Answer Key
ANSWER KEY (CHAPTER 1):
1) B
2) A
3) D
4) A
5) B
6) D
7) A
8) B
9) A
10) C
11) D
12) A
13) C
14) D
15) C
16) A
17) D
18) D
photovoltaics (page 1)
Grid Parity (page 4)
Renewable Energy Certificates (page 5)
Only about 10 years old (page 2)
that utilities allow solar energy system owners to sell excess energy to the utility (page 7)
occurs when the light from the sun is converted into electricity (page 10)
a photon (page 11)
Bell Labs (page 11)
electrons (page 11)
Antoine-César Becquerel (page 10)
renewable portfolio standard (RPS) (page 5)
distributed energy system (page 7)
interconnecting (page 7)
solar renewable energy certificate (page 6)
National Fire Protection Association (page 15)
Article 690 (page 15)
NFPA 70 (page 15)
feed-in tariff (FIT) (page 4)
19) A
20) B
21) D
22) A
23) A
the duck curve (page 9)
renewable portfolio standard (page 9)
a national recognized testing laboratory (NRTL) (page 18)
OSHA (page 20)
OSHA (page 20)
ANSWER KEY (CHAPTER 2):
1) B
2) D
3) A
4) C
5) B
6) D
7) A
8) D
semi-conductors (page 27)
band gap (page 28)
Monocrystalline cells (page 29)
Polycrystalline cells (page 29)
Multi-Junction photovoltaic cells (page 31)
Thin-Film (page 30)
1,000 watts per square meter (page 35)
atmospheric pressure equal 1.5, and cell temperatures of 25 degrees Celsius (page 36)
9) C
10) B
11) D
12) A
13) A
14) B
15) B
16) A
17) B
18) D
19) C
20) B
21) D
22) A
23) B
24) D
25) C
PTC addresses wind speed while STC does not (page 36)
the I-V curve (page 37)
maximum power point (page 38)
diode (page 28)
bypass diodes (page 40)
blocking diodes (page 41)
.55 volts (page 29)
ANSI/UL 1703 (page 35)
circuit (page 29)
the relative humidity (pages 38-40)
open circuit voltage (Voc) (page 38)
maximum power voltage (Vmp) (page 38)
1,300 watts/square meter (page 36)
nominal operating cell temperatures (page 37)
less (page 39)
temperature coefficient (page 39)
Answer A is not correct as temperatures lower than 20 degrees Celsius will result in an increase
in voltage, thus an increase in power output. Answer B will likely result in no more than a 10%
decrease in production if it falls off at an annual rate of 1% per year Answer C will produce at
minimum a 25% reduction in output (and likely greater given that unshaded cells will also be
affected). Answer D is simply comparing one of efficient panel with another and will have no
effect on production as they are both rated at 200 W.
26) D perovskite solar cells (page 31)
27) C a solar photovoltaic cell (page 34)
28) D busbar cells (page 33)
ANSWER KEY (CHAPTER 3):
1) D
2) B
3) A
4) A
5) C
6) A
7) D
8) C
off-grid system (page 51)
grid-interactive system (page 53)
grid-tied system (page 52)
grid-tied system (page 52)
hybrid system (page 55)
load requirements (page 57)
915 kWh (page 57)
watt-hours (Wh) (page 57)
Because the price of a watt-hour of electricity is so low, units are usually expressed per 1,000,
or kilowatt-hours. But this was not a choice, so watt-hour is the best answer.
9) D reduce the load demand through energy efficiencies (page 58)
10) A system load (pages 57-58)
11) C
12) A
13) C
400 Wh.
The mixer runs twice each week (for this problem it does not matter how long it runs each
time). Each time it is operated, it consumes 1,400 Wh. So two times per week times 1,400 Wh
equals 2,800 Wh per week. There are seven days in a week, so the weighted daily load is 2,800
Wh / 7 days = 400 Wh per day.
maximum power draw (page 60)
daily load demand (page 59)
The easiest way to calculate the answers to questions 14-16 is to create a table such as the one below with
the information provided:
Maximum Power Draw
Daily Load Demand
Water Heater
1,700 Wh/day
Milking Machine
900 Wh/day
Grain Mixer
1,200 W
Totals:
To calculate the daily load demand for the grain mixer: 1,200 W times 2 hours times 2 times per week = a
weekly load of 4,800 Wh per week. Divide by 7 days per week = 4,800 Wh/7 days = 686 Wh per day
To calculate the maximum power draw for the water heater: 1,700 Wh divided by 4 hours = 425 W
To calculate the maximum power draw for the milking machine: 900 Wh divided by 3 hours = 300 W
Now simply add the new information to the table and add up the results.
Maximum Power Draw
Daily Load Demand
Water Heater
425 W
1,700 Wh/day
Milking Machine
300 W
900 Wh/day
Grain Mixer
1,200 W
686 Wh/day
Totals:
1,925 W
3,286 Wh/day
14) A
15) B
16) C
17) C
18) D
19) C
3,286 Wh
1,925 W
1,200 W.
If only one machine can operate at a time, select the load that draws the most power as the
maximum power draw, in this case the grain mixer.
6OOV (page 55)
1,500 v (page 55)
1,000 v (page 55)
ANSWER KEY (CHAPTER 4):
1)
2)
3)
4)
5)
C
B
D
B
C
6) C
7) B
8) B
9) C
10) B
11) B
12) D
13) D
14) B
15) C
16) A
17) D
18) D
19) D
20) A
21) A
22) C
23) B
24) B
25) A
26) B
amps (page 70)
volts (page 70)
Always wear a respirator when working with exposed wiring. (page 71)
amps times volts (page 70)
regular household electrical (page 69)
1,000 watts
Electricity is sold in units known as a kilowatt hour (kWh). One kWh is equal to one thousand
watts of electricity consumed over a period of one hour.
240 watts consumed for 15 minutes. To determine watt-hours simply multiply watts times hours.
Option A = 6,000 W (6 kilowatts) x 10 hours = 60,000 Wh. Option B = 240 W x .25 hours = 60 Wh.
Option C = 60 W x 24 hours = 1,440 Wh. Option D = 120 W x 2 hours = 240 Wh
the voltage increases but the amps remain the same (page 80)
the voltage remains the same, but the amps increase (page 80)
hertz (page 82)
Since the waveform of a DC current is flat, there are no cycles. Therefore cycles per second
(hertz) is not relevant.
Use a smaller (in diameter) wire size. (page 84)
Using a larger diameter wire (smaller gauge) will lower voltage drop.
bonding and grounding (page 89)
bonding and grounding (page 90)
voltage drop (page 83)
the National Electrical Code (NEC) (page 13)
the grounding electrode (page 89)
ohms per 1,000 feet (page 84)
electrons can flow through it (page 79)
8.33 amps.
Current is measured in amps. Therefore only D can possibly be correct (all other options are
expressed in volts. When panels are hooked together in series, the voltage increases but the
amps stay the same. So the amp rating of one panel will be the amp rating of the entire string.
108 Vdc.
18 volts per panel times six panels = 108 volts. Since all solar panels generate DC current, the
output will be in volts – DC.
volts = current x resistance (V=I x R) (page 83)
resistance = 1 (or infinite) (page 87)
8.23 V
Use the voltage drop equation, Vd = Imp x R x 2d. Imp (amps) has been given as 8.3 amps. The
distance (d) is 400 feet. Looking at Table 4-1 find that the resistance of #10 AWG stranded
copper wire is 1.24 ohms per 1,000 feet. So Vd = 8.3 amps x 1.24 ohms/ 1,000 ft x 800 feet = 8.23
volts
17.15%.
If 8.23 volts are lost from a 48 volt system, then Vd% = 8.23 volts/48 volts = 17.15%
hertz (page 82)
harmonization (page 73)
ANSWER KEY (CHAPTER 5):
1) B
2) A
3) B
4) B
5) D
6) A
7) A
8) B
9) D
10) C
11) A
12) B
13) A
14) D
15) C
16) A
17) C
18) D
19) A
20) B
21) A
22) D
23) D
24) A
25) B
26) C
27) B
28) D
29) D
30) A
31) C
32) B
33) B
nominal voltage (page 98)
open-circuit voltage (page 38)
standard test conditions (page 35)
produce less power (page 36)
MC4 connectors (page 100)
0 AWG (page 103)
0 AWG (page 103)
its ampacity rating. Ampacity is a life/safety issue, while the price and resistance are
economic issues. Wire is generally not rated for voltage.
USE-2 (page 107)
pure sine wave (page 120)
bulk, absorption, floating (page 114)
70 amps (page 104)
35 amps (page 106)
According to Table 5-3, a conduit with 10 to 20 wires will be derated by 50%. So with 15 wires,
70 amps (original rating) derated by 50% = 35 amps.
anti-islanding feature (page 120).
Anti-islanding shuts down the system when the grid looses power. Since a stand-alone system
is not hooked to the grid, this feature is unnecessary.
the panels do not need to be replaced as often (page 126)
bimodal (page 125)
string (page 101)
increase the system voltage (pages 133-134)
Ah at C20 (page 136)
grid-tied
Only the grid-tied system (of these choices) connects to the utility, so it is the only one that
requires an interface (point of contact) with the utility.
islanding (page 120)
it is a semiconductor (page 27)
its Ah capacity (page 136)
PV panel (page 2)
Bicrystalline PV (pages 29-30)
at that rate of discharge, it will take 20 hours to fully drain the battery (page 136)
provide 20 amps of power for 15 hours to a 50% DOD.
Ah = amps x hours. So 20 amps x 15 hours = 300 Ah. 50% DOD of a 600 Ah battery = 300 Ah.
pulse width modulation (PWM) (page 113)
combiner box (page 101)
surge protection (page 103)
fixed voltage inverters with power optimizers (page 121)
easier to recycle (page 138)
lithium-ion (page 138)
34) A
35) B
36) D
will decrease by about 1% per degree C below 20°C (page 134)
they are very expensive compared to other alternatives (page 142)
they are extremely rugged and robust - can handle large amounts of
physical stress. (page 143)
37) C
combiner boxes connect PV source circuits in parallel while junction boxes do not (page
38) C
39) A
40) A
41) D
42) B
43) A
44) A
45) D
46) D
102)
a temperature of 30° C (86° F) (page 104)
black, red, or any other color ( other than green or white) (page 109)
black, red, or any other color ( other than green or white) (page 109)
six (page 111)
inverters must comply with Rule 21 and be listed as L1741-SA compliant (page 122)
smart inverters (page 123)
a transfer switch (page 128)
self-contained energy storage systems (page 139)
there is no voltage limit on these systems
The NEC Article 706:30 is silent as to how high the voltage can be if the ESS is self
contained. The wording of the article is as follows: 706.30 Installation of Batteries.
(A) Dwelling Units. An ESS for dwelling units shall not exceed 100 volts between
conductors or to ground. Exception: Where live parts are not accessible during routine
ESS maintenance, an ESS voltage exceeding 100 volts shall be permitted.
47) D
48) B
the size of the array (page 124)
the maximum power draw (page 119)
ANSWER KEY (CHAPTER 6):
1) A
2) A
3) A
4) B
5) D
6) C
7) D
8) D
9) D
10) C
11) D
12) D
13) D
14) D
Add additional panels to the system. Adding a tracking system would require a complete
redesign of the system. As the panels are only 5 years old, replacing all of them would not
likely be in the customer's best interests. Adding batteries only affects storage capacity,
not power output.
solar insolation map A solar insolation map was long used to give a vague idea of how much
irradiance a specific location might receive on an average day. But today most designers use a
much more accurate system such as PVWatts.
azimuth (page 156)
altitude (page 159)
sun-tracking (page 160)
passive air (page 166)
latitude (page 161)
latitude plus 15 degrees (page 161)
its demarcation point (page 165)
a sun chart (page 164)
all of the above (pages 165)
none of the above (page 167)
none of the above (page 175)
insolation (page 156)
15) C
16) D
17) D
18) B
19) A
20) D
21) D
22) B
23) A
24) C
The shipping weight of the racking and panels. (page 165)
This is largely determined by the racking system design. (page 167)
future proofing (page 174)
orient the array 11 degrees east of magnetic north
20% less (page 158)
Azimuth: southeast (135°), Altitude 45°
Imaging systems often rely on outdated information (page 171)
total solar resource (TSR) (page 172)
total solar resource fraction (TSRF) (page 172)
79.57 %
The total insolation is 4.68 hours per day. The site actually only receives 3.92 hours per day
due to the tilt and orientation of the array. So the tilt and orientation factor (TOF)= 3.92
hrs/4.68 hrs = 0.8376
The shading factor (SF) is 100% – 5% = 95% or 0.95
The total solar resource fraction (TSRF)= tilt and orientation factor (TOF) x shading factor
(SF) or TSRF = 0.8376 (TOF) x 0.95 (SF) = 0.7957 or 79.57 %
25) C
26) B
NEMA Type 3 (page 178)
3.84 kW
If the service panel is rated for 100 amps, then the 120% rule allows for 120 amps of power to
enter the box (a combination of the utility service and the PV array). Since the utility provides
100 amps of power, then the array can enter on a 20 amp breaker.
Since this is grid tied, the output from the inverter must match the grid – so it will be
240 Vac. 20 amps x 240 V = 4,800 W. But the NEC requires the wire and breaker be
sized with a 1.25 safety margin – so 4,800 W / 1.25 safety factor = 3,840 W
So the output from the inverter cannot exceed 3,840 W.
ANSWER KEY (CHAPTER 7):
1) C
2) C
3) D
4) C
5) D
6) A
7) D
8) C
9) A
10) A
11) B
Increase the size of the wires used. (pages 84-85)
Short Circuit Current (Isc) (page 38)
derate (page 198)
a ground fault (page 223)
amp-hours (page 136)
PV Input Circuit (page 228)
600 volts (page 55)
50 volts (page 55)
electrical metallic tubing (page 240)
The azimuth of the array. (pages 203-208)
be UL 1741 listed. (page 124)
12) B
13) A
14) D
15) D
16) B
17) B
18) B
19) C
20) D
21) C
22) A
23) B
24) C
25) A
#8 AWG with a resistance of .809.
Using the voltage drop equation, Vd = Imp x R x 2d, find the variables. Amps (Imp) is provided
at 8.5 amps. Distance is 100 feet. The resistance of each wire is given in the answer (.510/1,000
ft for #6 AWG, etc). The question asks for the smallest wire (largest AWG number) that will
allow for less than a 2% voltage drop from a 90 volt system. So calculate the allowable voltage
drop by 90 x .02 = 1.8 volts (below this amount of voltage drop will constitute less than 2%
system loss). Solving for A = 8.5A x .510Ω x 200ft/1000ft = .867 volts. This is less tan 1.8 volts,
but is it the smallest wire that meets this criteria? Solving for B = 8.5A x .809Ω x 200ft/1000ft =
1.375 volts, which also meets the criteria. Solving for C = 8.5A x 1.29Ω x 200ft/1000ft = 2.193
volts, which exceeds 2%. So B is the correct choice.
DC Voltage Converter (page 263)
PV Panel, DC Disconnect, Charge Controller, 12 Vdc Battery Bank, DC Voltage Converter.
There is no AC power in this system, so A, B, and C with AC Disconnects and Inverters are
not correct.
52.19 amps.
The source circuit wiring is sized by taking the Isc of a panel (since each string reflects the amps
of a single panel, regardless of how many are connected in series) x 1.25 (NEC safety factor) x
1.25 (solar variation). So 8.35 A x 1.25 x 1.25 = 13.05 A. But the question asks for the rating of
the PV Output Circuit, which is the circuit leaving the combiner box. Since four strings are
combined in the combiner box, then 13.05 A x 4 strings = 52.19 amps.
increase.
We have not been told how far off the surface of the roof the conduit is placed, or the
temperature at the site, so with this limited information we simply know that exposing the
conduit to sunlight will increase the required amp rating of the wire.
15 amps.
It has already been determined that the amp rating for each PV source circuit is 13.05 amps.
Since the wire is exposed to sunlight, it may have to be derated to carry this current, but the
current does not change. Since the combiner box is not exposed to sunlight, round up to the
next available circuit breaker size from the anticipated amps. Rounding up from 13.05 amps
leads us to select a 15 amp breaker.
Be located at least 10 feet away from the point where the PV system connects to the grid.
(page 165) It should be within 10 feet.
The inverter's power rating (in watts). (page 259)
redirects excess power from the array to a load dump, such as a hot water heater. (page
116)
Peak hours of sunlight (insolation) for the location. (pages 203-206)
2 kW.
When sizing a stand-alone system, the inverter is based on the maximum load demand. All the
rest of the information is unnecessary. (page 261)
They are easily accessible for repairs. (page 126)
UL 1741 listed string inverter (page 121)
8.33 kW.
Take the monthly load (960 kWh) / 30.5 (days per month) = 31.475 kWh per day. Divide this by
the hours of peak sunlight per day (insolation) = 31.475 kWh/4.2 hours = 7.494 kW. But since
the system is not 100% efficient, so derate this = 7.494 kW / .90 = 8.33 kW.
26) B
27) C
28) A
29) A
30) C
31) A
32) A
33) C
34) B
35) B
36) A
37) D
38) C
39) D
7 panels.
Begin with daily load (9.6 kWh) and divide this by the insolation = 9.6 kWh/ 6.54 h= 1.468 kW
system. Derate it for efficiencies, so 1,468 W / .95 = 1,545 W is the size of the array. The panels
and micro inverters are rated at 250 W, so 1,545 W / 250 W = 6.18 panels. Since the question
asks how many panels are required to fully meet load demand, round up to 7 panels.
8 batteries.
To size a battery bank system, begin with the daily load (3 kWh). Divide this by the voltage of
the system to get amp-hours = 3,000 Wh / 48 volts = 62.5 Ah. Multiply times days of autonomy
= 62.5 Ah x 2 days = 125 Ah. Derate by the depth of discharge = 125 Ah / .60 DOD = 208.33 Ah.
Now derate by the efficiency of the inverter = 208.33 Ah / .95 = 219 Ah. This is the required
storage capacity of the system. Divide the capacity of the battery selected by this amount =
250 Ah / 219 Ah = 1 string of batteries will successfully supply the system. Divide the voltage of
the system (48 V) by the voltage of a single battery (6 V) to determine how many batteries are
required in each string = 48 V / 6 V = 8 batteries. Since only one string is required, only 8
batteries are required.
total solar resource (page 199)
inverter load ratio (ILR) (page 215)
12 panels maximum.
To calculate the maximum panels in a string, begin with the temperature variation from STC
and the coldest day ever experienced at the site. -37°C – 25°C = Δ 62°C. Multiply this times the
temperature coefficient at Voc = 62°C x .30%/°C = 18.6%, meaning the panel may produce 18.6%
more voltage on the coldest day ever experienced on the site. Take the Voc (37.8 V) x the new
factor (118.6%) = 37.8 V x 1.186 = 44.83 V, finding that the panel now might produce 44.83 volts
on the coldest day. The maximum input voltage rating for the inverter selected is 550 Vdc, so
550 Vdc / 44.83 Vdc = 12.27 panels. Since we do not want to exceed the maximum rating, round
down to 12 panels in a string.
4 panels minimum.
To calculate the minimum panels in a string, begin again with the temperature variation from
STC and the hottest day. 43°C – 25°C = Δ 18°C. We must now add the temperature adjustment
based on where the array is mounted. Since this is a ground mounted system, add 25°C, so
18°C + 25°C = Δ 43°C. Multiply this times the temperature coefficient at Vmp, finding 43°C x .
45%/°C = 19.35% less voltage on the hottest day ever experienced on the site. Take the Vmp (31.1
V) x the new factor (80.65%) = 31.1 V x .8065 = 25.08 V, finding that the panel now might
produce only 25.08 volts on the hottest day. The minimum input voltage rating for the
inverter selected is 100 Vdc, so 100 Vdc / 25.08 Vdc = 3.99 panels. Since we must exceed the
minimum rating for the inverter to operate, round up to 4 panels in a string.
clipping (page 215)
ripple (page 235)
MPPT voltage input range (page 209)
18 inches (460 mm) deep (page 244)
the array is too small for a load side connection (page 252)
the 10-foot tap rule (page 253)
800 square feet (80 square meters) (page 255)
load shifting (page 270)
ANSWER KEY (CHAPTER 8):
1) C
2) A
3) D
4) B
5) C
6) A
7) D
8) A
9) C
10) A
11) A
12) B
13) B
14) A
15) A
16) A
17) D
18) D
flush-to-roof racking system (page 298)
helical piers (page 312)
all of the above (pages 298-300)
there is usually very little room to work behind the panels on a roof-top installation (page
306)
run perpendicular to the long side of the panel. (page 306)
the weight of the racking system. (page 300)
the Occupational Safety and Health Administration (OSHA) (page 20)
be made of fiberglass. (page 293)
Heavy Duty, Type I Blue (page 292)
top of the ladder should extend at least three feet above the point where it comes in
contact with the structure. (page 293)
extend below the frost line. (page 312)
subtract the take-up from the stub height. (page 317)
every 10 feet and within 3 feet of each box, fitting or cabinet. (page 316)
fish tape (page 318)
use the three-point climbing technique. (page 294)
Material Safety Data Sheet (MSDS) (page 297)
connections that become hot, and can lead to arcing or fires (page 307)
galvanic corrosion (page 308)
ANSWER KEY (CHAPTER 9):
1) C
an interconnect agreement (page 327)
2) A
one-line drawing (page 331)
3) B
array generating capacity (kW) (page 334)
4) D Commissioning forms (page 335)
5) A operation and maintenance documents (page 336)
6)
ANSWER KEY (CHAPTER 10):
1.
2.
3.
4.
5.
D
A
C
C
C
punch list (page 340)
Ensure all devices and disconnects are in the closed position.
reactance testing (pages 343-344)
System functioning tests. (page 345)
3.73 kW.
First determine the voltage at which the system should be operating. Take the Vmp (29.8 V) and
adjust it for temperature. The current cell temperature is 42°C. 42°C – 25°C (STC) = Δ 17°C. 17°C x
temperature coefficient (.485%/°C) = 8.245%. So the panel will generate 8.245% less voltage because
the actual temperature is greater than 25°C (STC). So Vmp (29.8 V) x 91.755% = 27.34 V. Next
calculate the amps under which the system will operate. 934 W/m2)/STC irradiance (1,000 W/m2) =
.934. Amps under STC (8.4 A) x .934 = 7.8456 A. Power output from the array is W = I x V, so W =
27.34 V x 7.8456 A = 214.5 W per panel. There are 20 panels in this array (5,000 W / 250W per panel
= 20 panels). So the output of the array will be 214.5 W per panel x 20 panels = 4,290 W. However
the question asks for the output from the inverter. So the system derate must be factored in. So
4,290 W x .87 (derate) = 3,732.3 W or 3.73 kW.
ANSWER KEY (CHAPTER 11):
1)
2)
3)
4)
5)
6)
C
A
A
C
C
D
periodic monitoring (page 353)
oversizing of wire and conduit (pages 354-355)
heat fade (page 355)
a ground fault (page 356)
too many batteries in series (page 356)
heat fade (page 355)