Junction,cutout boxes and enclosures ( NEC 300.14)
Quiz-summary
0 of 10 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
Information
Free practice NEC Code question and answers on Cabinets, enclosures, cutlet boxes.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
- Not categorized 0%
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- Answered
- Review
-
Question 1 of 10
1. Question
What is the minimum volume allowance required per a No 16 AWG conductor?
Correct
Incorrect
Refer: NEC Table 316.14(B)
-
Question 2 of 10
2. Question
A cabinet or enclosure cannot be used as a raceway.
Correct
Incorrect
Refer: NEC 312.8
An enclosure can be used as a raceway if the conductors does not fill 40% of the white space. -
Question 3 of 10
3. Question
Splices and taps can be installed in enclosures if they not fill more than …………… % of the white space with in the enclosure.
Correct
Incorrect
Refer: NEC 312.8
-
Question 4 of 10
4. Question
According to NEC, locknuts, bushings and internal cable clamps are not counted for box fill calculations.
Correct
Incorrect
Refer: NEC 314.16 (B)
locknuts and bushing are not considered for box fill calculations.Refer: NEC 316.16(B)(2)
Internal cable clams are considered as conductors. -
Question 5 of 10
5. Question
Which of the following conductors that will not be counted for box fill calculations?
Correct
Incorrect
Refer: NEC 314.16(B)(1)
-
Question 6 of 10
6. Question
A junction box contains two receptacles, four 12 AWG conductors and two 12 AWG equipment grounding conductors. What is the total volume of conductors and devices in the box?
Correct
Incorrect
Refer: NEC 314.16
conductors for two receptacles = 2 * 2 = 4 – 12 AWG conductors
four 12 AWG conductors = 4 – 12 AWG Conductors
Conductors for two equipment grounding conductors = 1-12 AWG conductor
Total 9 conductors.
Total volume for 9 conductors = 9 * 2.25 cu.in = 20.25 cu.in ( NEC table 314.16(B)) -
Question 7 of 10
7. Question
The remaining volume of outlet box is 14.40 cu.in. How many number of No 14 AWG conductors are permitted to use in the remaining volume?
Correct
Incorrect
Refer : NEC Table 314.16(B)
Volume required for No 14 AWG conductor = 2.00 cu.in
so number of 14 AWG conductors permitted = 14.40 / 2.00 = 7 conductors -
Question 8 of 10
8. Question
How many No 10 AWG conductors can be sliced in 11 cu in conduit body?
Correct
Incorrect
Refer: NEC 314.16(B)
Volume required for No 10 AWG = 2.50 cu.in
number of No 10 AWG conductors can be sliced = 11/2.50 = 4 conductos -
Question 9 of 10
9. Question
Which of the following systems a box must be installed at each termination point?
Correct
Incorrect
-
Question 10 of 10
10. Question
A 3 wire, single phase 120/240 volts circuit supplies 1000 W , 120 volts electric heater and 600 W, 120 volts electric dryer. What is the operating voltage of the heater?
Correct
Incorrect
Resistance of heater = V2 / P = (120 * 120) / 1000 = 14400/1000 = 14.4 ohms
Resistance of dryer = V2 / P = 14400/ 600 = 24 ohmsCurrent = 240 / (14.4 + 24)= 6.25 amperes
Operating voltage of heater = 6.25 * 14.4 = 90 volts