Journeyman Electrician Practice Test24
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Question 1 of 10
1. Question
1 pointsWhat is the general lighting unit load for a 6000 square foot hospital?
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Sol : Hospital general lighting load = 2 V – A /Sq.Foot (NEC 220.12)
6000 * 2 = 12000 VoltAmpere

Question 2 of 10
2. Question
1 pointsThe calculated lighting load with corresponding demand factors shall not be applied to determine number branch circuits required for wiring.
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Refer: NEC 220.42

Question 3 of 10
3. Question
1 pointsDwelling unit has a floor area of 2000 sq ft, what are the minimum number of branch circuits required for lighting load at 115 volt ?
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Refer : NEC 220.22
2000 * 3 = 6000 Volt Ampere
6000/ 115 = 52.17 amperes
Four 15 amp 2wire circuits or three 20amp 2wire circuits. 
Question 4 of 10
4. Question
1 pointsWhat is the minimum volume of a 4 inch round/octagonal metal box having 1/2 inch depth?
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Refer: NEC Table 314.6

Question 5 of 10
5. Question
1 pointsThe conductors shall be protected from abrasion at both entry and exit points of a boxes, conduit bodies and fittings.
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Refer: NEC 314.17 (Entry of the boxes only)

Question 6 of 10
6. Question
1 pointsWhich of the total maximum bend that an EMT conduit can be bend?
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Refer: NEC 358.36

Question 7 of 10
7. Question
1 pointsThe electrical wire has a resistance of 1.5 ohms has a current flow of 4 amp. If supply end of the electrical wire indicates 120 volts what will the voltmeter reading of at the end of the wire?
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120 – 1.5*4 = 114 volts

Question 8 of 10
8. Question
1 pointsAirfield lighting cable used in electrical circuit rated over 5000 volts and powered by electrical regulator shall be permitted to be nonshielded.
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Refer : NEC 310.E

Question 9 of 10
9. Question
1 pointsA 120/240 single family dwelling unit has a calculated demand load of 120 amperes, what is the minimum type of THW copper conductor size allowed?
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Refer : NEC Table 310.15(B)

Question 10 of 10
10. Question
1 pointsWhat is the equivalent resistance offered by parallel connection of two resistors each having 2 ohms of resistance?
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( 2 * 2) (2+2) = 1 ohm