Journeyman Electrician #5 Exam Questions With Verified Answers

Unless listed and identified otherwise, terminations and conductors of sizes 14 AWG through 1 AWG shall be rated at _____. a) 75°C b) 75°F c) 90°C d) 60°C - Answer D: 60°C Ref: 110.14(C)(1)(a)(1) The MINIMUM size copper equipment bonding jumper on the supply side of a service consisting of parallel copper phase conductors with a total circular mil area of 2000 kcmil shall be ____. a) 2/0 AWG b) 3/0 AWG c) 4/0 AWG d) 250 kcmil - Answer D: 250 kcmil Ref: 250.102(C) Sol: 2,000 kcmil x 12.5% = 250 kcmil Which of the following grounding electrodes is NOT permitted to be used as an effective ground-fault current path? a) the metal frame of a building or structure b) a metal underground water pipe in direct contact with the earth for 10 feet or more c) an electrode encased by at least 2 inches of concrete d) the earth - Answer D: the earth Ref: 250.4(A)(5); 250.54 Primary overcurrent devices, impedance limiting means, or other inherent protective means shall be permitted to protect a motor control circuit transformer when the transformer is rated no more than _____ VA. a) 50 b) 60 c) 75 d) 100 - Answer A: 50 Ref: 430.72(C)(3) When installing electric space heating cables in ceilings, which of the following statements is/are true? a) The heating cables are permitted to be installed over walls. b) The non-heating leads of the cables shall be at least 6 inches in length c) The heating cables shall be kept free from contact with other electrically conductive surfaces. d) All of these statements are true. - Answer C: The heating cables shall be kept free from contact with other electrically conductive surfaces. Ref: 424.38(B)(1); 424.34; 424.41(H) Given: An unbroken length of size 12 AWG conductor is looped inside a junction box so that there is more than 12 inches of free conductor. The conductor is not spliced and does not terminate on a device. What volume, in cubic inches, must be counted for the looped conductor? a) 2.25 cubic inches b) 3.25 cubic inches c) 4.50 cubic inches d) 4.00 cubic inches - Answer C: 4.50 cubic inches Ref: 314.16(B)(1); Tbl. 314.16(B) Sol: 2.25 cubic inches x 2 wires = 4.50 cubic inches