Question 1 of 270
Choose the proper ammeter and voltmeter connection in this circuit.
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
A
B)
B
C)
C
D)
D
Points: 0 out of 1
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The voltage is always taken in parallel and the amperage is always taken in series in an electric circuit.
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Question 2 of 270
What is the equivalent resistance of this series circuit?
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
400 Ohms
B)
350 Ohms
C)
300 Ohms
D)
100 Ohms
Points: 0 out of 1
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Question 3 of 270
What is the equivalent resistance of this parallel circuit?
Correct answer: D)
Selected answer: No answer given
Selected answer: No answer given
A)
80 Ohms
B)
25 Ohms
C)
26.3 Ohms
D)
8.57 Ohms
Points: 0 out of 1
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Question 4 of 270
What is the equivalent resistance of this parallel-series circuit?
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
100 Ohms
B)
180 Ohms
C)
108.57 Ohms
D)
1000 Ohms
Points: 0 out of 1
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Question 5 of 270
Calculate the total current of this series circuit and the voltage at terminals 1 and 2, as well as at terminals 2 and 3.
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
Total current: 300mA
Voltage between 1 and 2: 60V
Voltage between 2 and 3: 60V
Voltage between 1 and 2: 60V
Voltage between 2 and 3: 60V
B)
Total current: 300mA
Voltage between 1 and 2: 30V
Voltage between 2 and 3: 90V
Voltage between 1 and 2: 30V
Voltage between 2 and 3: 90V
C)
Total current: 3A
Voltage between 1 and 2: 60V
Voltage between 2 and 3: 60V
Voltage between 1 and 2: 60V
Voltage between 2 and 3: 60V
D)
none of these answers
Points: 0 out of 1
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Question 6 of 270
Calculate the total current of this parallel circuit and the voltage at terminals 3 and 4, as well as at terminals 2 and 4.
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
Total current: 14A
Voltage between 3 and 4: 120V
Voltage between 2 and 4: 120V
Voltage between 3 and 4: 120V
Voltage between 2 and 4: 120V
B)
Total current: 7A
Voltage between 3 and 4: 60V
Voltage between 2 and 4: 60V
Voltage between 3 and 4: 60V
Voltage between 2 and 4: 60V
C)
Total current: 7A
Voltage between 3 and 4: 120V
Voltage between 2 and 4: 120V
Voltage between 3 and 4: 120V
Voltage between 2 and 4: 120V
D)
none of these answers
Points: 0 out of 1
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The voltage for terminals 3 and 4 as well as for 2 and 4 is identical to that of the source.
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Question 7 of 270
Calculate the total current of this parallel-series circuit and the voltage at terminals 3 and 4, as well as at terminals 1 and 3.
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
Total current: 1.80A
Voltage between 1 and 3: 82V
Voltage between 3 and 4: 38V
Voltage between 1 and 3: 82V
Voltage between 3 and 4: 38V
B)
Total current: 5A
Voltage between 1 and 3: 60V
Voltage between 3 and 4: 60V
Voltage between 1 and 3: 60V
Voltage between 3 and 4: 60V
C)
Total current: 1.105A
Voltage between 1 and 3: 110.5V
Voltage between 3 and 4: 9.5V
Voltage between 1 and 3: 110.5V
Voltage between 3 and 4: 9.5V
D)
none of these answers
Points: 0 out of 1
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Question 8 of 270
Complete the following table:
V = Volt
I = Ampere
R = Resistance in Ohm
V = Volt
I = Ampere
R = Resistance in Ohm
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
1) 12V
2) 10mA
3) 100000 ohms
2) 10mA
3) 100000 ohms
B)
1) 120V
2) 10A
3) 100 ohms
2) 10A
3) 100 ohms
C)
1) 12V
2) 10mA
3) 1000 ohms
2) 10mA
3) 1000 ohms
D)
none of these answers
Points: 0 out of 1
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Question 9 of 270
Complete the following table:
P = Power in Watt
V = Volt
I = Ampere
R = Resistance in Ohm
P = Power in Watt
V = Volt
I = Ampere
R = Resistance in Ohm
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
1) 100W
2) 1V
3) 3.75A
4) 1562.5W
2) 1V
3) 3.75A
4) 1562.5W
B)
1) 1KW
2) 10V
3) 37.5A
4) 15625W
2) 10V
3) 37.5A
4) 15625W
C)
1) 23W
2) 120V
3) 20A
4) 180W
2) 120V
3) 20A
4) 180W
D)
none of these answers
Points: 0 out of 1
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Question 10 of 270
What happens to It (total intensity) of this circuit, if the resistance of 300 ohms is replaced by a resistor of 100 ohms?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
it decreases
B)
it increases
C)
she does not change
D)
none of these answers
You have not selected all the correct options available.
Points: 0 out of 1
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Question 11 of 270
What happens with the voltage at terminals 2 and 3 if the 300 ohm resistor is replaced by a 100 ohm resistor?
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
it decreases
B)
it's increasing
C)
he does not change
D)
none of these answers
You have not selected all the correct options available.
Points: 0 out of 1
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Question 12 of 270
Complete the following sentence:
The total resistance of a parallel circuit is always:
The total resistance of a parallel circuit is always:
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
less than the smallest resistance of this circuit
B)
greater than the greatest resistance of this circuit
C)
equal to the sum of the resistances of this circuit
D)
equal to the subtraction of the resistances of this circuit
Points: 0 out of 1
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Question 13 of 270
What happens to It (total intensity) of this circuit, if the 20 ohm resistor (1) burns and becomes open?
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
it decreases
B)
it increases
C)
it vanishes
D)
she does not change
You have not selected all the correct options available.
Points: 0 out of 1
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Question 14 of 270
What is the voltage at terminals 3 and 4, if the 20 ohm resistor is replaced by a 100 ohm resistor?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
it decreases
B)
it increases
C)
she does not change
D)
none of these answers
You have not selected all the correct options available.
Points: 0 out of 1
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Question 15 of 270
If the voltage of the source is doubled (240V), what happens to the resistance of 20 ohms (1)?
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
do not change
B)
becomes zero (0)
C)
double (x 2)
D)
decreases (-)
Points: 0 out of 1
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Intensity or Current = Voltage / Resistance
I = V / R
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Question 16 of 270
Two resistors (R1 and R2) are connected in series at a voltage of 240V. You measure 30V across R1, how much do you measure across R2?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
30V
B)
210V
C)
240V
D)
none of these answers
Points: 0 out of 1
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In a series circuit, the voltages add up.
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Question 17 of 270
The total intensity of a serial circuit:
Correct answer: D)
Selected answer: No answer given
Selected answer: No answer given
A)
it adds up
B)
she gets away
C)
it vanishes
D)
is the same everywhere
Points: 0 out of 1
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Question 18 of 270
Find the power in watts of a 40 ohm soldering iron, knowing that its current is 3 A.
Correct answer: B)
You have chosen: B)
You have chosen: B)
A)
300 W
B)
360 W
C)
400 W
D)
100 W
Points: 1 out of 1
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P = R x I 2
40 ohms x (3A x 3A) = 360 Watts Noticed, if there is 2 times more amperage, there is 4 times more power for the same resistance. I is squared. Take for example, this same soldering iron of 40 ohms with now a current of 6 A.
40 ohms x (6A x 6A) = 1440 W
40 ohms x (3A x 3A) = 360 Watts Noticed, if there is 2 times more amperage, there is 4 times more power for the same resistance. I is squared. Take for example, this same soldering iron of 40 ohms with now a current of 6 A.
40 ohms x (6A x 6A) = 1440 W
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Question 19 of 270
What is the electrical capacitance unit (capacitor)?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
coulomb
B)
farad
C)
henry
D)
ohm
Points: 0 out of 1
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Farad = unit of capacitance (capacitor)
Henry = unit of inductance (coil or winding)
Coulomb = is the unit derived from electric charge
Ohm = unit of resistance
Henry = unit of inductance (coil or winding)
Coulomb = is the unit derived from electric charge
Ohm = unit of resistance
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Question 20 of 270
An alternative source supplies a capacitor, what happens to the voltage with respect to the current?
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
The voltage is in phase with the current.
B)
The voltage is ahead of the current.
C)
The voltage is lagging behind the current.
D)
none of these answers
Points: 0 out of 1
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The main property of the capacitor is to be able to
store electrical energy. The unit of measure for the capacity of the capacitors is the farad.
In an alternating circuit (AC) a capacitor influences the current to take an advance with respect to the voltage. To solve this problem we must add a winding (opposite effect) to this circuit to improve the power factor.
The power factor is a measure of how efficiently the electrical system converts electricity (supplied by the utility) into useful output power.
The power factor is on the electricity bill. When the power factor is low (in%). This means that the company does not fully use the electricity it pays for, which results in significant penalties on the electricity bill and an overload of the electricity distribution system.
Here is a checklist with the word ELICE: In a capacitive circuit "C", the intensity "I" is before, so ahead of the voltage "E".
store electrical energy. The unit of measure for the capacity of the capacitors is the farad.
In an alternating circuit (AC) a capacitor influences the current to take an advance with respect to the voltage. To solve this problem we must add a winding (opposite effect) to this circuit to improve the power factor.
The power factor is a measure of how efficiently the electrical system converts electricity (supplied by the utility) into useful output power.
The power factor is on the electricity bill. When the power factor is low (in%). This means that the company does not fully use the electricity it pays for, which results in significant penalties on the electricity bill and an overload of the electricity distribution system.
Here is a checklist with the word ELICE: In a capacitive circuit "C", the intensity "I" is before, so ahead of the voltage "E".
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Question 21 of 270
An alternative source feeds a winding (coil), what happens to the voltage with respect to the current?
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
The voltage is in phase with the current.
B)
The voltage is lagging behind the current.
C)
The current is late compared to the voltage.
D)
none of these answers
Points: 0 out of 1
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In an alternating circuit (AC) a winding (coil) influences the current to lag behind the voltage. To solve this problem we must add a capacitor (opposite effect) to this circuit. That's why in most plants that use multiple motors (coil or coil), a capacitor bank system is added to improve the power factor.
The power factor is a measure of how efficiently the electrical system converts electricity (supplied by the utility) into useful output power.
The power factor is written on the electricity bill. When the power factor is low (in%). This means that the company does not fully use the electricity it pays for, which results in significant penalties on the electricity bill and an overload of the electricity distribution system.
The unit of measurement of the inductance of a winding is the "henri".
Here is a reminder with the word ELICE: In a capacitive circuit "L", the voltage "E" is before, so in advance on the intensity "I".
The power factor is a measure of how efficiently the electrical system converts electricity (supplied by the utility) into useful output power.
The power factor is written on the electricity bill. When the power factor is low (in%). This means that the company does not fully use the electricity it pays for, which results in significant penalties on the electricity bill and an overload of the electricity distribution system.
The unit of measurement of the inductance of a winding is the "henri".
Here is a reminder with the word ELICE: In a capacitive circuit "L", the voltage "E" is before, so in advance on the intensity "I".
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Question 22 of 270
Identify the illustration of a forward-biasing diode that passes the current.
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
A
B)
B
C)
C
D)
D
You have not selected all the correct options available.
Points: 0 out of 1
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The diode lets the current flow only in one direction, and not in the other.
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Question 23 of 270
You have several batteries 12 volts, 10 A / h. You want to produce a source 48 volts, 10 A / h. Select the appropriate configuration.
Correct answer: A)
You have chosen: A)
You have chosen: A)
A)
B)
C)
D)
Points: 1 out of 1
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Batteries connected in series add up the voltages. The total current is the same as that of the batteries separately.
Caution: Check the polarity of the batteries in series. (+ - + - + - + -)
Batteries connected in parallel add currents. The total voltage is the same as that of the batteries separately. Caution: Check the polarity of the batteries in parallel. (+ + + + - - - -)
Caution: Check the polarity of the batteries in series. (+ - + - + - + -)
Batteries connected in parallel add currents. The total voltage is the same as that of the batteries separately. Caution: Check the polarity of the batteries in parallel. (+ + + + - - - -)
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Question 24 of 270
You have several batteries 12 volts, 10 A / h. You want to produce a source 24 volts, 20 A / h. Select the appropriate configuration.
Correct answer: C)
You have chosen: C)
You have chosen: C)
A)
B)
C)
D)
Points: 1 out of 1
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Batteries connected in series add up the voltages. The total current is the same as that of the batteries separately.
Caution: Check the polarity of the batteries in series. (+ - + - + - + -)
Batteries connected in parallel add currents. The total voltage is the same as that of the batteries separately. Caution: Check the polarity of the batteries in parallel. (+ + + + - - - -)
Caution: Check the polarity of the batteries in series. (+ - + - + - + -)
Batteries connected in parallel add currents. The total voltage is the same as that of the batteries separately. Caution: Check the polarity of the batteries in parallel. (+ + + + - - - -)
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Question 25 of 270
If each battery produces 3 volts and the resistor " R " equals 3 ohms. By neglecting the internal resistance of the battery, what is the current supplied by the batteries C and D?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
0.5 A
B)
1 A
C)
2 A
D)
3 A
Points: 0 out of 1
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2 batteries in series: 2 x 3 volts = 6 volts
I = V / R therefore 6 volts / 3 ohms = 2 A total
There are two sets of batteries connected in parallel, so 2 A / 2 = 1A
I = V / R therefore 6 volts / 3 ohms = 2 A total
There are two sets of batteries connected in parallel, so 2 A / 2 = 1A
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Question 26 of 270
What does the term equalization charge on the storage batteries mean?
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
is used to recharge the batteries in cycle environments, applying a short voltage of high amperage.
B)
is used to fully recharge the batteries, applying them a higher current.
C)
is used to fully recharge the batteries, applying them a higher voltage.
D)
none of these answers
Points: 0 out of 1
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An equalization charge is intended to fully charge each battery to avoid imbalances between the different accumulators that make up the batteries. It consists in charging the batteries with a low current, but at a voltage higher than the voltage generally applied at the end of charging.
The battery area must have good ventilation to allow the diffusion of gases and the prevention of explosions. Most batteries contain 35% sulfuric acid. Always wear splash goggles and protective clothing (gloves and aprons). A face shield (with goggles) may also be required. Avoid short circuits when working on batteries.
The battery area must have good ventilation to allow the diffusion of gases and the prevention of explosions. Most batteries contain 35% sulfuric acid. Always wear splash goggles and protective clothing (gloves and aprons). A face shield (with goggles) may also be required. Avoid short circuits when working on batteries.
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Question 27 of 270
Associate the appropriate symbol for each of the following definitions.
Diode
| Right answer | |
Deacon
| Right answer | |
Thyristor (SCR)
| Right answer | |
Triac
| Right answer | |
Points: 1 out of 1
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The thyristor is often used as a rectifier and is sometimes called SCR ( Silicon Controlled Rectifier ).
The triac is an electronic component equivalent to the paralleling of two thyristors, the anode of one of which would be connected to the cathode of the other, the respective gates being controlled simultaneously.
These graphic symbols refer to the Technical Guide, corporation of the master electricians of Quebec.
The triac is an electronic component equivalent to the paralleling of two thyristors, the anode of one of which would be connected to the cathode of the other, the respective gates being controlled simultaneously.
These graphic symbols refer to the Technical Guide, corporation of the master electricians of Quebec.
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Question 28 of 270
What is the shape of the wave, at the output "A" of this bridge diodes?
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
A
B)
B
C)
C
D)
none of these answers
Points: 0 out of 1
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A diode bridge or rectifier is an assembly of four diodes mounted in a bridge which, in a single-phase regime, rectifies the alternating current in direct current.
Here is a three phase rectifier.
Here is a three phase rectifier.
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Question 29 of 270
This sine wave has a peak voltage of 169.7 volts, determine the effective voltage and the frequency in hertz of this power supply.
Correct answer: D)
Selected answer: No answer given
Selected answer: No answer given
A)
169.7 5 and 60 hertz
B)
5 120 and 120 hertz
C)
5 108 and 60 hertz
D)
5 120 and 60 hertz
Points: 0 out of 1
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This AC circuit has a sine wave of 169.7 volts peak CA or 120 volts effective AC is the voltage used by Hydro-Quebec which is already expressed in
Hertz effective voltage : number of complete cycle per second. If 6 cycles are performed in 0.10 seconds, there will be 60 cycles per second.
Peak value or Vmax : Maximum value encountered in an alternation.
Average value or Vmoy : For a sine wave, the average value of a period or a cycle is zero. On the other hand, that of the half-cycle is equal to 63.6% of the maximum value.
RMS or Veff: We associate with the word efficient, the terms of performance and power. The rms value of a variable current or voltage over time corresponds to the value of a direct current or a DC voltage. The rms value is 70.7% of the peak value.
Hertz effective voltage : number of complete cycle per second. If 6 cycles are performed in 0.10 seconds, there will be 60 cycles per second.
Peak value or Vmax : Maximum value encountered in an alternation.
Average value or Vmoy : For a sine wave, the average value of a period or a cycle is zero. On the other hand, that of the half-cycle is equal to 63.6% of the maximum value.
RMS or Veff: We associate with the word efficient, the terms of performance and power. The rms value of a variable current or voltage over time corresponds to the value of a direct current or a DC voltage. The rms value is 70.7% of the peak value.
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Question 30 of 270
Associate the appropriate symbol for each of the following definitions.
Level switch
| Right answer | |
Thermal switch (thermostat)
| Right answer | |
Pressure switch (pressure switch)
| Right answer | |
Flow switch
| Right answer | |
Points: 1 out of 1
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These graphic symbols refer to the Technical Guide, corporation of the master electricians of Quebec.
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Question 31 of 270
Find the equivalent of 1 microfarad.
Correct answer: D)
Selected answer: No answer given
Selected answer: No answer given
A)
0.1 farad
B)
0.01 farad
C)
0.001 farad
D)
0.000001 farad
Points: 0 out of 1
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Conversion des centaines
kilo = 1000 (ex. 1kV = 1000 Volts)
milli = 0.001 (ex. 1 mA = 0.001 A ) 10-3
micro = 0.000001 (ex. 1uF = 0.000001 Farad) 10-6
nano = 0.000000001 (ex. 1nF = 0.000000001) Farad 10-9
pico = 0.000000000001 (ex. 1pF = 0.000000000001) Farad 10-12
kilo = 1000 (ex. 1kV = 1000 Volts)
milli = 0.001 (ex. 1 mA = 0.001 A ) 10-3
micro = 0.000001 (ex. 1uF = 0.000001 Farad) 10-6
nano = 0.000000001 (ex. 1nF = 0.000000001) Farad 10-9
pico = 0.000000000001 (ex. 1pF = 0.000000000001) Farad 10-12
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Question 32 of 270
What is the unit of measurement of the luminous flux?
Correct answer: D)
Selected answer: No answer given
Selected answer: No answer given
A)
Watt
B)
Candle
C)
Luxury
D)
Lumen
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LUMEN : is based on human perception or the luminous flux of light.
LUX : is the illumination of a surface which receives, in a uniformly distributed way, a luminous flux of one lumen per square meter.
CANDELA : It is used to measure the luminous intensity or brightness perceived by the human eye from a light source.
WATT : is a unit derived from the international system to quantify the power of a circuit .
LUX : is the illumination of a surface which receives, in a uniformly distributed way, a luminous flux of one lumen per square meter.
CANDELA : It is used to measure the luminous intensity or brightness perceived by the human eye from a light source.
WATT : is a unit derived from the international system to quantify the power of a circuit .
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Question 33 of 270
Which of these voltages is not three-phase?
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
120V-240V
B)
120V-208V
C)
277V-480V
D)
347V-600V
Points: 0 out of 1
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A three-phase current system (or voltage) consists of three sinusoidal currents (or voltages) of the same frequency of the same amplitude which are phase shifted by one third of a turn. A network whose effective voltage between phase and neutral is 120 V will have a voltage of 208 V between 2 phases. (120 V x 1.732 = 208 V)
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Question 34 of 270
Determine the value of this resistance.
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
110 ohms
B)
1 100 ohms
C)
11 000 ohms
D)
100 000 ohms
Points: 0 out of 1
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Question 35 of 270
Identify the fuse (s) in bad condition (open) of this 600-volt three-phase power supply?
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
L3
B)
L2
C)
L1
D)
L1 and L2
Points: 0 out of 1
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There is no potential difference, therefore 0 volts, across a fuse in good condition.
In the case of a three-phase 600-volt supply, the voltage between two phases is always 600 volts and the voltage from one phase to earth is always 347 V (600V / 1.732 = 347V).
In the case of a three-phase 600-volt supply, the voltage between two phases is always 600 volts and the voltage from one phase to earth is always 347 V (600V / 1.732 = 347V).
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Question 36 of 270
As the ambient temperature rises, the resistivity (ohm) of a conductor increases.
Correct answer: A)
You have chosen: A)
You have chosen: A)
A)
True
B)
False
Points: 1 out of 1
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Question 37 of 270
Which marking must have a weatherproof engine?
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
WP
B)
ZP
C)
TP
D)
type 12
Points: 0 out of 1
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You must know the different markings.
WP: W eather P roof
ZP and TP : Engines with the "ZP" and thermally protected "TP" impedance protected marking may be installed without the overload protection required by article 28-300.
Type 12 : See Table 65, selection of envelopes for non-hazardous locations
WP: W eather P roof
ZP and TP : Engines with the "ZP" and thermally protected "TP" impedance protected marking may be installed without the overload protection required by article 28-300.
Type 12 : See Table 65, selection of envelopes for non-hazardous locations
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Question 38 of 270
What is the full-load current of a 30HP 208HP three-phase wound rotor motor?
Correct answer: C)
Selected answer: No answer given
Selected answer: No answer given
A)
72A
B)
80A
C)
88A
D)
100A
Points: 0 out of 1
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Determine the full-load current of a 30HP 208V three-phase motor , Table 44 (see note 1)
Table 44 of some versions of the electrical code, do not include
208V full-load motor currents, add 10% of current at full load of the corresponding 230 V motor. 80 A + (80 x 10%) = 88 A Table 44 : 30HP at 230V = 80 A (Attention choose the column "Wound rotors")
ATTENTION:
To determine the full load current of DC motors , use table D2.
* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
Table 44 of some versions of the electrical code, do not include
208V full-load motor currents, add 10% of current at full load of the corresponding 230 V motor. 80 A + (80 x 10%) = 88 A Table 44 : 30HP at 230V = 80 A (Attention choose the column "Wound rotors")
ATTENTION:
To determine the full load current of DC motors , use table D2.
* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
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Question 39 of 270
What is the required size of copper conductors (RW90), for the derivation of a 115V 3 HP motor, used in continuous service ?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
10 AWG
B)
8 AWG
C)
6 AWG
D)
4 AWG
Points: 0 out of 1
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Section 28 - Engines and Generators, sections 28-106, 28-010 and 28-704 of the Quebec Construction Code, Chapter V - Electricity.
Step 1: Determine the full load current of the AC single-phase motor in Table 45
Table 45: 3HP = 34
Step 2: Determine the ampacity of the conductors
The bypass lines for a motor used in continuous duty must have a rated current of at least 125% of the rated full load motor current (Clauses 28-106)
34 A x 125% = 42.5
Step 3: Select copper conductors in Table 2
Table 2: 42.5 A (90 degree column) = 8 AWG RW90
CAUTION:
Bypass conductors powering an engine used inNon-continuous service shall have a current rating not less than the value of the current obtained by multiplying the nominal full load motor current and the relevant percentage of Table 27 for the service in question.
* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
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It is important, when modifying an installation, to add new specifications to existing electrical plans if required. In this case, the full load current or the '' HP '' number of the new engine must be changed on the map.
Answered correctly Answered incorrectly Option of good multiple choices missed
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* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
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Step 1: Determine the full load current of the AC single-phase motor in Table 45
Table 45: 3HP = 34
Step 2: Determine the ampacity of the conductors
The bypass lines for a motor used in continuous duty must have a rated current of at least 125% of the rated full load motor current (Clauses 28-106)
34 A x 125% = 42.5
Step 3: Select copper conductors in Table 2
Table 2: 42.5 A (90 degree column) = 8 AWG RW90
CAUTION:
Bypass conductors powering an engine used inNon-continuous service shall have a current rating not less than the value of the current obtained by multiplying the nominal full load motor current and the relevant percentage of Table 27 for the service in question.
* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
40 of 270
What is the maximum rating of non-timed fuses supplying a single-phase 115hp 2-HP AC motor (resistance start)?
Correct answer: D)
You have chosen: C)
You have chosen: C)
A)
30A
B)
40A
C)
60A
D)
70A
Points: 0 out of 1
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Section 28 - Engines and Generators, sections 28-010 and 28-704 of the Quebec Construction Code, Chapter V - Electrical
1st Step: Determine the Full Load Current of the AC Single Phase Motor in Table 45
Table 45: 2HP = 24A
Step 2 : Determine the maximum rated ampacity of protections in Appendix D, Table D16
Table D16: 24A = Unscheduled Fuses 70A (Resistance Start)
Appendix D, Table D16 or Table 29 can be used to determine the size of the protective devices. overcurrent protection (Article 28-200)
Note:
If the value of the full load current of the motor is between 2 values in Table D16 (Appendix D), always take the lower value to determine the overcurrent protection devices.
Take, for example, a motor with a full load current of 23 A, so you have to use 22 A. * Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
1st Step: Determine the Full Load Current of the AC Single Phase Motor in Table 45
Table 45: 2HP = 24A
Step 2 : Determine the maximum rated ampacity of protections in Appendix D, Table D16
Table D16: 24A = Unscheduled Fuses 70A (Resistance Start)
Appendix D, Table D16 or Table 29 can be used to determine the size of the protective devices. overcurrent protection (Article 28-200)
Note:
If the value of the full load current of the motor is between 2 values in Table D16 (Appendix D), always take the lower value to determine the overcurrent protection devices.
Take, for example, a motor with a full load current of 23 A, so you have to use 22 A. * Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
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Question 41 of 270
What is the required size of copper (RW90) conductors, for the derivation of a 575V 2HP motor, used in temporary duty 15 minutes?
Correct answer: B)
You have chosen: D)
You have chosen: D)
A)
16 AWG
B)
14 AWG
C)
12 AWG
D)
10 AWG
Points: 0 out of 1
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Section 28 - Engines and Generators , sections 28-106, 28-010 and 28-704 of the Quebec Construction Code, Chapter V - Electricity.
Step 1: Determine the full load current of the AC 3-phase motor in Table 44
Table 44: 2HP = 2.7
Step 2: Determine the ampacity of the conductors
The bypass lines supplying a motor used in non-continuous operation must have a non-continuous current less than the value of the current obtained by multiplying the nominal full load motor current and the relevant percentage in Table 27 for the service in question (clauses 28-106)
Refer to the temporary service column, operating 15 minutes .
Table 27: 120% x 2.7 = 3.24 A.
Step 3: Select Copper Conductors in Table 2
Table 2: 3.24 A = 14 AWG RW90 (minimum 14 AWG for one installation)
* Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
Step 1: Determine the full load current of the AC 3-phase motor in Table 44
Table 44: 2HP = 2.7
Step 2: Determine the ampacity of the conductors
The bypass lines supplying a motor used in non-continuous operation must have a non-continuous current less than the value of the current obtained by multiplying the nominal full load motor current and the relevant percentage in Table 27 for the service in question (clauses 28-106)
Refer to the temporary service column, operating 15 minutes .
Table 27: 120% x 2.7 = 3.24 A.
Step 3: Select Copper Conductors in Table 2
Table 2: 3.24 A = 14 AWG RW90 (minimum 14 AWG for one installation)
* Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
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Question 42 of 270
What is the maximum rating of a circuit breaker, feeding a 575V three-phase AC motor of 2HP (star-delta start)?
Correct answer: A)
You have chosen: C)
You have chosen: C)
A)
15A
B)
20A
C)
10A
D)
2A
Points: 0 out of 1
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Section 28 - Engines and Generators , sections 28-106, 28-010 and 28-704 of the Quebec Construction Code, Chapter V - Electricity.
Step 1: Determine the full load current of the AC 3-phase motor in Table 44
Table 44: 2HP = 2.7
Step 2: Determine the maximum rating of the protections in Appendix D, Table D16 Appendix D16: 2.7 A = Breaker 15 A (star-delta start) Appendix D, Table D16 or Table 29 can be used to determine the size of the overcurrent protection devices (Item 28-200) Note:
If the value of the full load current of the motor is between 2 values in Table D16 (Appendix D), always take the lower value to determine the overcurrent protection devices.
Take, for example, a motor with a full load current of 23 A, so you have to use 22 A. * Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
Step 1: Determine the full load current of the AC 3-phase motor in Table 44
Table 44: 2HP = 2.7
Step 2: Determine the maximum rating of the protections in Appendix D, Table D16 Appendix D16: 2.7 A = Breaker 15 A (star-delta start) Appendix D, Table D16 or Table 29 can be used to determine the size of the overcurrent protection devices (Item 28-200) Note:
If the value of the full load current of the motor is between 2 values in Table D16 (Appendix D), always take the lower value to determine the overcurrent protection devices.
Take, for example, a motor with a full load current of 23 A, so you have to use 22 A. * Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
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Question 43 of 270
What is the required size of copper conductors type TWU, for an arterial of 3 AC motors, 575V (or 600V) of 10 HP?
Correct answer: B)
You have chosen: C)
You have chosen: C)
A)
10 AWG
B)
8 AWG
C)
6 AWG
D)
4 AWG
Points: 0 out of 1
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Section 28 - Engines and Generators , sections 28-106, 28-108, 28-010 and 28-704 of the Quebec Construction Code, Chapter V - Electricity.
Step 1: Determine the full load current of AC 3-phase motors in Table 44
Table 44: 10 HP = 11 A
Step 2: Determine the ampacity of the conductors
The calculation is done for a conductor supplying 2 or more motors, 125% of the full load current of larger motor plus full load current of other motors (article 28-108)
(11 x 125%) + 11 + 11 = 35.75 A
3rd Step: Determine the maximum permissible temperature for TWU conductors at Table 19.
Table 19: TWU = 60 degrees Celsius
4e Step: Select Copper Conductors (60 degrees Celsius) in Table 2
Table 2: 35.75A (60 Degree Column) = 8 AWG TWU
* Articles, tables and others applicable to this question refer to the Construction Code of the Quebec, Chapter V - Electricity
Step 1: Determine the full load current of AC 3-phase motors in Table 44
Table 44: 10 HP = 11 A
Step 2: Determine the ampacity of the conductors
The calculation is done for a conductor supplying 2 or more motors, 125% of the full load current of larger motor plus full load current of other motors (article 28-108)
(11 x 125%) + 11 + 11 = 35.75 A
3rd Step: Determine the maximum permissible temperature for TWU conductors at Table 19.
Table 19: TWU = 60 degrees Celsius
4e Step: Select Copper Conductors (60 degrees Celsius) in Table 2
Table 2: 35.75A (60 Degree Column) = 8 AWG TWU
* Articles, tables and others applicable to this question refer to the Construction Code of the Quebec, Chapter V - Electricity
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Question 44 of 270
What is the required size of the copper conductors (TW75), for this line of 3 single-phase 115V (or 120V) AC motors?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
1 AWG
B)
0 AWG
C)
00 AWG
D)
000 AWG
Points: 0 out of 1
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Section 28 - Engines and Generators , sections 28-106, 28-108, 28-010 and 28-704 of the Quebec Construction Code, Chapter V - Electricity.
Step 1: Determine the full load current of single-phase AC motors in Table 45
5HP = 56A
3HP = 34A
Step 2: Determine the ampacity of the conductors
The calculation is done for a conductor supplying 2 or more motors, 125% of the current to full load of larger motor plus full load current of other motors (items 28-108)
(56A x 125%) + 34A + 34A = 138A
Step 3: Select Copper Conductors (75 degrees Celsius) on the board 2
Table 2 (75 degree column): 138A = 0 AWG TW75
CAUTION:
The temperatures in Table 2 are the maximum permissible temperatures at which one, two or three pipe conductors, or a cable with two or three conductors may be exposed. To determine the permissible current of the types of conductors other than those mentioned in this table, installed under the same conditions as in the latter, it is necessary to determine the temperature indicated in Table 19 for the type in question and then refer to the column corresponding table 2.
* Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
Step 1: Determine the full load current of single-phase AC motors in Table 45
5HP = 56A
3HP = 34A
Step 2: Determine the ampacity of the conductors
The calculation is done for a conductor supplying 2 or more motors, 125% of the current to full load of larger motor plus full load current of other motors (items 28-108)
(56A x 125%) + 34A + 34A = 138A
Step 3: Select Copper Conductors (75 degrees Celsius) on the board 2
Table 2 (75 degree column): 138A = 0 AWG TW75
CAUTION:
The temperatures in Table 2 are the maximum permissible temperatures at which one, two or three pipe conductors, or a cable with two or three conductors may be exposed. To determine the permissible current of the types of conductors other than those mentioned in this table, installed under the same conditions as in the latter, it is necessary to determine the temperature indicated in Table 19 for the type in question and then refer to the column corresponding table 2.
* Articles, tables and others applicable to this question, refer to the Quebec Construction Code, Chapter V - Electricity
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Question 45 of 270
You must replace a defective motor with a new one. An identical engine is not available. Which of these specifications is not mandatory to consider when choosing the replacement engine?
Correct answer: D)
Selected answer: No answer given
Selected answer: No answer given
A)
The class of insulation
B)
The type of case (sealed, open, ventilated)
C)
The designation of the case
D)
The name of the manufacturer
Points: 0 out of 1
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When choosing an engine, it is important to consider the following specifications:
- The nominal voltage
- The number of phases
- The class of insulation
- The type of case (sealed, open, ventilated)
- The nominal current
- The designation of the case
It is important, when modifying an installation, to add new specifications to existing electrical plans if required. In this case, the full load current or the '' HP '' number of the new engine must be changed on the map.
Exam preparation Electrician non construction Emploi Québec
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Question 46 of 270
Protection against overcurrent is a device that provides protection against excess current, but not necessarily against short circuits. It is able to automatically open an electrical circuit.
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
True
B)
False
Points: 0 out of 1
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Refer to the definitions in Section 0 of the Quebec Construction Code, Chapter V - Electricity.
Overload protection
device Device that protects against overcurrent, but not necessarily against short circuits, and is able to automatically open an electrical circuit, by melting metal or by electromechanical means.
Overcurrent protection
device Device for automatically opening an electrical circuit, under specified conditions of overload or short circuit, by metal melting or by electromechanical means.
Thermal circuit breaker
Protection device against overcurrent, but not necessarily against short circuits, and which contains in addition to a fuse element, a thermal element on which depends the fuse element for opening the circuit.
Mechanical Protection (Section 10-000)
Mechanical protection refers to the provision of additional protection for a service connection or electrical installation.
* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
Overload protection
device Device that protects against overcurrent, but not necessarily against short circuits, and is able to automatically open an electrical circuit, by melting metal or by electromechanical means.
Overcurrent protection
device Device for automatically opening an electrical circuit, under specified conditions of overload or short circuit, by metal melting or by electromechanical means.
Thermal circuit breaker
Protection device against overcurrent, but not necessarily against short circuits, and which contains in addition to a fuse element, a thermal element on which depends the fuse element for opening the circuit.
Mechanical Protection (Section 10-000)
Mechanical protection refers to the provision of additional protection for a service connection or electrical installation.
* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
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Question 47 of 270
What is the setting of the overload devices , for the derivation of a three-phase AC motor, 600V of 5 HP with an overload coefficient 1.15?
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
6.1 A
B)
7.63 A
C)
10 A
D)
15 A
Points: 0 out of 1
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Step 1: Determine the full load current of the AC 3-phase motor in Table 44
Table 44: 5 HP = 6.1
Step 2: Determine the maximum rated current of protections
If the overload coefficient is equal to or greater than 1.15, the setting of Overload protection devices must be 125% of the rated full load motor current (Clause 28-306)
If the overload coefficient is not marked or is less than 1.15, the setting of the protective devices against overloads must be 115% of the nominal current at full load of the motor (article 28-306)
Therefore 6.1 A x 125% = 7.63 A
ATTENTION:
It is important to understand the different abbreviations of an engine nameplate.
Table 44: 5 HP = 6.1
Step 2: Determine the maximum rated current of protections
If the overload coefficient is equal to or greater than 1.15, the setting of Overload protection devices must be 125% of the rated full load motor current (Clause 28-306)
If the overload coefficient is not marked or is less than 1.15, the setting of the protective devices against overloads must be 115% of the nominal current at full load of the motor (article 28-306)
Therefore 6.1 A x 125% = 7.63 A
ATTENTION:
It is important to understand the different abbreviations of an engine nameplate.
| HP or CV, Power = Horsepower HP or HP for horsepower |
| Speed or RPM = Speed in RPM Rotation per minute |
| Temp. Ambiante = Temperature of the isolation of classes A, B, F et H |
| V = Voltage or voltage permissible Volts |
| AMP or FLA (Full load amperage) = Amperage (current at full load) |
| ENCL enclosure = Type of construction, engine open or closed |
| Time Rating = Type of service, continuous, temporary, periodic. |
| HP at startup or KVA = starting power according to NEMA standards. |
| NEMA FL EFF. = Full load efficiency in% |
| Frame (FR) = Type of motor housing according to NEMA standards |
| Service Factor (SF) = Service Factor |
| NEMA design = Torque-torque according to NEMA standards |
| HZ = frequency in Hertz |
| PH = number of Phase |
* The articles, tables and others applicable to this question, refer to the Construction Code of Québec, Chapter V - Electricity
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Question 48 of 270
Overload protection devices are not required for:
Correct answer: A)
Selected answer: No answer given
Selected answer: No answer given
A)
a manual starting motor with a power of not more than 1 hp, monitored continuously.
B)
a manual starting motor with a power of not more than 5 hp, monitored continuously.
C)
a manual starting motor with a power of not more than 10 hp, monitored continuously.
D)
All these answers
Points: 0 out of 1
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Section 28 - Engines and Generators
Refer to Appendix B and Article 28-308 of the Quebec Construction Code, Chapter V - Electrical
ATTENTION:
Designed engines that are protected by "ZP" impedance protected and thermally protected "TP" may be installed without the overload protection required in item 28-300.
Refer to Appendix B and Article 28-308 of the Quebec Construction Code, Chapter V - Electrical
ATTENTION:
Designed engines that are protected by "ZP" impedance protected and thermally protected "TP" may be installed without the overload protection required in item 28-300.
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Question 49 of 270
What is the maximum fit of the overload protection device from a 3000 watt motorcycle to 240 V?
Correct answer: D)
Selected answer: No answer given
Selected answer: No answer given
A)
12.5
B)
14,375
C)
15.625
D)
17.5
Points: 0 out of 1
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Section 28 - Engines and Generators, article 28-710 of the Quebec Construction Code, Chapter V - Electricity
Step 1: Determine the Nominal Compressor Current
I = P / V
3000W / 240V = 12.5 A
2nd Step: Select the setting overload relays
The nominal current or the setting of the overload relays must not exceed 140% of the nominal load current indicated on the motor-compressor (article 28-710)
12.5A x 140% = 17.5
Step 1: Determine the Nominal Compressor Current
I = P / V
3000W / 240V = 12.5 A
2nd Step: Select the setting overload relays
The nominal current or the setting of the overload relays must not exceed 140% of the nominal load current indicated on the motor-compressor (article 28-710)
12.5A x 140% = 17.5
Question 50 of 270
The overcurrent protections of a three-phase motor open instantly when powering on. What is the best measuring instrument to check this engine.
Correct answer: B)
Selected answer: No answer given
Selected answer: No answer given
A)
Ammeter
B)
megger
C)
Wattmeter
D)
Voltmeter
Points: 0 out of 1
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The megohmmeter generates a significant DC voltage (from a few hundred volts to several kV).
This device is used to measure the insulation of electrical equipment such as, for example, an electrical machine, power cables, insulators, surge arresters and any equipment or installation that may pose a safety hazard in the event of of default.
This device is used to measure the insulation of electrical equipment such as, for example, an electrical machine, power cables, insulators, surge arresters and any equipment or installation that may pose a safety hazard in the event of of default.
