FINAL-IT 21 – ELECTRICAL INSPECTOR EXAMINATION
ANSWER SCHEDULE
Notes:1.
(1 mark) means that the preceding statement/answer earns 1 mark.
2.
This schedule sets out the expected answers to the examination questions. The marker can
exercise their discretion and decide on the overall accuracy of any answer that is presented in
the candidate’s own words.
3.
Symbols and terms - alternatives
Power
W or P
Voltage
V or E or U
Phase
Active
Question 1
Marks
(a)
That people and property are protected
from dangers arising from the work.
(1 mark)
ESR 13(3) Accept answers
that generally
relate to safety
out of ESR 13
(b)
That the installation is electrically safe if
installed to the design requirements.
(1 mark)
ESR 14(1)
(c)
The owner or occupier of the property
where the danger exists.
(1 mark)
ESR 19 (2)
(d)
(i)
For a period of 3 years.
(½ mark)
ESR 68(2)(a)
(ii)
The employer
(½ mark)
ESR 68(2)(b)
●
Yes. A grade A offence has been
committed.
(½ mark)
ESR 71(5)(b)
●
Regulation 71(3)(c) prohibits a
person inspecting work they have
certified.
(½ mark)
●
The work will be electrically safe
when connected to the power
supply.
The work has been carried out in
accordance with the Act and
regulations
(½ mark)
ESR 71(1)(a)
(½ mark)
ESR 71(1)(b)
(e)
(f)
●
(g)
(h)
Reference
●
Suitable
warning
notices
are
attached to the point of isolation.
(½ mark)
ESR 106(1)
●
If a locking facility is available, it
must be used.
(½ mark)
ESR 106(2)
(1 mark)
ESR 119(1)
An arbitrator
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
Marking notes
1
Question 1
Marks
(i)
A combined protective earthing and
neutral (PEN) conductor shall not be
isolated or switched.
(1 mark)
They shall not be installed through any
space formed between the wall-lining
and roofing material and its immediate
supporting member.
(1 mark)
Any TWO of:
(1 mark)
(j)
(k)
●
●
●
(l)
Reference
Marking notes
AS/NZS 3000:
2.3.2.1.1(c)
AS/NZS 3000:
3.9.4.3.1
Shall not be connected to the
protective earthing conductor of the
source of supply.
Shall not be connected to the
exposed conductive parts of the
source of supply.
Shall
be
connected
to
the
equipotential earthing conductor of
the isolated supply.
AS/NZS 3000
7.4.3(c)
AS/NZS 3000
7.4.3(c)
AS/NZS 3000
7.4.6(b)
The fuses or circuit-breakers shall be
installed in all unearthed conductors.
(1 mark)
(m) Shall be low enough to permit the
passage of current necessary to operate
the circuit protective devices.
(1 mark)
AS/NZS 3000:
8.3.5.2
(n)
To ensure that the three-phase motors
operated from any of the outlets will run
in the same direction of rotation:
(1 mark)
GK
(o)
(i)
General purpose.
(½ mark)
GK
(ii)
Anywhere between 1.25 and 1.6
(½ mark)
GK
(i)
Motor applications
(½ mark)
GK
(ii)
Anywhere between 1.6 and 2
(½ mark)
GK
ISC
=
(½ mark)
GK
(p)
(q)
=
230
1.21
190A
AS/NZS 3000
7.5.9.1
(½ mark)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
2
Question 1
Marks
(r)
The tester induces a leakage current on
the RCD and measures the time taken
for it to trip
(1 mark)
GK
(s)
ISC
(½ mark)
GK
(½ mark)
GK
=
=
(t)
2000 kVA
√3 x 400V
2887A
Reference
Marking notes
(i)
Type A, 30mA
(½ mark)
GK
(ii)
Type A,100mA
(½ mark)
GK
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
3
Question 2
Marks
(a)
(1 mark)
GK
NZECP 51 – New Zealand Electrical
Code
of
Practice
for
Homeowner/Occupier’s
Electrical
Wiring
Work
in
Domestic
Installations.
(ii)
Lighting circuit.
1 mm2
(½ mark)
NZECP 51:
Table 3
Socket outlet circuit
2.5 mm2
(½ mark)
NZECP 51:
Table 3
hot
(½ mark)
NZECP 51:
Table 3
The permanently connected room
heater
2.5 mm2
(½ mark)
NZECP 51:
Table 3
Minimum size of 0.75 mm2
(½ mark)
NZECP 51:
5.3.3
Flexible cord of the sheathed,
rounded type
(½ mark)
NZECP 51:
5.3.3
(½ mark)
NZECP 51:
5.4.1
(1 mark)
ESR 64(b)(ii)
AS/NZS 3000:
8.3.5
AS/NZS 3000:
8.3.6
AS/NZS 3000:
8.3.8
(iii) ●
●
connected
1.5 mm2
(iv) 15A
(c)
Marking notes
(i)
The permanently
water cylinder
(b)
Reference
(i)
AS/NZS 3000
(ii)
●
Earth continuity test
(½ mark)
●
Insulation resistance test
(½ mark)
●
Correct circuit connections test
(½ mark)
Lighting circuit.
RCCB and MCB
(1 mark)
Socket outlet circuit
RCCB and MCB
(1 mark)
The permanently connected hot water
cylinder
MCB
The permanently connected room heater
MCB
(½ mark)
(½ mark)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
GK Accept
MCB
current
ratings
GK
as
alternative
GK answers
GK
4
Marks
Question 3
Reference
Marking notes
Total kW of the installation
=
10+ (12 x 0.25) + 20
(1 mark)
=
33 kW
(1 mark)
(b)
10,000
3,000
0.8 = 36.870
20,000
0.5 = 600
Q1
Q2
QT
S
=
ą x tan ǿ
(½ mark)
=
3000 x tan 36.87
(½ mark)
=
2250 VAr
=
ą x tan ǿ
=
20000 x tan 60
=
34641 VAr
=
34641 + 2250
(½ mark)
=
36891
(½ mark)
=
√ 330002 + 368912
(½ mark)
=
49496.9 VA
(1 mark)
(½ mark)
(1 mark)
(1 mark)
49.5 kVA
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
5
Marks
Question 3
Reference
Marking notes
Alternative solution 1
Phase angle of lighting load
=
cos-1 0.8 lag
=
-36.87 degrees
Phase angle of machines load
=
cos-1 0.5 lag
=
-60 degrees
Vertical component of lighting load
=
3kW tan -36.87 degrees
=
2.25KVAr
Vertical component of machines load
=
20kW tan -60 degrees
=
34.64 KVAr
Total load
=
(10kW + j0) + (3kW - j2.25)
+ (20kW - j34.64) KVA
(Rectangular Form.....R +jX )
Total load
=
(33kW- j36.89)
(Rectangular Form Total)
Total load = 49.5KVA
/_ 48.1 degrees
(Polar Form Total)
Alternative solution 2
Lighting
Cos-1 0.8 = 36.86 0
Tan 36.86 0 x 3 Kw
Q1 =
2,250 Kvar
Machines
Cos-1 0.5 = 60 0
Tan 60 0 x 20 Kw.
Q2 =
34.65 Kvar
Qt = Q1 + Q2
=
2,250 + 34.65
=
36 . 9 Kvar
Kw t from (a) = 33 Kw.
S
=
=
√33
=
√P
2
2
+ Q
+ 36 .9
2
2
49 .5 Kvar
Increase in load
=
49.5 + (49.5 x 33.33%)
=
66 kVA
Suitable size transformer
(½ mark)
(½ mark)
=
100kVA
(1 mark)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
6
Question 4
Marks
(a)
(c)
AS/NZS 3112
(½ mark)
ESR 61(1)
(ii)
●
To ensure that any apparatus
or accessory that operates at
12V
Is not able to be connected to
the 230V supply.
(½ mark)
GK
(1 mark)
GK
Each conductor of a multi-core cable
must be insulated for the highest voltage
present.
(1 mark)
AS/NZS 3000
●
The low voltage cables shall be of a
type providing the equivalent of
double insulation.
The low voltage cables shall be of a
shall be installed in a separate
compartment of the trunking.
(1 mark)
When viewed from the front of the
socket-outlet,
the
order
of
connection commencing at the slot
in the radial line shall be earth,
active and neutral in a clockwise
direction.
(1 mark)
The switch shall interrupt all live
conductors.
(1 mark)
They shall have their
conspicuously marked.
(1 mark)
●
(d)
(i)
(ii)
(e)
(i)
(ii)
(f)
Marking notes
(i)
●
(b)
Reference
voltage
Be of a form that prevents it being
inserted into a circuit of greater
than extra-low voltage.
Any ONE of:
●
●
3.9.8.3(b)
AS/NZS 3000
3.9.8.3(a)
(1 mark)
AS/NZS 3000
3.9.8.3(c)
AS/NZS 3000
4.4.5
AS/NZS 3000
4.4.4.4
AS/NZS 3000
4.4.1.2(a)
(1 mark)
AS/NZS 3000
4.4.1.2(b)
7.5.10(a)
(1 mark)
Installation of the luminaire in a
suitable fire-resistant enclosure
AS/NZS 3000
Provision of default clearances from
combustive and thermal insulating
material.
AS/NZS 3000
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
4.5.2.3(b)
4.5.2.3(d)
7
Marks
Question 5
Reference
Marking notes
Marking Notes:
There are two possible solutions to this question.
●
Solution 1 is derived from column 2, Table C1, AS/NZS 3000. This is based on treating
Domestic installation Group 1 and Domestic installation Group 2 as separate groups. That
is, one installation (individual living unit) per phase from each group.
●
Solution 2 is derived from column 3, Table C1, AS/NZS 3000. This is based treating
Domestic installation Group 1 and Domestic installation Group 2 as one group. That is 2
installations/phase – one installation from each group.
The solution must come from either column (not a combination of the two).
The Communal Group maximum demand is the same for both solutions.
Solution 1
(a)
Domestic installation Group 1.
Load Group
Calculation
Group A
Lighting
3
Group B
Socket outlets
10
Group C
Oven
Load (A)
3
(½ mark)
10
(½ mark)
(2500 x 0.5) ÷ 230 = 5.43
5.43
(½ mark)
Group D
Air conditioning
3000 ÷ 230 x 0.75
9.78
(½ mark)
Group F
Storage water heater
2000 ÷ 230 = 8.7
8.7
(½ mark)
36.91
(1 mark)
Maximum demand
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
8
(b)
Domestic installation Group 2.
Load Group
Calculation
Group A
Lighting
3
Group B
Socket outlets
10
Group C
Range
(5000 x 0.5) ÷ 230 = 10.87
10.87
Group D
Sauna
(6000 ÷ 230) x 0.75 = 19.58
19.58
Group F
Storage water heater
(3000 x 0.33) ÷ 230 = 4.35
4.35
3
10
47.8
Maximum demand
(c)
(½ mark)
(1 mark)
Communal Group
Load Group
Calculation
Group H
Lighting
100 ÷ 230 x 10
Group J(i)
Clothes dryer
Group J(ii)
Air conditioning
(1 mark)
4000 ÷ 230 x 0.05
8.7
(1 mark)
6000 ÷ 230 x 0.75
19.58
(1 mark)
32.63
(1 mark)
Maximum demand of the heaviest loaded
phase
=
36.91 + 47.8 + 32.63
(½ mark)
=
(½ mark)
117.34A
Load (A)
4.35
Maximum demand
(d)
Load (A)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
9
Solution 2
(a)
Domestic installation Group 1.
Load Group
Calculation
Group A
Lighting
5 + 0.25
Group B
Socket outlets
15 + 3.75
Group C
Oven
1 x 2.8
Group D
Air conditioning
3000 ÷ 230 x 0.75
Group F
Storage water heater
1x6
Maximum demand
(b)
Load (A)
5.25
(½ mark)
18.75
(½ mark)
2.8
(½ mark)
9.78
(½ mark)
6
(½ mark)
42.58
(1 mark)
Domestic installation Group 2.
Load Group
Calculation
Group A
Lighting
5 + 0.25
Group B
Socket outlets
15 + 3.75
Group C
Range
1 x 2.8
Group D
Sauna
6000 ÷ 230 x 0.75
Group F
Storage water heater
1x6
Load (A)
5.25
18.75
2.8
19.58
6
52.38
Maximum demand
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
(½ mark)
(1 mark)
10
(c)
Communal Group
Load Group
Calculation
Group H
Lighting
100 ÷ 230 x 10
Group J(i)
Clothes dryer
Group J(ii)
Air conditioning
4.35
(1 mark)
4000 ÷ 230 x 0.05
8.7
(1 mark)
6000 ÷ 230 x 0.75
19.58
(1 mark)
32.63
(1 mark)
Maximum demand
(d)
Maximum demand of the heaviest loaded
phase
=
42.58 + 52.38 + 32.63
(½ mark)
=
(½ mark)
127.59A
Load (A)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
11
Marks
Question 6
(a)
(i)
Is dated not less than 6 months
before the proposed connection.
(½ mark)
(ii)
Any FOUR of:
(2 marks)
●
●
●
The installation is electrically
safe
The
work was
done in
accordance with the Act and
regulations
The installation has been
installed in accordance with a
design prepared in accordance
with Part 1 of AS/NZS 3000.
Reference
Marking notes
ESR 73(2)(b)(ii)
ESR 67(1)(a)
ESR 67(1)(b)
ESR 67(2)(a)(i)
Or
The installation has been
installed in accordance with a
design prepared in accordance
with Part 2 of AS/NZS 3000
●
The declaration of conformity
has been sighted.
ESR 67(2)(a)(ii)
●
Tests have been satisfactorily
completed.
ESR 67(2)(b)
●
The rating of the
system is correct.
earthing
ESR 67(2)(c)
●
Fittings are safe to connect to
a power supply.
ESR 67(2)(d)
●
The installation is safe to be
connected
to
the
power
supply.
ESR 67(2)(e)
(iii) Any ONE of:
●
Confirm that there is a
declaration
of
conformity
signed by the installer.
●
Verify its electrical safety
before checking and testing.
(1 mark)
(iv) ●
What tests were carried out
(1 mark)
ESR 73(5) Accept answers
relating
to
testing
information
being
on
a
certificate
of
compliance
The results of the tests
(1 mark)
ESR 73(5)
●
ESR 73(2)(a)(v)
ESR 73(2)(a)(v)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
12
Marks
Question 6
(b)
Reference
Marking notes
(1 mark)
ESR 73(3)
AS/NZS 3019
(½ mark)
ESR 74(3)
Verification certificate
(½ mark)
AS/NZS 3019:
(v)
The person who did the tests.
(i)
(ii)
2.9
(iii) ●
It was issued no earlier than 6
months before the date of
reconnection
It certified that the installation
is suitable for use.
(½ mark)
ESR 74(3)(a)
(½ mark)
ESR 74(3)(b)
●
It was given by a person
authorised to certify mains
work.
(½ mark)
ESR 74(3)(c)
(iv) ●
Shall
be
verified
by
measurement of the fault loop
impedance.
At the point of connection of
the protective device.
(½ mark)
AS/NZS 3019:
●
●
2.5
(½ mark)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
AS/NZS 3019:
2.5
13
Question 7
Marks
Reference
(a)
(1 mark)
AS/NZS 3000
(i)
AS/NZS 3000, requires that the
device
must
automatically
disconnect the supply before the
overcurrent:
could cause injury
-
(ii)
(b)
●
Marking notes
2.5.1
damage the installation
The rating of each circuit breaker
shall not exceed the rating of the
socket outlet.
Three-pin flat pin
(1 mark)
AS/NZS 3001
2.2.7.4(b)
(½ mark)
AS/NZS 3001
2.2.7.2(a)
●
Complying with AS/NZS 3112
(½ mark)
Minimum rating of 15A
(½ mark)
Round pin
(½ mark)
AS/NZS 3001
2.2.7.2(c)
(c)
●
Complying with IEC 60309
(½ mark)
Minimum rating of 16A
(½ mark)
Three-pin flat pin
(½ mark)
AS/NZS 3001
2.3.2(a)
●
Complying with AS/NZS 3112
(½ mark)
Round pin
(½ mark)
AS/NZS 3001
2.3.2(b)
●
Complying with AS/NZS 3123
(½ mark)
Round pin
(½ mark)
AS/NZS 3001
2.3.2(c)
Complying with IEC 60309
(d)
●
●
Shall
be
protected
by
overcurrent circuit breaker.
(½ mark)
an
The rating of the circuit breaker
shall not exceed the rating of:
the supply lead; or
-
(1 mark)
AS/NZS 3001
3.3.3.1
(1 mark)
AS/NZS 3001
3.3.3.1
the supply lead fittings,
Whichever is the smaller.
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
14
Reference
Question 8
Marking notes
Marks
(a)
(i)
a.c. voltage
(½ mark)
(ii)
A range that will read 230V
(½ mark)
(iii) ●
●
A temporary earth stake
(½ mark)
An earth lead
(½ mark)
(b)
Meter box
Neutral bar
V
V
Main switch
Earth bar
Single-phase consumer
●
●
(c)
(i)
Testing between the temporary earth
and the line side of the main switch
Testing between the temporary earth
and a point on the MEN connection.
(1 mark)
●
About
230V
between
the
temporary earth and the line side
of the main switch.
About 0V between the temporary
earth and the MEN connection.
(1 mark)
About 0V between the temporary
earth and the line side of the main
switch.
About
230V
between
the
temporary earth and the MEN
connection.
(1 mark)
●
(ii)
●
●
(1 mark)
(1 mark)
(1 mark)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
15
Reference
Question 8
Marking notes
Marks
(d)
●
●
The fault current is too low to blow the
fuse.
Because the impedance of the fault
loop is too high.
(1 mark)
(1 mark)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
16
Marks
Question 9
(a)
Maximum volt drop permitted = 230 x
0.05 = 11.5V
(½ mark)
VD
(½ mark)
=
V/A.m x amps x metres
Reference
Marking notes
1000
V/A.m
=
VD x 1000
(½ mark)
amps x metres
=
11.5 x 1000
(½ mark)
10 x 300
=
3.83 V/A.m
(1 mark)
From table 42 - converted to threephase
=
3.83
(½ mark)
1.155
=
3.316
(1 mark)
From table 42, a 16 mm2 cable, with an
operating temperature of 75 0C has an
mV/A.m of 2.43
Therefore, a 16 mm2 cable will satisfy
the voltage drop requirements.
(b)
I(FL) =
(½ mark)
(1 mark)
P
√3 x V
(½ mark)
=
70000
√3 x 400
(½ mark)
=
101.04A
(1 mark)
From table 12, a 25 mm2 cable is rated
for 104 amps.
(½ mark)
The correction factor from table 27(1) is
0.94. The cable rating would be 104 x
0.94 = 97.76 amps.
The 25 mm2 cable is too small for the
load, so the correct size is 35 mm2.
(½ mark)
(1 mark)
FINAL-IT 21 ANSWERS – 20 NOVEMBER 2010
17