 2 W
20. How much electrical energy in kilowatt hours is consumed in operating ten 50 watt bulbs for 10 hours
per day in a month (30 days) ?
A. 1500
B. 15000
C. 15
D. 150 E=Power (in KW) x Time (in hours)
Total kilowatts = 10 x50  1000 = 0.5 kW
Total time = 10 x 30 = 300 hours
Therefore E = 0.5x300 = 150 kWH
While multiple-choice exams may seem challenging, you can prepare yourself by understanding the structure
of a multiple-choice test and how to approach these types of exams. Remember, the correct answer is right
there in front of you! By carefully analyzing each question and choices offered, you can increase your
chances of performing well on each multiple-choice test you take.
The Structure of a Multiple-Choice Question
Each question consists of just three parts:
1. The first part a multiple-choice problem is the basic section that asks a question, gives and
incomplete sentence, or poses a problem that you are expected to solve.
2. The next part of the question is a number of distracting alternatives. These are the incorrect
answers that are designed to test your true knowledge of the subject. Some of these
alternatives may seem correct, so it is important to know the topic well to avoid selecting an
incorrect answer.
3. The final part of a multiple-choice problem is the correct answer to the question or problem
that is posed.
Techniques For Exams
In addition to following good study habits, there are steps you can take during an exam to ensure you choose the correct answer.
Year 10 Advanced Science – Electricity – Page 3
12
Part B: Fill in the Blanks [ 15 marks ]
Use these words to complete the following sentences:
(not all words need to be used, some may be used more than once)
1. Electric current is a measure of the amount of electrons passing a particular point in an
electric circuit every second. Electric current can be measured with an ammeter .
2. Voltage is a measure of the amount of electrical push in a circuit. The voltage can be
measured with an voltmeter .
3. Electrical resistance is a measure of how difficult it is for electricity to move around a
circuit. It is measured in
ohms.
4. Electricity supplied by a battery is called direct current.
5. The greater the voltage in a circuit, the higher the current.
6. The greater the resistance in a circuit the less current it will
carry.
7. A generator changes kinetic energy into electrical energy.
8. An electric motor changes electrical energy into kinetic
energy.
9. A fuse is an electrical device that can interrupt the flow of electrical current when it is
overloaded.
10. An electrical resistor, with two terminals, whose resistance is continuously variable by
moving a knob or slider, is known as a rheostat .
Rheostat is inside the dimmer.
Part C: Short answered Questions
Year 10 Advanced Science – Electricity – Page 4
15
All
calculations must show formulas, steps and units clearly.
1.
Draw circuit diagrams for the following electrical circuits. You need to include the
direction of current in each circuit.
[ 3+3 ]
NB: Direction of electron flow is from
negative to positive; the current is from
positive to negative.
2 bulbs in series
2.
Use circuit diagrams to show 4 different ways of connecting 3 light bulbs in a circuit. [4]
All in series
Parallel and series
All in parallel
Many students repeat
the same circuit !
Series and parallel
Year 10 Advanced Science – Electricity – Page 5
3. .
a) Find VT of the circuit. [2]
VT of the circuit means the total voltage
supplies to all loads, ie. R1, R2 and R3 in
series.
VT = V1 + V2 + V3 = 10 + 24 + 36 = 70V
(Straight forward question)
b) Find the value of R1, R2 and R3 if the
current is 2A [6]
If the ? A is given to be 2A, you can
find the individual Resistance using
R=V/I
R1 = 10/2 = 5
R2 = 24/2 = 12
R3 = 36/2 = 18
(This seems to be a very easy question, most students did well)
4.a) Find the Total Equivalent Resistance of the circuit if [3]
R1 = 10 , R2 = 20  and R3 = 40 
b) If the battery is 12V, find the maximum current given by
the battery.
[3]
Maximum current means Io
c) Find the value of I1, I2 and I3
[6]
Since all resistors are in parallel, the voltage supplying each of them is the same as the
original Vo= 12V, using I = V/R
I1 = 12/10 = 1.20A
I2 = 12/40 = 0.30A
I3 = 12/20 = 0.60A
d) What is the relationship between I, I1, I2 and I3?
Adding all current in parallel branches = 1.2 + 0.3 + 0.6 = 2.1A, same as Io
Therefore Io = I1 + I2 + I3
(You need to prove that the above equation is correct!)
[2]
5. The diagram shows a circuit diagram with 3 resistors - 15, 25 and an unknown resistor C.
Year 10 Advanced Science – Electricity – Page 6
a) What is the current in A and B?
[2]
Most students found this question difficult. It is atypical of
the set up. Given two known resistance with an unknown
resistance C in series.
The traditional way of drawing this circuit may be:
100V
2A
b)
B
C
25
Unknown
A
15
This diagram seems to be
clearer. The current at Resistor
C is 2A, so did all the other
resistors as this is a series
circuit.
No calculation is needed. Many
students did calculations for
individual resistor – WRONG!
The Answers are:
A = 2A, B = 2A and C = 2A
Find the value of the Resistor C.
[4]
There are 2 methods:
Method (1) Let C be the unknown resistance, the total equivalent resistance = 25+15+C = 40+C
ohms
Given the voltage =100V and the Current = 2A, applying V=IR  R = V/I
40 + C = 100/2  40 + C = 50  C = 50 – 40 = 10
Method (2) Find the Total equivalent resistance of the circuit: R = V/I = 100/2 = 50
Take away B and A: C = 50 – 25 – 15 = 10
This is an excellent question to test student’s thinking and mathematical skills.
Congratulation to those who got full marks.
6. Use the circuit diagram below to answer the questions below
Give the formula(s), show calculations and give the
answer with the correct unit.
R1 = 100 Ohm, R2 = 240 Ohm, R3 = 60 Ohm
a) Connect an ammeter that measures the current
through R2 and connect a voltmeter that
measures the potential difference across R1.
(Draw them on the diagram)
[2]
Draw the Voltmeter in parallel and the
Ammeter in series in front of R2.
b) Find the combined resistance (Total Equivalent Resistance) of R1, R2 and R3.
[4]
In order to find the Total Equivalent Resistance, you need to analyse the arrangement of
the resistors. Many students simply used the formula:
1 1 1 1
= + +
R R R R
T
1
2
3
However, R2 and R3 are not parallel to each other.
You need to find the Equivalent of R2 and R3 in series first. RE = R2 + R3 = 240 + 60 = 300
Year 10 Advanced Science – Electricity – Page 7
Now, R1 and RE are parallel to each other, you can use
c) Find the current from the battery (labelled A) in two decimal places.
[3]
That is the Ammeter next to the battery. If you have trouble finding the correcting answer in Part
b), you will have the wrong answer in this part. However, marks will be given to correct steps.
V=IR  I=V/R (R is the Total Equivalent Resistance)
I = 10/75 = 0.13A
d) Find the current through R3 in two decimal places.
[3]
The current in the parallel branch with R1 is different from the current in parallel branch with R2
and R3. Since the current in R2 and R3 is affected by both resistors added together. In order to find
the current, you need to use the RE of R2 and R3.
e) Find the voltage over R2.
[3]
100
I1
I2 = V/RE = 10/300 = 0.03A
NB: The current passing through R2 and R3 must be the same
Since we have found the current in the R2 and R3 branch, we can find the
individual voltage across each resistor. In this question, you need to find
Voltage across R2 (240).
Don’t use the answer of I2 as 0.03A (not accurate), use the fraction answer.
I2
R1
300
RE
10V
V2 = I2R2 = 1/30 x 240 = 8V (if you’ve used 0.03A, then V
=0.03x240 = 7.2V inaccurate!)
I 2= 1/3 A
240
10V
f) Show how you calculate another data (not mentioned before) from this circuit setup. [3]
You can find the Voltage across R3 or Current flowing through R1
V3 = I3R3 = 1/30 x 60 = 2V
NB: V2 + V3 = V0 = 8+2 = 10V
or
I1 = V1/R1 = 10/100 = 0.10 A
Overall, students found this question very difficult as they ddin’t know how to simply combined
circuit. Once they knew, they should find this type of questions (VCE) easy!
Year 10 Advanced Science – Electricity – Page 8
7. a) An hair drier is attached to 240V electric plug. If the heater draws 15 A,
calculate the Power of this heater.
P = VI = 240 x 15 = 3600 Watts
b) What does the value of Power mean?
Power is the amount of energy in Joules used in 1 second.
This hair drier actually used 3600 Joules in 1 second.
[2]
[2]
8.
What current is drawn from two 150 Watt – halogen light tubes in series at 240V? [3]
Most students did this question carelessly as they forgot to multiple the 150 Watt by 2 as
there were 2 light tubes.
Total Power = 150 x 2 = 300 Watts
P = VI  I = P/V = 300/240 = 1.25 A
9. An electric hotplate draws 5 amps of current at 240 V when it is heating up at maximum power.
a) a) Find the Power of this hotplate.
[2]
P = VI = 240 x 5 = 1200 Watts (very straight forward question)
b) If the hotplate is used at full power for 30 minutes, find the amount of energy used in
Kilowatt-hour.
[3]
Electricity Energy can be in Joules or KWh.
E (Joules) = Power (Watts) x Time (seconds) or
E (KWh) = Power (KW) x Time (hours) – this one is used in this question!
Power = 1200 W = 1.2 KW Time = 30 min = 0.5 hour
E = 1.2 x 0.5 = 0.6 KWh
c) if an electricity authority charges 25.5c per KWh, how much will it cost for the use of the
same hotplate for 45 minutes.
[3]
The time is 45 mins = 0.75 hour
Cost = E x Rate = 1.2 x 0.75 x 25.5cents = 22.95 cents
9. The symbol for electric charge is written as Q and its unit is coulomb “C”. The charge of
one electron is equal to the charge of one proton, which is 1.6 x 10 - 19 C .
a) Find the number of electrons in one coulomb of charge.
[2]
1electron 1.6 ´10 C
=
x
1 C
-19
The following equation may be able to help you:
Cross multiplication: 1.6x10-19 (x) = 1
1
1´10 10 ´10
x=
=
=
= 6.25 ´10
1.6 ´10
1.6
1.6
19
18
18
-19
b) Using the formula: Charge = Current x Time Q = IT
[2]
Find out the current of the circuit if 4 Coulomb of charges (electrons) are passing through
in 3.5 sec.
One ampere means one Coulomb of charge or 6.25 x 1018 electrons are passing
through a point in a second.
Q=IT  I = Q/T = 4/3.5= 1.14A
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Year 10 Advanced Science – Electricity – Page 9