Upgrading, Service and Maintenance CEPE 2015 set is 6 out of 6 sets on the CEPE (Canadian Electrical Practice Exam) assessment tool – Block F. Please keep in mind that not everything is in the code book and not everything on the real exam should be looked up in the code book otherwise if we do so we will miss the last 30 questions on the exam.
Click Advanced study mode select Construction and select 50 questions then set all the other blocks to 0 so you’re focusing only on block E. We will focus on one block at a time.
Click Start. The questions in these sets are all randomized. Spend only one minute to answer the question if within 30 seconds you have not answered the question from your general trade knowledge then click feedback and if you still can’t answer the question or don’t understand if it, then take a picture of the question and text to 416 841-1399 or email to info@ElectricalExam.ca and you will get a quick response.
Find the apparent power first. Multiply 240 volt by 23 amp which equals 5520 VA. The power factor equals 80% of the VA (Apparent Power) which equal 80% 5520 VA. 4416 W total. To find the size of the capacitor, we take the square root of 5520 squared minus 4416 squared which yields 3.3 kVAR. Please see module 12 in your full membership.
Rule 14-612. A transfer switch separates the normal from emergency power.
Supply means transformer. Transformer authority which is Hydro.
Synchronous motors are AC motors and the speed (RPM) depends on the frequency and number of poles.
This is a 3 step process. First we find the difference between 1,000,000 and 50,000 which is 950,000 Ohms. Next we find the difference between 20 mA and 4 mA which is 16 mA. The third step, we follow the X. Since the unknown is the Ohms, we divide 950,000 by 16 mA which is 59,375 Ohms per 1 mA. On the 4 mA to 20 mA scale, there is a climb of (11.2 mA minus 4 mA) 7.2 mA which is worth 59,375 Ohms per 1 mA (each) on the Ohms scale. Therefore, we multiply 59, 375 by 7.2 mA. 427500 total. However, the Ohms scale starts from 50, 000 not zero Ohms so we need to add 50,000 Ohms to the calculated 427,500 Ohms to find what the unknown is reading. 50,000 plus 427,500 is 477,500 Ohms. It is not an option. Csa will update the CEPE.
Find the apparent power first. Multiply 250 volt by 40 amp which equals 10,000 VA. True power is given at 8,000 W total. To find the size of the capacitor, we take the square root of 10,000 squared minus 8000 squared which yields 6000 VAR. Please see module 12 in your full membership.
GTK (General Trade Knowledge).
This is a full wave rectifier. The output equals 90 percent of the input. 240 V x .9 equals 216 V.
GTK (General Trade Knowledge).
600 to 5 is 120 to 1 ratio. Since we have 240 A on the primary, then 240 A divided by 120 would yield 2 A.
One way to solve this, is to look for an answer that needs 600 ohms to get to a whole number. Out of those options B is the only answer. Another way is to use logic. If 600 Ohms produces 3 V then 150 V should produce 300 Ohms X 150 V then divided by 3 V. 30,000 total. We need to add 29400 Ohms to the 600 Ohms to get to 30, 000 Ohms.
This is a Delta transformer (3 phase, 3 wire). When one phase touches the ground, the delta keeps outputting the same voltage. If two phases touch the ground, then nothing works. In a Wye/Star transformer, if one phase touches the ground, a fuse attached to that line would blow.
When we buy a brand new DC motor, the commutator is dull copper. After the motor runs for a while, the brushes are riding on the commutator would leave carbon build up changing the colour to chocolate brown.
GTK (General Trade Knowledge)
GTK (General Trade Knowledge)
Proximity switch is too far away from the chain. It is used for very close range detection. Answer B.
Line to Line will always read no matter which fuse is blown.
An inverter is required to convert DC back to AC. Answer B
It is the voltage drop caused by the starting surge of the motor and the increased load on the phase conductors. Answer D.
The arrow always towards the negative. Conventional current flow is plus to minus. Answer D.
An SCR is a silicon controlled rectifier. It is a diode with a gate. It should be shorted one way and open the other way. Zero Ohms means it is shorted both ways. Answer B.
Table 24 displays the minimum insulation resistances for installations and it depends on amperage of the conductor. Since this is AL (Aluminium), we will use T4. Applying Rule 4-006 and 4/0 is larger than No. 1 awg, we need to use T4, 75 degree column. 180 A total. Using T24, the answer is 50,000 Ohms.
Do not look at the symbol, look at the product instead. If it is AB, then it is an AND gate. If it is A + B then it is an OR gate.
Please see module 12 on www.electricalexam.ca. To find the impedance (Z), we take the square root of R squared plus Xc squared. Z = square root of (500 squared plus 400 squared) equals 640 Ohms.
Minimum is 1 minute. 60 seconds.
If we break the delta triangle at any point, we would have two resistors that are in series with each other and then in parallel with the third resistor. The two resistors are in series would measure 10 Ohms since in series circuits, resistance adds. To find the R total, we would inverse the 10 Ohm and then add it to the inverse of 5 Ohms and we would total 3.3 Ohms. R total =(1/10 + 1/5 ).
An example would be a range (stove) connected in an apartment (condo) and fed from a 3 phase 4 wire panel. The neutral (identified) conductor would carry the same current as the line current in the line conductors. Zero if the circuit is balanced ONLY if it is a 3 phase load connected to a 3 phase transformer. Answer A.
For half wave rectifiers, it is 240 V X 45 percent which is 108 V.
This is new. After installing neon lights for ground detection devices, 1 light would go out and the two would be brighter out of these options.
The circuit feeds a junction box or a receptacle box then feeds the EXIT sign. A plug attachment is plugged into the junction box or receptacle and there is a return conduit to feed DC into the EXIT sign once we lose the AC power from the unit equipment (battery Pack).
Contactors have build-in laminations to stop the chatter of AC power. Once the laminations become loose, the chatter (humming) comes back.
To find the disconnect size, we need to find what the conductors will carry through the lugs of the disconnect. Using T44 and reading down the 575 V column and across the 50 HP row, the FLA/FLC of the motor is 52 A then we multiply it by 1.25 to make it Csa ok. 65 A total. The disconnects come in size 30 A, 60 A, 100 A. We need 100 A disconnect. The only mistake Csa made here is that they chose the FLA from T44 using the induction and not the synchronous motor FLA (44 x 1.25)=55A. The answer should be 60A disconnect.
Which contact is the forward interlock means which contact is the normally closed contact. Answer B.
Variable Frequency drives are considered noisy so that’s why we always ground the supply and isolate the load to drain the electrical noise to ground.
Fault currents produce bending and twisting of switchgear or MCC bus bars.
Since there’s no need to protect an operator, we would use low voltage release with such as switches. To protect a motor operator, we would need to use low voltage protection with a start-stop station and a contactor. Answer is A.
A 3 phase balanced feeder means the current in all three phases is the same.
For interference reasons.
Every question counts and every word counts. Rule 2-3 24 Appendix B. 1 mm from the natural gas and 3 meters from propane gas.
A rectifier that is center tapped would produce 45% of the input AC in DC. If one diode cells the frequency will remain the same.
A SCR is a silicon controlled rectifier. The only way to shut off the current is to disconnect the power supply.
Rules 20-102 (1) and 86-404. 2 inches above the floor level.
23 kilos from the ceiling. 13 kilos for the wall box and 2.3 kilos for the flexible cord.
A capacitor consists of two plates that build up electrons on them and they’re separated by air or some sort of insulating material. Reading infinity means the capacitor is good.
Answer B. The purpose of using ground counterpoise in an airport Runway lighting system is to interconnect the system ground electrodes. General trade knowledge.
Answer D. Rule 60-700.
The answer is in the question. The real problem is the fact that the robotic arm cannot find the next position so we troubleshoot the server motor to make sure it’s finding its position.
A Voltmeter would read 70.7% of peak voltage. Therefore a peak voltage of 849 V AC X .707 would result in 600 volts AC.
To remove the water from the oil, we would run the oil through a separator. Answer C.
Rule 2-308 (1) and 2-310 (2). Answer A.
First we find the area of 53 mm conduit using table 9A rigid conduit and reading down the 40% column. That would be 879 mm squared. Then we subtract the area of No. 4 awg conductors without a jacket from table 10A and that would be 52.46 mm squared. Three 120 /208 with the neutral circuits means we have three times multiplied by four conductors which is 12 conductors. We multiply 12 by 52.46 mm squared (the area of number 4 AWG conductors). 629.52 total. What is available for the No.10 is 879 minus 629.52 which is 249.48 mm squared. To find how many number 10 we can pull in this conduit we divide 249.48 by the area of 10 AWG without a jacket from table 10 A which is 15.67 mm squared and the answer is 15.
General trade knowledge.
Waveform Clipper – Two Zener diodes facing each other in series will act to clip both halves of an input signal. Waveform Clippers can be used to not only reshape a signal, but also to prevent voltage spikes from affecting circuits that are connected to the power supply. Voltage Shifter – A Zener diode can be applied to a circuit with a resistor to act as a voltage shifter. This circuit lowers the output voltage by a quantity that is equal to the Zener diode’s breakdown voltage.Voltage regulator – A Zener diode can be applied in a voltage regulator circuit to regulate the voltage applied to a load, such as in a linear regulator.
All normally closed contacts would be ON and normally open contacts will be OFF. Please pay attention to the word engaged. The last contact is normally closed float switch engaged which means it’s normally open and therefore it’s light will not be ON so the only lights that will be ON are the two normally closed push buttons.
Between the neutral and the ground, there should be 0 volts because the neutral is grounded.
Answer A. W
hen all the numbers are the same, that would be considered a balanced load and since these numbers are different, then this would be an unbalanced load.
This is a grounded system. Therefore, if a hot conductor touches the ground that would blow the circuit fuse. Answer C.
A good fuse will always read 0 V.
This question deals with pressure so therefore the actuator has a damaged diaphragm.
Rule 26-700 (8).
Answer C makes sense. So the forced air gas furnace does not blow cold air, the burner will ignite first then the chamber heats up then the fan blows the hot air into the rest of the house via heating ducts
To find the power factor, we divide the small number by the big so therefore 2 horsepower is 746 x 2 which is 1492 Watts. To find the power factor we divide 1492 by 2000 VA and that will give us 74.6%.
Rule 14-200(2).
General trade knowledge.
Always read at the very far right side of the symbol. If it’s AB, that means it’s an AND gate. If it’s A + B, that’s an OR gate. If it’s an AB with a bar on top, that’s a NAND gate and if it is A + B with a bar on top, that’s a NOR gate.
The line to line voltage would be 208 V.
Never work live.
Single phase has two legs. Three phase three legs. If we go from 3 to 2 we lose one which means no current is flowing in one of the line.
AC is noisy so we install shading coils on either side of the forward – reverse controller to keep the controller quiet. When the controller goes back to noisy or there’s an uneven wear on the forward contact poles, this means the shading coils are missing or broken.
The capacitor is considered a filter in a rectifying system so it will smooth out the DC pulsating voltage.
In a Delta system, the voltage will always be the same if one of the phases goes to ground. If two phases go to ground then we’ll have 0 V.
This graph indicates a good insulation at the beginning of measurement but overtime the insulation of the motor stator windings has completely dried out.
This is the end of Block F. The next step is to run tests from test portion of the CEPE called PRACTICE EXAM and score 90% or better before you can write and pass the real exam- The First Time. Your support is 24/7. Thank you. Choose CONSTRUCTION then Click Start.
